GOMATHGR12W1 by admin Bookbuzz

Compiled by Chez Nell

2 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Forward: Welcome to “GO MATH WORKBOOKS”. This workbook is designed to be a text book and class work book in one. There are sufficient exercises to ensure that learners get the required practice. A detailed memorandum booklet is available for each workbook. The statement “You get out what you put in.” is very apt where maths is concerned. To succeed in mathematics one must be prepared to invest the time and effort to achieve that success. The partnership that you as a learner and this GO MATH WORKBOOK develop will be profitable if you allow it to be. Chez Nell : Mathematics Educator : Northwood School  Norma

Nell 2011

3 GO MATH WORKBOOKS

Grade 12 Core Mathematics

GRADE 12 CORE MATHEMATICS Contents: Paper One: Topic:

Page:

Calculus

(4 – 74)

Number Patterns

( 75 – 104)

Financial Maths

(105 – 145)

Functions & Graphs

(146 – 208)

Linear Programming

(208 – 229)

Probability Theory

(230 – 242)

4 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Calculus:

Average Gradient between two points on a curve. Development of the average gradient formula:

f(x+h)

Δy = f(x+h) – f(x) Gradient AB = f(x) A

Δx = h

Ave Gradient =

y x

f ( x  h)  f ( x ) h

h x1

xh

Example: Find the average gradient between x = 2 and x = 5 for a curve defined by f ( x )  x 2 ( 2;4) (5;25) f ( x  h)  f ( x ) y AveM  mAB  h y  25  4 x 25  4  21 21 AveM  mAB  3 3 x  5  2 21 AveM ( AB )  7 AveM  3 3 AveM  7

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Exercise 1.1: 1.

Find the average gradients for f ( x )  x 2 between the following points: 1.1 x = 3 and x+h = 5 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 1.2

x= 2 and x+h= 7 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.3

x= -2 and x+h = 3. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

Find the average gradient on f ( x )  x 2  4 2.1 x = 1 and x+h = 4 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 2.2

x= 3 and x+h= 8 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2.3

x= -1 and x+h = 5. _____________________________________________________ _____________________________________________________ _____________________________________________________

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Find the average gradients between the following points for f ( x )  x 3 : 3.1 x = -3 and x+h = 1 _____________________________________________________

_____________________________________________________ _____________________________________________________ _____________________________________________________ 3.2

x = -5 and x+h = -2. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.3

x = 2 and h = 3 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

Find the average gradients between the following points for f ( x )  x 3  2 : 4.1 x = 2 and x+h = 4. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 4.2 x = 3 and x+h = 5 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 4.3 x= -2 and x+h =3 _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

7 GO MATH WORKBOOKS

Grade 12 Core Mathematics Example: Method 1 : Substitution into Average Gradient =

y f (a  h)  f (a ) or AveM ( AB )  x h

Example: Find the average gradient between the following points on f(x) = x2 -x x = 2 and 5 If x = 2 then y = 2 If x = 5 then y = 20 Average Gradient

y x

20  2 3 18  3 6 

Method 2 : By First Principles (a)

Find the average gradient between any two points on f(x) = x2 -x f(x) = x2 -x Average Gradient

= = = =

f ( x  h)  f ( x ) h [( x  h) 2  ( x  h)]  [ x 2  x ] h 2 2 x  2 xh  h  x  h  x 2  x h 2 2 xh  h  h h

= 2x + h - 1 Find the average gradient between the following points on f(x) = x2 -x (1) x = 2 and 5 Ave M = 2x + h - 1 = 2(2) + 3 - 1 = 6 (2)

x = -2 and 1 Ave M = 2x + h - 1 = 2(-2) + 3 - 1 = -2

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Exercise 1.2: 1.1

Find an equation for the average gradient between any two points on y = x2 + 3x + 2. ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.2

Now find the specific average gradients between the following points: 1.2.1

x = 2 and 5

2.1

Find an equation for the average gradient of y = 2x2 –x – 1 between any two points. ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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Use the equation above to find the average gradients between the following points: 2.2.1 x = -5 and –2 ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2.2.2. x = 3 and 7 ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

Derive an equation that will help you to find the average gradients between any two given points on the following curves: 3.1

f ( x)  x 3 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.2

f ( x)  x 3  x

10 Grade 12 Core Mathematics 3.3

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f ( x)  2 x 3  2 x 2 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.4

f ( x)  x 3  2 x 2  3 x  2

11 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Limit Concept:

y y = 3x

In the sketch above: As x moves from the left towards 2 , y in turn moves up towards 6 As x moves from the right towards 2 , y in turn moves down towards 6. It can thus be deduced that as x tends (moves) towards 2, so y will tend towards 6. We can thus say that the limit reached as x tend towards 2 will be 6. [the arrow  means “tends towards ”] This is written as : lim3 x  6 x2

This is read as the limit of y = 3x as x tends towards 2 is 6 There two important exceptions to remember: 1. Case

0 In this case first factorise the given expression and then simplify before finding the 0

relevant limit. e.g.  x2  4   ( x  2)( x  2)    lim lim  x2 x  2 x2    x2   lim x  2 x2

4

12 GO MATH WORKBOOKS

Grade 12 Core Mathematics 2. Case

ď&#x201A;Ľ ď&#x201A;Ľ

The graph represents the function

1 x

yď&#x20AC;˝

As x tends towards infinity: so y tends towards zero By deduction: If x tends towards infinity thus fraction with x in the denominator will tend to zero.

ď&#x201A;Ľ

NB: lim x ď&#x201A;Žď&#x201A;Ľ

2 ď&#x20AC;˝0 x3

Example: lim 4 x 3 ď&#x20AC; 2 x 2 ď&#x20AC; 3 x ď&#x20AC; 4 xď&#x201A;Žď&#x201A;Ľ

Divide each term by đ?&#x2018;Ľ 3

4x3 2x2 3x 4 ď&#x20AC;˝ lim 3 ď&#x20AC; 3 ď&#x20AC; 3 ď&#x20AC; 3 xď&#x201A;Žď&#x201A;Ľ x x x x 2 3 4 ď&#x20AC;˝ lim 4 ď&#x20AC; ď&#x20AC; 2 ď&#x20AC; 3 xď&#x201A;Žď&#x201A;Ľ x x x ď&#x20AC;˝ 4ď&#x20AC;0ď&#x20AC;0ď&#x20AC;0

Tend đ?&#x2018;Ľ â&#x2020;&#x2019; â&#x2C6;&#x17E; thus đ?&#x2018;&#x2DC; =0 đ?&#x2018;Ľđ?&#x2018;&#x203A;

ď&#x20AC;˝4 By Inspection: Simply write down the value of the coefficients of the highest power of the variable. Exercise 1.3 1. Find the following limits: 1.1

lim( x ď&#x20AC;Ť 5)

x ď&#x201A;Ž3

1.2

x2 ď&#x20AC;Ť 3x ď&#x20AC;Ť 2 x ď&#x201A;Ž ď&#x20AC;2 xď&#x20AC;Ť2 lim

13 GO MATH WORKBOOKS

Grade 12 Core Mathematics 1.3

lim

3x2  x  1

1.4

 x2  9   lim x3 x  3  

2x2  x  6 ______________________________________________ x 

1.5

lim ( 3 x  6)

x 2

1.6

5x  2 x  x  1

lim

14 GO MATH WORKBOOKS

Grade 12 Core Mathematics Limits of the form: lim h 0

f ( x  h)  f ( x ) h

This is the limit concept applied to the average gradient formula: This in fact gives the gradient of a tangent to the curve at any point on the curve.

f(x+h)

C f(x) A A 1 h0

xh

In the figure above: by limiting h to zero the line AB is shifted to the position A1 B1 . This line is tangential to the curve at point C. This new position gives the gradient at a point ( the gradient of a tangent at the point) which is referred to as the derived function or the gradient function. Def: The derivative of a function at any point on a given function is given by the following formula: f ( x )  lim h o

f ( x  h)  f ( x ) h

Alternative notations used for the derived function are: f ( x ) 1. 2. Dx f 3.

dy dx

The process whereby the derived function is arrived at is called differentiation. We say that f ( x ) or D x f or

dy is obtained by differentiating f(x) with respect to x. dx

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There are two methods used for differentiation. A. By First Principles: This is a formal method which arrives at a general formula for the derivative of a specific function. This derived formula can be utilized to obtain the gradient function at a specific point on the curve by substituting the points x – value into it. Examples: 1, Find the gradient function of the following equation. f ( x)  2 x 2 f ( x  h)  f ( x ) h 2 2( x  h)  2 x 2  lim h 0 h 2 2 x  4 xh  2h 2  2 x 2  lim h 0 h  lim 4 x  2h f ' ( x )  lim h 0

h 0

f '( x)  4 x

Find the derivative of the following equation.

2.1

f ( x)  2 x 3  3 x 2  3 x  2 f ( x  h)  f ( x ) h 0 h 2( x  h) 3  3( x  h) 2  3( x  h)  2  2 x 3  3 x 2  3 x  2 f ( x )  lim h 0 h 3 2 2 3 2 x  6 x h  6 xh  2h  3 x 2  6 xh  3h 2  3 x  3h  2  2 x 3  3 x 2  3 x  2 f ( x )  lim h 0 h 2 2 f ( x )  lim6 x  6 xh  2h  6 x  3h  3 f ( x )  lim

h 0

f ( x )  6 x 2  6 x  3

Now find the gradient of the tangent to the curve in 2.1 at the point (3 ; 34)

f '( x)  6 x 2  6 x  3 f ( 3)  6( 3) 2  6( 3)  3  54  18  3  39

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Exercise 1.4: Use First Principles to differentiate the following: 1.

f ( x)  x 2 ______________________________________________

f ( x)  x 2  2 ______________________________________________

17 Grade 12 Core Mathematics 3.

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f ( x)  3 x 2 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

f ( x)  x 2  3 x  4 ______________________________________________

18 Grade 12 Core Mathematics B.

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Standard Rules for Finding Derivatives:

This is a short – cut method to find the derivatives of given functions by inspection. D x [cons tan t ]  0 NB f (constant) = 0 ie If f(x) = 8 then f „(x) = 0

Proof: f ( x)  k

f ( x )  lim h 0

f ( x  h)  f ( x ) h

kk h 0 h 0  lim h 0 h  lim 0  lim

h 0

0 Method: f ( x)  x n f ( x)  nx n 1

MULTIPLY THE COEFFICIENT BY THE EXPONENTIAL VALUE AND THEN SUBTRACT ONE UNIT FROM THE EXPONENT. REMEMBER : D x [cons tan t ]  0 Example: 1.

f ( x)  4 x 5 f ( x)  20 x 4

f ( x)  3x 4 f ( x)  12 x 5 3.

f ( x)  3 x 3  2 x 2  x  9 f ( x )  9 x 2  4 x  1

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BEFORE APPLYING THE STANDARD RULE THE FOLLOWING MUST BE DONE: 1.

e.g

THE EXPRESSION CANNOT CONTAIN ANY BRACKETS. IE CANNOT BE IN FACTORISED FORM. THE EXPRESSION MUST BE MULTIPLIED OUT FIRST. f ( x)  ( x  1)( x 2  2 x  3) f ( x)  x 3  2 x 2  3x  x 2  2 x  3 f ( x)  x 3  3x 2  5 x  3 f ( x)  3x 2  6 x  5

e.g

THE ORIGINAL EXPRESSION MUST NOT HAVE ANY VARIABLES IN THE DENOMENATORS. USE NEGATIVE EXPONENTS TO CONVERT THESE TO WHOLE VALUES. 4 f ( x)  3  x 2  1 x f ( x )  4 x 3  x 2  1

f ( x)  12 x  4  2 x 3.

THE ORIGINAL EXPRESSION MUST NOT HAVE ANY ROOT VALUES AT ALL. CONVERT THESE TO EXPONENTIAL VALUES FIRST:

f ( x)  3 x 2 2

e.g.

f ( x)  x 3 f ( x) 

2  13 x 3

Exercise 1.5: Find the derivatives of the following expressions using standard rules. 1.

f ( x)  x 3  3 x 2  5 x  7 ______________________________________________

20 Grade 12 Core Mathematics 2.

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f ( x)  ( x  3)(2 x  5) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

( x  4)( x 2  4 x  16) f ( x)  x4 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

4x 2  9 f ( x)  2x  3 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

8 x 3  27 f ( x)  2x  3 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

f ( x) 

3 x3

______________________________________________ _____________________________________________________ _____________________________________________________

21 Grade 12 Core Mathematics 7.

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f ( x)  33 x

2x 3  4 x2 ______________________________________________ f ( x) 

f ( x) 

1 15 x 5

f ( x)  5x 4  6 x 3  2 x 2  7 x  20 ______________________________________________

22 GO MATH WORKBOOKS

Grade 12 Core Mathematics Equations of Tangents to curves:

The following information is required to find the equation of a tangent to a given curve. 1. The equation of a specific curve. 2. The x or y- values (or both) of the point of tangency. Method: 1. Find the derivative ( gradient function ) of the given curve. 2. Substitute the x- value of the point of tangency into the derivative to calculate the gradient of the tangent at that specific point. 3. Substitute this gradient into y = mx + c or y – y1 = m(x – x1 ) 4. Now substitute the x and y value from either the point of tangency or any other point that is said to lie on the tangent to calculate the „c- value‟( y-intercept of the tangent.) 5. Write down the equation of this tangent. Example: 1. Find the equation of a tangent that touches the curve, f ( x)  x 2  2 x  3 , at (2 ; -3) Solution: f ( x)  x 2  2 x  3 f ( x)  2 x  2 Gradient of the tangent at point (2 ; -3) f (2)  2(2)  2  m=2 Equation of tangent : y = mx+ c y = 2x + c Substitute point (2; -3) -3 = 2(2) + c c = -7  y = 2x – 7 2. Find the equation of a tangent that touches the curve f ( x)  x 3  6 x 2  3x at the point (-2 ; a). f ( x)  x 3  6 x 2  3 x f ( x)  3x 2  12 x  3 Gradient of the tangent at (-2 ; a) f (2)  3 2  12 2  3 2

= 12 – 24 + 3 m = -9 Equation of tangent: y = mx + c y = -9x + c substitute ( -2 ;15) 15 = -9(-2) + c c = -3 Exercise 1.6:

N.B. to calculate the y- value of the point substitute the x- value of the into y = -9x 3 point the -ORIGINAL equation

23 Grade 12 Core Mathematics 1.

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In each of the following determine the equation of the tangent at the point indicated 1.1 y  2 x 2  x  3 point(-1 ; 4) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 1.2 y  x 3  x 2 point (1; 2) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 1.3 y  x 2  2 x  3 point ( -2 ; y) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

Determine the equation of the line which touches the parabola y  x 2  2 x  3

24 GO MATH WORKBOOKS

Grade 12 Core Mathematics and is parallel to the line 4 x  2 y  4 . ______________________________________________

Determine the equation of the line which touches f ( x )  3 x 2  8 x  6 and which is perpendicular to y  

1 x 3. 2

25 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Factorising equations to the 3rd degree [NB it would be to your advantage to purchase a calculator that can factorise equations to the 3rd degree] Synthetic Division with Polynomials Step 1:

Use the Remainder Theorem to find a factor of the given polynomial. (If one is not supplied)

Step 2:

(a)

Write down the constant values in the polynomial with their respective + or – signs. The expression must decrease in order of exponential value IF NOT put a 0 ( zero) in the place of the missing order value.

(b)

To the left of these constants on the line below, write down the value of x which gives the zero of the chosen factor.

Step 3:

(a)

Start the synthetic division by writing down the same value of the first term on the third line below the first term

(b)

Multiply the first term in the third row by the zero value of x on the left and place the answer on row two under the 2nd term. Add rows 1 and 2 together and place the answer on row 3 . Repeat the procedure until all values utilized NB The last value on the bottom line is the remainder after the division . i.e. if the divisor is a factor this should be zero and if not the constant value obtained is the remainder.

(c)

write these constant values obtained on the bottom line with their appropriate x value alongside the original factor and factorise further.

(a)

Factorise 3x2 – 7x + 4 fully if (x-1) is a factor.

Examples:

3 3

-7 3 -4

0 -4 -4

+4 +4 0

f(x) = (x-1)(3x2 – 4x – 4) = (x-1)(3x+2)(x-2)

ie x – 1 = 0 x = 1

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Grade 12 Core Mathematics f ( x)  x 3  x 2  22 x  40 Test (x-2) f (2)  8  4  44  40 =0 (x-2) is a factor

(b)

1 1

-1 +2 1

-22 +2 -20

x-2 = 0 x=2 +40 +40 0

f ( x)  ( x  2)( x 2  x  20) = ( x  2)( x  5)( x  4)

Factors of equations to the 3rd degree : Using Synthetic Division NB: The terms must decrease in order of exponential value. i.e. If an exponential value is not represented then use a zero in its place. E.G.

f ( x)  3x 3  4 x  6 the 2nd power is missing so use a zero in its place

-4

(Use a zero for the x2 position)

Example: 1.

f ( x)  x 3  x 2  22 x  40 f (2)  8  4  44  40 Test (x-2) =0 (x-2) is a factor

x-2 = 0 x=2

1 2 1

-1 +2 1

-22 +2 -20

f ( x)  ( x  2)( x 2  x  20) = ( x  2)( x  5)( x  4)

Exercise 1. 7:

+40 +40 0

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The following have one factor given use this to factorise the expression completely 1.

f ( x)  3x 3  7 x 2  4 and (x –2) is a factor ______________________________________________

f ( x)  x 3  x 2  22 x  40 and (x +5) is a factor

f ( x)  4 x 3  19 x  15 and (x+1) is a factor.

f ( x)  x 3  6 x 2  11x  6 and (x – 3) is a factor

28 Grade 12 Core Mathematics

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f ( x )  x 3  2 x 2  5x  6

and (x + 2) is a factor ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

f ( x)  x 3  3x 2  x  3 and (x + 3) is a factor ______________________________________________

f ( x)  x 3  3x 2  6 x  8 and (x + 4) is a factor

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f ( x)  x 3  6 x 2  3x  10 and (x – 5) is a factor ______________________________________________

f ( x)  x 3  x 2  x  1 ______________________________________________

f ( x)  x 3  2 x 2  9 x  18 ______________________________________________

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f ( x)  2 x 3  x 2  13x  6 ______________________________________________

f ( x)  2 x 3  5x 2  23x  10 ______________________________________________

f ( x)  4 x 3  8 x 2  x  2 ______________________________________________

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Grade 12 Core Mathematics

Substitution using the derivative: Find the derivative of the equation and equate it to zero (0) to find the axis of symmetry of the curves. Substitute the x – value(s) into the ORIGINAL equation to find the corresponding y- value(s). The calculated point(s) are the stationary , turning , maximum/minimum co-ordinates of the curve graph. To test which point is a max/min point the following procedure can be used. NB direction on any graph is left to right i.e. from negative to positive on the x – axis. f‟(x) = 0 +ve

-ve

(-ve)

+ve

(+ve)

f‟(x) = 0

Use the x – value ( axis of symmetry) in the test. Take a whole value smaller than the x- value and substitute it into the derivative and ascertain whether a positive or negative gradient is found. Do the same procedure with a value larger than the x- value. Use a simple sketch to illustrate the result. Example:

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Grade 12 Core Mathematics

Find the co-ordinates of the maximum and minimum stationary point of f ( x)  x 3  4 x 2  11x  30 and show that they are max or min points. f ( x)  x 3  4 x 2  11x  30 f ( x)  3x 2  8x  11 Turning points at f‟(x) = 0 3x 2  8x  11  0 (3x – 11)(x +1)= 0 x = -1 or

11 3

y = 36 or 

400 27

 11 400  ;  27  3

TP‟s (-1 ; 36) and 

Test for max/min tp’s at f ( x)  3x 2  8x  11 (-1 ; 36)

f (2)  17 f (2)  15

 11 400   ;  27  3

0 +VE

f (0)  11 f ( x)  5

-VE

-ve

+ve

0 2. Double derivative test: Find the second derivative (Cannot use this method for equations of the second degree ). Substitute the x- value (axis of symmetry) of the turning point into the 2nd derivative. a) If the answer is greater than zero (0) then it is a minimum turning point. b) If the answer is less than zero (0) then it is a maximum turning point. Example: f ( x)  x 3  4x 2  11x  30 f ( x)  3x 2  8x  11 (Ist derivative) f ( x)  6 x  8 (2nd derivative) Test:

f (axis of symmetry )  6x  8 f (1)  6(1)  8  14 Maximum turning point at (-1 ; 36)  11   11   11 400  f    6   8  14 Minimum turning point at  ;  27  3 3 3

Tangents; Normals and Points of Inflection on Curves

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Grade 12 Core Mathematics

A tangent to a curve is a straight line that touches the curve at only one point. The gradient of this point of tangency is derived from the gradient function and the x-value of the tangential point. A Normal is a straight line that is perpendicular to the tangent and passes through the point of tangency. Their gradients are inverse i.e. (m1)(m2) = -1. Example:

Tangent

Normal Points of inflection. Inflection refers to the case where a graph does not change direction after the maximum or minimum point but continues in the same direction .i.e. It maintains the same gradient after the point of inflection (Stationary point)‟ Points of inflection normally occur when there is only one stationary point in an equation to the 3rd degree ( f ( x)  x 3 ). They can be detected in the test for maximum or minimum turning points using the Ist derivative test i.e. substituting a value less and then larger than the axis of symmetry. The gradients will not change. i.e. They will both be negative or both be positive. Example: f ( x)  x 3  6 x 2  12 x  7 f ( x)  3x 2  12 x  12

Turning points at f (x)  0

3x 2  12 x  12  0 x 2  4x  4  0 ( x  2) 2  0 x= 2 y = 1 TP(2 ; 1) Test f (1)  3  12  12  3 positive gradient f (3)  27  36  12  13 positive gradient NB A point of inflection occurs at this stationary point Double derivative test: f (x)  It can also be noted when using this test. If the answer in a double derivative substitution is zero (0) then a point of inflection occurs. Example:

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Grade 12 Core Mathematics f ( x)  x 3  6 x 2  12 x  7 f ( x)  3x 2  12 x  12 f ( x)  6 x  12

Test:

f (2)  6(2)  12  0 The value is zero thus a point of inflection occurs. Sketch: 1

Example: Sketch the graph of

Point of inflection

f ( x)  ( x  2) 2  8

f ( x)  ( x  2) 2  8 f ( x)  x 3  6 x 2  12 x f ( x)  3x 2  12 x  12 f ( x)  6 x  12 3x 2  12 x  12  0

Turning Point at

x 2  4x  4  0 ( x  2) 2  0 x=2 y = 8 TP(2 ; 8 )

f (2)  6(2)  12  0 Point of Inflection occurs. Sketch:

Concavity: Concave up: – this is when all the tangents to a curve lie below the curve. The gradient function is increasing i.e.  f ( x )  0.

Concave down:- occurs when all the tangents to a curve lie above the curve. The gradient function is decreasing . i.e.  f ( x )  0.

35 Grade 12 Core Mathematics

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Sketching curves of Equations of the 3rd degree. Steps. 1. Find the first derivative of the equation. 2. Equate the derivative to zero and solve the equation to find the axes of symmetry of the stationary points. 3. Find their corresponding y- values by substituting the axes of symmetry into the ORIGINAL equation. 4. If necessary use the 2nd derivative to ascertain which of the points is the maximum/minimum one. 5. Draw a set of axes and start by plotting the axes of symmetry using a dotted line. 6. Plot the roots and the y-intercept and sketch a neat curve through these points. 7. Label the necessary points and the graph. Example: 1.

Sketch the graph of f ( x)  x 3  12 x 2  36 x using calculus methods.

f ( x)  x 3  12 x 2  36 x f ( x)  3x 2  24 x  36

f ( x)  6 x  24 Stationary points at f ( x)  0

3x 2  24 x  36  0 3( x 2  8 x  12)  0 3( x  2)( x  6)  0 x  2 or x  6 y  32 or y  0 TP‟s (2 : 32) & (6 ; 0) Check for max/min TP’s NB Use the 2nd derivative ( f (x) ) NB substitute the x – values of the turning points into f (x) . If the answer is –ve then you have a maximum stationary poinr If the answer is +ve than you have a minimum stationary point.

f ( x)  6 x  24 f (2)  6(2)  24  12 max TP. f (6)  6(6)  24  12 min TP f (4)  6(4)  24  0  a point of inflection

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Grade 12 Core Mathematics

Sketching the graph: NB these are sketches and are NOT drawn to scale. 1. 2. 3.

Draw a set of axes. First draw in the axes of symmetry of the turning points using dotted lines Using the knowledge which turning point is max or min ( 2nd derivative test) Sketch the correct shape, making sure that the y – intercept is on the correct side of the x – axes( i.e. above or below) Label the diagram with the values calculated.

y (2 ; 32)

f ( x)  x 3  12 x 2  36 x

Point of Inflection

(0 ; 6)

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Exercise 1.9 : Sketch the following using the methods in applied in the example above. NB the shape is important (remember these are sketches and are not to scale.) 1.

f ( x)  ( x  1)( x  2) 2 ______________________________________________

f ( x)  (2  x)( x  1) 2 ______________________________________________

38 Grade 12 Core Mathematics 3.

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f ( x)  x 3  3 x 2 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

f ( x)  6 x 2  x 3 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

39 Grade 12 Core Mathematics 5.

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f ( x)   x 3  6 x 2  9 x ______________________________________________

f ( x)  x 3  3 x  2 ______________________________________________

40 Grade 12 Core Mathematics 7.

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f ( x )  ( x  1) 3  8  0 ______________________________________________

f ( x)  x3  x 2  8x  12

Solve for x if x3  x 2  8x  12  0 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.2

Find f (x) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

41 Grade 12 Core Mathematics 1.3

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Determine the coordinates of the turning points of f. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.4

Draw a neat sketch of f and label it correctly. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2 2.1

f ( x)  3 x 2  2 x 3

Solve for x if 3x 2  2 x 3  0 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

42 Grade 12 Core Mathematics 2.2

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Find f ď&#x201A;˘(x) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2.3

2.4

43 Grade 12 Core Mathematics 3. 3.1

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f ( x)  x 3  6 x 2  9 x  4 Solve for x if

x3  6x 2  9x  4  0

3.2

Find

f (x)

3.3

3.4

Draw a neat sketch of f and label it correctly. ______________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

44 Grade 12 Core Mathematics 4. 4.1

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f ( x)  x 3  3 x 2

Solve for x if x 3  3x 2  0 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

4.2

4.3

4.4

45 Grade 12 Core Mathematics 5 5.1

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f ( x)  2 x 3  3 x 2

Solve for x if 2 x 3  3x 2  0 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

5.2

5.3

5.4

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Grade 12 Core Mathematics Examples of increasing and decreasing functions

f ( x)   x 2  6 x  5

f ( x)  2 x  6

Questions: (a)

Find out whether f(x) is increasing or decreasing at : (i) 2 (ii) 4

(b) (c)

Find the stationary points of f. Find the interval on which f is: (i) increasing (ii) decreasing.

Solutions:

f ( x)   x 2  6 x  5 f ( x)  2 x  6 N.B. IF f ( x)  0 increasing AND f ( x)  0 decreasing f (2)  2(2)  6  2 Increasing a) (i) f (4)  2(4)  6  2 Decreasing (iii) (b)

(d)

Stationary points occur at f ( x)  0 -2x +6 = 0 x=3 y=4 (3 ; 4) is the stationary point. (I) f(x) is increasing when f ( x)  0 : -2x +6  0

x3

f(x) is increasing for x  (;3]

(ii) (iii)

f(x) is decreasing f ( x)  0 -2x +6  0 x 3 f(x) is decreasing for [3; )

increasing 3 decreasing

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Grade 12 Core Mathematics

Example: Equation: f ( x)  6 x 2  x 3

increasing Derivative: f ( x)  12 x  3x 2

i) (ii)

decreasing

f(x) is increasing when f ( x)  0 ie where x [0 ; 4] f(x) is decreasing when f ( x)  0 c ie where x  (;0] and x  [4; )

y Questions on interpretation of Curve graphs : equations to third degree ( x 3) Q 1.

x P(-2 ;0)

The graph( not to scale) represents the function f , where f ( x)  ax 3  bx  8 P(-2 ; 0) and Q are turning points. 1 2

1.1

Deduce that a  

1.2

Determine the co-ordinates of Q.

1.3

Use your graph to write down the value(s) of t for which the equation 

and b = 6.

1 3 x  6 x  t  0 will have only one root. 2

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Grade 12 Core Mathematics Answers: 1.1 0  a(2) 3  b(2)  8

0  8a  2b  8

f ( x)  ax 3  bx  8

2b  8a  8

f ( x)  3ax 2  b f (2)  12a  b  0

b  4a  4

b  12a

b  4a  4

b  12a

 12a  4a  4

 1 b  12    2 b6

 8a  4 a

1 2

N.B. You may not simply substitute the given values back into the equation as proof. The method used is to use the original expression and its derivative and then solve using simultaneous equations to show that the values are true. 1.2 3 f ( x)   x 2  6 2 3  x2  6  0 2 2 x 4 x2

or  2

y  16 or

1.3



1 3 x  6x  t  0 2

t > 8, since the graph will have to move vertically to ensure that it cuts the x- axis only once. 2.

The figure represents the graph of y  f (x) with f ( x)  ax 3  bx 2  cx . Show that a 

2 11 ; b and c  4 . 3 3

f ‟

2 3

x 3

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Grade 12 Core Mathematics

Solution: The method to use : Find the derivative of the equation of the curve and then use the xvalues on the given graph and substitute these into the derivative to get two equations and then solve these simultaneously (NB. The x-values on the graph of the derivative are the axes of symmetry for the curve graph i.e the equation of y = ax3 +bx2 + cx f ( x)  ax 3  bx 2  cx f ( x)  3ax  2bx  c 2

At x 

2 and 3 f ( x)  0 i.e. m = 0 3

2 2 3a   2b   4  0 3 3 4a 4b  40 3 3 a  b  3  0 equation 1

3a(3) 2  2b(3)  4  0 27a  6b  4  0 equation 2

Simultaneous solution using equations 1 & 2 a  b  3 from equation 1 Substitute equation 1 into equation 2  11  27(b  3)  6b  4  0 a    3   3  27b  81  6b  4  0 11 9 a   21b  77 3 3 77 2 b a 21 3 11 c4 b 3 3.

The graph of f ( x)  ax 3  bx 2  cx is a decreasing or increasing function when x-values are as follows: y

Increases for x < 2 3

2 Decreases for < x < 3 3

f‟ Increasing for x > 3 x

2 3

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Exercise 1.11: 1.

The curve of y  ax 3  24 x  b has a local minimum point at (2 ; -17) Calculate: 1.1 The values of a and b and The co-ordinates of the maximum turning point on the curve. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

2 For a given function f ( x ) the derivative is f ( x )   x  x  2 2.1 What is the gradient of the tangent to the function f ( x ) at x  0 ? ______________________________________________

____________________________________________________ ____________________________________________________ ____________________________________________________

2.2 Where is f ( x ) increasing? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Grade 12 Core Mathematics 3.

f ( x )  ax 3  bx 2  cx .

The figure below shows the graph of y  f ( x )

y=f‟((x )

3.1 Prove that a = 2; b = -9 and c = 12. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________

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Grade 12 Core Mathematics 4, Sketched is the graph of f ( x )  x 3  4 x 2  11 x  30 , A and B are turning points of f .

y C

x B

4.1 Determine the coordinates of A and B.

53 Grade 12 Core Mathematics

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4.2 Determine the turning points of g if g( x )  f ( x  2).

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Grade 12 Core Mathematics

4.5 Determine the x-coordinate of the point at which the tangent in 3.4 cuts the graph of

again.

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Grade 12 Core Mathematics Solving Problems using Calculus :

Calculus can be utilized to solve problems in mathematics. These are problems that involve maximum and minimum concepts. i.e. finding dimensions in a problem that give maximum or minimum volumes or areas etc. The important concept is the occurrence of max/min stationary points on curves where the derivatives are equal zero. An equation must either be available or be formulated in order to find its derivative. The stationary points help us to calculate the maximum or minimum values as required. f‟(x) = 0 Maximum stationary point

x2 x - axis

f‟(x) = 0

Minimum

stationary point Example: A solid cylinder is cast from brass so that its total surface area is 50000 mm2 . Calculate the dimensions of its radius and height in order for the cylinder to have a maximum volume. NB. Volume of a sphere (V) =  r 2 r=x 2 Total surface area (S) = 2r  2rh S = 50000 mm2 S = 50 000mm2 2 2r  2rh = 50000 50000  2r 2 h 2r h 2 V=  r h r 2 50000  2r 2 V (r )  2r V (r )  25000r  r 3 V r   25000  3r 2 r=x





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Grade 12 Core Mathematics Maximum volume @ V ( x)  0

25000  3r 2  0

V ( x)  6r check: V 51,5  6(51,5) V 51,5  970,75

 3r 2  25000 r  51,5

50000  2(51,5) 2 Radius = 51,5 mm and h   103mm for maximum volume. 2(51,5) 2.Find the maximum distance ( PQ) between the two graphs depicted below.

y =2x2-8x+5

y = -x2 +7x - 7

NB: to get an equation subtract the equation of the lower point from the equation of the upper point

Method 1: As for grade 11: Completing the square:

PQ   x 2  7 x  7  (2 x 2  8 x  5) PQ =

 3x 2  15 x  12  3[ x 2  5 x  4]  3x  52   6 14 2

Maximum distance is 6 14 units.

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Grade 12 Core Mathematics Method 2 Using calculus.

PQ   x 2  7 x  7  (2 x 2  8 x  5)  3x 2  15 x  12 PQ( x)  6 x  15 Max distance at f ( x)  0  6 x  15  0

x

5 2 2

5 5 y  3   15   12 2 2 1 y64

(Substitute x – value back into original PQ equation to find y – value)

3. A big open top rectangular container with a square base has to be made of metal plate. The volume of the container should be 108 m3 . Hint: Let the length = breadth = x and the height = h x

Volume (V) = l . b . h Surface area (S) = x2 + 4xh V= x2h

108 h 2 x

108 S  x 2  4 x( 2 ) x S  x2 

432 x

S ( x)  2 x 

432 x2

432 0 x2 432 2x  2 x 3 x  216 x6 2x 

108 x2 108 h 36 h3 h

h x

Solution: The length of the box must be 6 m and the height must be 3 m to minimize the amount of metal plate needed.

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Grade 12 Core Mathematics Exercise 1.12:

1. A rectangular piece of cardboard has sides of 50 cm and 30cm Equal squares of x cm are cut from the corners to make an open box by folding up along the dotted lines as shown in the sketch below.

50cm

30cm

x x

1,1

Write down , in terms of x the length, breadth and height of the box. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

1.2

Determine an equation for the volume of the box. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

59 Grade 12 Core Mathematics 1.3

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Determine the value of x for which V is a maximum. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

1.4

Calculate the maximum volume of the box. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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Grade 12 Core Mathematics 2.

If y  x 2  4 x  3 and y   x 2  6 x , calculate the maximum length of PQ.

y  x2  4x  3

y   x2  6x

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Grade 12 Core Mathematics

3. A cylinder has a height of (40 â&#x20AC;&#x201C; x) cm, its base has a radius of x cm 3.1 Derive a formula to calculate the volume ( V = ď °r 2 h ) 3.2 Calculate the radius that will give a maximum volume.

40-x m

r=x ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Grade 12 Core Mathematics Calculus of Motion

If the particle P at C were moving from left to right at 10 m/s , its velocity would be +10 m/s even though s = -2. For a particle at A moving from left to right at 10 m/s; v = -10 m/s but the speed would be 10 m/s. For speed , like distance we do not take direction into account. + DIRECTION 10 m/s C

-2

-1

5 -10m/s

- DIRECTION

If s  f (t ) is the equation of motion, which gives the position at time t  ; Then v 

d (s ) OR v  f (s) gives the velocity at time t  . dt

The symbol (s) thus denotes the displacement or movement of a particle. d Then a  2 ( s) OR a  f (s) gives the acceleration at time t  . d t Example: 1. 1.1 1.2 1.3 1.4

The displacement ( motion) of a particle is given by : s  9  6t  t 2 find s at t  0;1;2;3 . What is the particle doing during the interval (0 to 3) Find s at t  4;5;6 Describe the motion of the particle relative to 0 after t = 3.

2.1 Find an expression for v in terms of t. 2.2 Find v at t  1 & t  2 . v should be negative. What does this imply about the motion of the particle? [ compare with 1.2 ] 2.3 Find v at t  4 & t  5. What do the positive values abtained imply about the motion of the particle? [ compare with 1.4 ] 2.4 Find v at t  3 . What does the value obtained tell you about the motion of the particle?

Sketch the graph of v and t on separate systems of axes.

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Grade 12 Core Mathematics

Answers: 1.1 v  9;4;1;0. 1.2 The particle is moving back to 0 (zero) from a point 9 to the right of 0. 1.3 1 ; 4 ; 9. 1.4 The particle is moving away from 0 in a positive direction. 2.1 2.2 2.3 2.4

v  6  2t –4 ; -2; Particle is moving in a negative direction. 2 ; 4; particle is moving in a positive direction. 0 ; Particle is momentarily at rest.

y y S(t

V(t)=f‟(s)

-6

Example 2.

A particle is moving in a straight line so that at t seconds from the start its displacement, s metres from a fixed point 0 is given by s  4t 3  12t 2  9t  28 . Find: 2.1 the velocity of the particle when t = 2 s. 2.2 when the velocity of the particle is 0. 2.3 How far the particle is from 0 when its velocity first becomes 0. Answers: v  12t 2  24t  9 2.1 v  12(2) 2  24(2)  9

v  48  48  9 v  9m / s 2.2

If v = 0 then 12t 2  24t  9  0 4t 2  8t  3  0 (2t  1)(2t  3)  0 t

2.3

1 3 s or t  s 2 2

s  4t 3  12t 2  9t  28

64 Grade 12 Core Mathematics 3

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ď&#x192;Ś1ď&#x192;ś ď&#x192;Ś1ď&#x192;ś ď&#x192;Ś1ď&#x192;ś s ď&#x20AC;˝ 4ď&#x192;§ ď&#x192;ˇ ď&#x20AC; 12ď&#x192;§ ď&#x192;ˇ ď&#x20AC;Ť 9ď&#x192;§ ď&#x192;ˇ ď&#x20AC;Ť 28 ď&#x192;¨2ď&#x192;¸ ď&#x192;¨2ď&#x192;¸ ď&#x192;¨2ď&#x192;¸

s ď&#x20AC;˝ 30m

Exercise 1.13: 1. The motion of a particle is given by: đ?&#x2018; = 9 â&#x2C6;&#x2019; 6đ?&#x2018;Ą + đ?&#x2018;Ą 2 . 1.1 Find s at đ?&#x2018;Ą = 0 ; 1 ; 2 ; 3. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 1.2 What is the particle doing during the interval (0 ; 3) for t ? _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 1.3 Find s at đ?&#x2018;Ą = 4 ; 5 ; 6. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 1.4

Describe the motion of the particle relative to 0 after đ?&#x2018;Ą = 3. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Find an expression for v in terms of t. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ Find v at đ?&#x2018;Ą = 1 ; đ?&#x2018;Ą = 2. v should be negative. What does this imply about the motion of the of the particle? (compare with 1.2) _____________________________________________________ 1.6

A cricket ball is thrown vertically up into the air. After x seconds, its height is y metres where đ?&#x2018;Ś = 50đ?&#x2018;Ľ â&#x2C6;&#x2019; 5đ?&#x2018;Ľ 2 . Determine: 2.1 the velocity of the ball after 3 seconds _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

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2.2 the maximum height reached by the ball. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 2.3 the acceleration of the ball. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 2.4 the total distance travelled by the ball when it returns to the ground. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ Exercise 1.14: Further questions: Question 1. The function f ( x)  2 x 2  1 is given. Determine the average gradient between x  2 and x  5 ? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

67 Grade 12 Core Mathematics Question 2.

From First Principles determine f (x) of :

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f ( x)  4 x 2

Determine:

lim

x 2  4 x  21 x3

3.1

x 3

3.2

lim 2 x

 5x  3

x 1

68 Grade 12 Core Mathematics 3.3

lim x x 3

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x3 2  3x ______________________________________________

Determine using standard rules.

f (x) if f ( x)  3x 3  2 x 2  3x 1  6 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

4.2

dy if y  (2 x  5)( x  3) dx

f (x) if f ( x) 

2  3x 2  4 3 x

69 Grade 12 Core Mathematics 5:

If g ( x)  3x 2

5.1

Determine g (x)

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Calculate the value of g (2) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

5.3

Explain what g (2) represents . ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

5.4

Find the co-ordinates of the point on the curve of g where the gradient is equal to 6. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

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Question 6: The equation, f ( x)  x 3  6 x 2  9 x , represents a curve graph. 6.1

Find the intercepts on the axes. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

6.2

Find the co-ordinates of the stationery points. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

6.3

Make a neat sketch of the graph. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Find the equation of the tangent to the curve when x = 2 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

Question 7: If f ( x)  x 3  4 x 2  4 x state the value for x when f is: 7.1 Increasing ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 7.2 Decreasing ______________________________________________ _____________________________________________________ _____________________________________________________

CALCULUS : TERMINOLOGY ETCETERA: 1. DERIVATIVE: This is the gradient function (or gradient at a particular point on a specific curve) and is denoted using the following symbols: (a) (b) (c) (d) 2.

f ( x ) Dx dy dx d dx

DIFFERENTIATION: Finding the derivative or gradient at a point of a function.

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Grade 12 Core Mathematics 3.

LIMIT FUNCTION: This is a point to which a function moves i.e. the furtherest extent that can be reached. When dealing with limits we only care about where we are going and not whether we get there. lim( x  2) [ means the limit of (x+2) when x tends towards 3 is] x 3

4. FIRST PRINCIPLES : This is a formal method of differentiating.[Finding the derivative ] 5. STANDARD RULES FOR DIFFERENTIATION: A short quick method of finding the gradient function (derivative). f ( x)  x n f ( x)  nx n1 6. TANGENTS: Lines touching at one point to a curve: 7. NORMALS: A line that is perpendicular to the tangent at the point of tangency. 8. LOCAL MAXIMUM/MINIMUM POINTS: Turning points on the curve. Also called STATIONARY POINTS. 9. POINTS OF INFLECTION: These are points where the curve does not turn BUT veers away in the same direction:



f ( x)  0

10. SECOND DERIVATIVE: derive the equation twice and use this equation to test for maximum or minimum turning points. f (x) 10.1 10.2

d2y dx 2

11.

INCREASING OR DECREASING FUNCTIONS:

f (c)  stationary value If the f (c)  ve then it shows an increasing function at x = c If the f (c)  ve then it shows a decreasing function at x = c. If the f (c)  0 then it has a stationary point at x = c. 12.

AVERAGE GRADIENT:

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Grade 12 Core Mathematics

This is the gradient measured between two (2) given points on a curve. Average gradient =

f ( x  h)  f ( x ) where x represents smaller x –value h

and (x + h) represents larger x – value.

f(x)=x2

f(x+h)

Ave MAB=

f ( x  h)  f ( x ) h

f(x)

13.

x+h

GRADIENT FUNCTION OR GRADIENT AT A POINT OR

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Grade 12 Core Mathematics DERIVATIVE FUNCTION

This is the gradient of a tangent at any particular point on a curve and is thus referred to as the gradient at that point of tangency. To calculate this value it is necessary to apply a limit to the average gradient formula. This moves the straight line between two (2) points to a single point on the curve i.e. a point of tangency. f ( x)  lim it h 0

f ( x  h)  f ( x ) h

[Using First Principles)

f ( x)  x n f ( x)  nx n1

[Using Standard Rules]

f(x) A

f(x+h)

Average Gradient between two points A & B Ave M AB = C f(x)

Gradient at one point C, point of tangency f ( x)  lim it

h 0

f ( x  h)  f ( x ) h

x+h

14. CONCAVITY: Concavity refers to the shape of the curve when approaching the stationary (turning) points. Concave up – refers to the minimum turning point.

Concave down – refers to the maximum turning point.

2. Number Patterns:

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Grade 12 Core Mathematics Arithmetic Progressions. 1.

Sequences: An arithmetic sequence is a string of values which increase or decrease by a constant value. This value is referred to as the “common difference” ( d). The first term of the sequence is referred to as “a”.The last term (nth) in the sequence is referred to as Tn. ( N.B. There are an infinite number of terms ,however you choose which one you want to use as the last term) The formula for the nth (last) term ( Tn) in a sequence is formulated as follows. If a sequence of numbers is : 7 ; 10 ; 13 ; 16 ; …..(to the nth term). T1 = 7 T2 = 7 + 3 = 10 T3 = 7 + 3 + 3 = 13 T4 = 7 + 3 + 3 + 3 = 16 T5 = 7 + 4(3) = 19 T10 = 7 + 9(3) = 34

Tn  a  (n  1)d last term

first term

common difference

second last term

Calculations in Arithmetic Progressions (AP’s). It is a good idea to develop the formula for the nth term of a progression prior to any calculation. It is an easier formula to work with and it is relevant to the progression in question. Example: A progression is given and you want to develop the nth term formula for it. 5;9;13;17...... i.e Simply use the general nth term formula for an arithmetic progression and substitute the „a‟ and „d‟ values into it and simplify. Tn  a  (n  1)d

Tn  5  (n  1)4 Tn  5  4n  4 Tn  4n  1 NOW if I need to find the 20th term simply substitute 20 for n in the formula above: i.e.

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Grade 12 Core Mathematics Tn  4n  1 T20  4(20)  1 T20  81

Further Examples: 1.

If a sequence of numbers is : 7 ; 10 ; 13 ; 16 ; …..(to the nth term).

1.1

Find the 20th term in the sequence. a = 7; d = 3; n = 20 ;T20 = ? Tn = a + (n-1)d T20 = 7 + 19(3) = 64 64 is the 20th term in the sequence.

1.2

If 94 is the nth term in the sequence find out the number of terms (n). Tn = 94 ; a = 7 ; d = 3 ; n = ? Tn = a + (n-1)d 94 = 7 + (n-1)3 94 = 7 +3n – 3 90 = 3n n = 30

1.3

94 is the 30th term in the sequence. If 94 is the 30th term in the sequence Find the first term. Tn = 94 ; ; d = 3 ; n = 30 ; a = ? Tn = a + (n-1)d 94 = a + 29(3) a=7 7 is the first term in the sequence.

1.4.

If 94 is the 30th term in the sequence and 7 is the first term find the common difference. Tn = 94 ; ; n = 30 ; a = 7 ; d = ? Tn = a + (n-1)d 94 = 7 + 29d 29d = 87 d=3 the common difference is 3

Finding an general formula that satisfies the nth term of a given sequence. i.e. you must be given or can calculate the 1st term and the common difference.

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Grade 12 Core Mathematics Example: If the first three terms of an arithmetic sequence is 3x  1 ; 2 x  3 ; 2 x  1 …… NB the constant concept is the common difference. Thus T2 – T1 = T3 - T2

2x  3  3x  1  2x  1  2x  3 x= 8 T1 = 23; T2 = 19 and T3 = 15

Tn = a + (n-1)d Tn = 23 +(n-1)d Tn = 23 –4d This is the general term representing the nth term of this specific sequence. 3.

Simultaneous Equations in AP’s Given that Tn = 25 and T11 = 81 find the arithmetic progression. i. e. find the first three terms of the sequence. Start with the nth term formula and write down a specific formula for the terms supplied. Tn = a + (n-1)d T11 = a + 10d = 81 T4 = a + 3d = 25 ( subtract the 2 equations to solve for „d‟) 7d = 56 d=8 a = 1 ( by substitution) AP = 1 ; 9 ; 17 ;….

Sum of an Arithmetic Progression N.B.

Theorem to be learned for testing. Number of terms Sn 

Sum of series

Example 1.

n 2a  (n  1)d  2

common difference Ist term

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In the series 1; 9; 17 find the sum of the first 20 terms. a = 1 ; d = 8 S20 = ? Sn 

n 2a  (n  1)d  2

S20 = 10[ 2 + 19(8)] = 1740 Example 2: If the first term of a arithmetic series is 1 and the sum of the first 20 terms is 1740. Find the common difference. Sn 

n 2a  (n  1)d  2

1740 = 10[2 + 19d] 1740 = 20 + 190d 1720 = 190d d = 8. Exercise 2.1: 1. Determine which term in the arithmetic sequence 3; 5; 7;……is equal to 27. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ 2. In the sequence, 23; 16; 9;… 2.1 Determine the 13th term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

79 Grade 12 Core Mathematics 2.2

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Which term in the sequence is -131?

3. Given the arithmetic sequence: 2; 3Â˝; 5;â&#x20AC;Ś 3.1 determine the 53rd term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________

3.2

Which term in the sequence is 53?

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4. Determine the 1st 3 terms in an arithmetic sequence with the 4th term equals 25 and the 11th term is 81. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________

5. The 5th term of an arithmetic progressions is 2 and the sum of the first 10 terms is 30. Determine the sum of the first 60 terms. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ 6. The first term of an arithmetic progression is 5 and the common difference is 2 . Find the number of terms that give a sum of 140. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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7. Evaluate the sum of the series: 1 â&#x20AC;&#x201C; 4 â&#x20AC;&#x201C; 9 - â&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Ś- 239. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 8. In an arithmetic series, the sum to n terms đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = đ?&#x2018;&#x203A;2 â&#x2C6;&#x2019; 2đ?&#x2018;&#x203A;. Determine: 8.1 The sum to 8 terms. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ 8.2

The eighth term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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9. The first 3 terms of an A.P. are : k + 1; k + 4 ; 4k + 1. Find the value of k and the sum of the first eighty terms. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 10. The second term of an arithmetic progression is 4 and the sixteenth term is 25. Find the first term and the common difference ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 11. In an arithmetic progression 23; 19; 15; â&#x20AC;Ś. 11.1 Determine the twelfth term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

83 Grade 12 Core Mathematics 11.2

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Which term in the sequence is –53?

If x+4; 3x – 1; 4x – 3 are the first three terms in an arithmetic progression determine: 12.1 the value of x ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

12.2 the first three terms of the sequence. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Grade 12 Core Mathematics

Geometric Progressions. SEQUENCES A geometric sequence is a string of values which increase or decrease by a constant value. This value is referred to as the “common ratio” ( r). The first term of the sequence is referred to as “a”. The last term (nth) in the sequence is referred to as Tn. ( N.B. There are an infinite number of terms ,however you choose which one you want to use as the last term) The formula for the nth (last) term ( Tn) in a geometric sequence is formulated as follows. If a sequence of numbers is : 7 ; 14 ; 28 ; 56 ; …..(to the nth term). T1 = 7 T2 = 7 .2 = 14 T3 = 7 .2.2 = 28 T4 = 7.2.2.2 = 56 T5 = 7 .24 = 112 T10  7.29  3584

second last term Tn  ar n1

last term

common ratio

first term

Calculations in Geometric Progressions (GP’s) Examples: 1 If a sequence of numbers is : 7 ; 14 ; 28 ; 56 ; …..(to the nth term). 2.

Find the 20th term in the sequence. a = 7; r = 2; n = 20 ;T20 = ? Tn  ar n1

T20 = 7.219 = 3670016 3670016 is the 20th term in the sequence. 3.

If 3584 is the nth term in the sequence find out the number of terms (n). Tn = 3584 ; a = 7 ; r = 2 ; n = ? Tn  ar n1

3584 = 7.2n-1 512 = 2n-1 29 = 2n-1 n–1=9 n = 10 3584 is the 10th term in the sequence.

85 Grade 12 Core Mathematics 4.

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If 3584 is the 10th term in the sequence Find the first term. Tn = 3584 ; ; r = 2 ; n = 10 ; a = ? Tn  ar n1

3584 = a .29 a=

3584 512

a=7 7 is the first term in the sequence. 5.

If 3584 is the 10th term in the sequence and 7 is the first term find the common ratio. Tn = 84 ; ; n = 10 ; a = 7 ; r = ? Tn  ar n1

3584 = 7.r9 512 = r9 29 = r9 r=2 the common ratio is 2

Finding an general formula that satisfies the nth term of a given sequence. i.e. you must be given or can calculate the Ist term and the common difference.

Example: If the first three terms of an Geometric sequence is 3x  1 ; 2 x  3 ; 2 x  1 …… NB the constant concept is the common ratio. T T Thus 2 = 3 T1 T2 2x  3 2x  1  3x  1 2 x  3

(2 x  3) 2  (3x  1)(2 x  1)

4 x 2  12 x  3  6 x 2  5x  1 Tn = 23 –4d This is the general term representing the nth term of this specific sequence.

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Grade 12 Core Mathematics

SERIES

Sum of a geometric progression. N.B. learn the theorem for testing. a(r n  1) Sn  for r 1

Sn 

a(1  r n ) for 1 r

r > 1 or r < -1 i.e. r is a whole number

-1 < r < 1 i.e. r is a fraction.

Example 1: In the series 2 + 6 + 18+…… find the sum of the first 5 terms. a = 2; r = 3; S5 = ?

a(r n  1) Sn  r 1 2(35  1) S5 = 2 S5 = 80.

Example 2: If 2 + 6 + 18……(to n terms) = 80. Find the number of terms in the series. i.e. Find n.

a(r n  1) Sn  r 1 2(3 n  1) 80 = 3 1 80 = 3n – 1 81= 3n 34 = 3n n = 4.

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Grade 12 Core Mathematics Exercise 2.2: 1.

Calculate the tenth term of a sequence: 81; 27; 9; â&#x20AC;Ś ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

2. If the first term of a sequence is

3 2

and the fourth is â&#x20AC;&#x201C;12, determine:

2.1 the second and third terms if the sequence is arithmetic. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 2.2 The second and third terms if the sequence is geometric. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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3. Find the 10th term in a geometric progression where 1st term is 5 and the common ratio is 3. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 4.

If the 7th term is 192 and the 2nd term is 6 find the geometric sequence. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

Three consecutive terms of a geometric sequence are 3x-2; 2x+2 and 4x+1. 5.1 Determine the value of x, if x is a natural number. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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Grade 12 Core Mathematics 5.2

Determine the common ratio of the sequence ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

SIGMA NOTATION Sigma notation refers to the sum of a series and will always have a sum total as an answer. Number of terms in the series

n  (2K  1)  S n k 1 Start by substituting from 1 (consecutively) to get the components of the series.

Sum of the series

series

If there is no statement as to the type of progression (i.e. AP or GP) then a test must be carried out to ascertain whether the series has a common difference or a common ratio To do this simply substitute from value 1 into series and get the 1st three terms and then do the check. T i.e. d = T2 – T1 and r = 2 T1

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Grade 12 Core Mathematics Examples: 5

Expand

 2k  3  1  1  3  5  7  15 k 1

Evaluate:

 4k  2

a = 6 T2 = 10 T3 = 14 AP d = 4

k 1

n 2a  (n  1)d  2 10  2(6)  (10  1)4 2

Sn  S10

= 5[12 + 36] = 24 8

Evaluate:

 2(3)

r 1

a = 2 T2 = 6 T3 = 18 GP r = 3

r 1

a(r n  1) r 1 2(38  1) S8  3 1 8 = 3 -1 = 6561

Sn 

Exercise 2.3: 1.

Evaluate: 8

1.1

 3k  1 k 1

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 5k  2

1.2

k 1

1.3

 4k  2 k 3

1.4

 3(2) k 1

k 1

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Grade 12 Core Mathematics 10

1.5

 5(3) n1 n 1

Converging series: i.e.

A geometric series that has a common ratio that lies between 1 and -1 –1 < r < 1 is called a converging series or a series that sums to infinity.

A series with a r – value that is a whole number i.e r >1 or r < -1 is a diverging series. S 

a 1 r

Example: In a converging series , 1 +

1 1  ……… 2 x  1 (2 x  1) 2 -1 <

1 <1 2x  1

1  1 2x  1

1 1 2x  1

1 1  0 2x  1

1 1  0 2x  1

1 + 2x- 1 > 0 2x-1

1 –2x + 1 < 0 2x - 1

2x 0 2x  1

x < 0 and x >

2  2x 0 2x  1

1 2

0 1 2

x< 0

1 2

Solution :

1 x < 0 or x > 1

1 2

and x > 1

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Grade 12 Core Mathematics Alternatively:

Due to the fact that inequalities are absolute values , the above example can be set out in a simpler way: -1 <

1 <1 2x ď&#x20AC; 1

If you invert the fraction between the inequality signs you must then invert all values including the inequality signs. e.g. 1 Invert 2đ?&#x2018;Ľâ&#x2C6;&#x2019;1; Do not 2x ď&#x20AC; 1 forget to invert the ď&#x20AC;1 ď&#x20AC;ž ď&#x20AC;ž1 inequality signs: 1 Add 1 to each side and 0 > 2x > 2 simplify. 0 > x > 1 OR

Solution :

x < 0 and x > 1

Exercise 2.4: Mixed Progressions: 1.

In a GP with first 3 terms: 5k + 1 ; 2k + 2; k + 1â&#x20AC;Śâ&#x20AC;Ś.. Find the the value of k. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Grade 12 Core Mathematics 2.

Use a suitable formula to find which term in an Arithmetic Series –61 – 58 – 55 ----- is the first term to exceed 10. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ n

Find the largest number for

 (2r  3)  48

r 1

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Grade 12 Core Mathematics n

If T2 = 8 and T6 = 24 determine n if

 Tk

 480

k 1

Find n of and AP 15 + 13 + 11 ---- whose sum = - 36. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ______________________________________________

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Given the sequence 23 ; 27 ; 31 ------- Find: 6.1 The number in the sequence which will be greater than 5000. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 6.2

How many terms must be added in the sequence so that the sum is greater than 5000.

If T3 = 8 and T8 = 7.1

1 in a GP Find: 4

The common ratio ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ 7.2

The sum of the first 8 terms.

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The sum of the first n- terms of an arithmetic sequence is Sn = n2 + 4n. 8.1

Calculate the first 4 terms of the sequence.

_____________________________________________________ _____________________________________________________ _____________________________________________________

The sum of the first n â&#x20AC;&#x201C; terms of an AP is given by 2n2 â&#x20AC;&#x201C; n. 9.1

Calculate the first 3 terms of the sequence. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

98 Grade 12 Core Mathematics 9.2

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Determine a formula for the n –th term of this sequence. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

10.

The first 3 terms of an arithmetic sequence are 2x – 4 ; x – 3 and 8 – 2x . Determine the value of x and hence the sum of the first 20 terms. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

11.

The sum of the first 7 terms of an arithmetic series is 126 and the 20th term is 130. Determine the tenth term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

99 Grade 12 Core Mathematics 12.

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The sum of the first 12 terms of an arithmetic progression is 186. The 6th term is 14. Calculate the first 3 terms of the progression. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

13.

The sum of three consecutive terms of an AP is 18. Their product is 192. Calculate the numbers. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

14.

Calculate the value the value of n if : n

14.1

 (4k  6)  240

k 1

100 GO MATH WORKBOOKS

Grade 12 Core Mathematics n

_14.2

 5.3

r 1

 605

r 1

15.

The 3rd term of a geometric series is

2 2 and the 8th term is  . 9 2187

Find the first term and the common ratio. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

101 GO MATH WORKBOOKS

Grade 12 Core Mathematics 16.

The first term of an arithmetic series is 2 The sum of the 3rd and 11th term is 40. 16.1 Find n if the n-th term of the series is 212. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ______________________________________________ 16.2

Determine the sum of the first 71 terms of the series.

______________________________________________ _____________________________________________________ _____________________________________________________

1  4 2  k 1 n

17.

Solve for n , the number of terms, if

k 1

 7 63 64

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Grade 12 Core Mathematics  1  27 3  k 1 n

18.

Calculate the value of n if:

k 1

 21

19.

The sum to infinity of a geometric series is 81 and the sum of the first 3 terms of this series is 57. Find the first term and the common ratio. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

103 Grade 12 Core Mathematics 20.

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8 . The sum of the first 3 terms of a geometric sequence of which the terms are positive is 1 49

If the first term is 1 , find : 20.1 the common ratio. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 20.2 The sum to infinity. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

21.

Consider the infinite geomtric series: ( x  2) 2  ( x  2) 3  ( x  2) 4  ........

21.1 Write down the common ratio in terms of x. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

104 GO MATH WORKBOOKS

Grade 12 Core Mathematics 21.2

Determine the value(s) of x for which the series will converge.

21.3

If the sum to infinity of this series is

xď&#x20AC;2 , calculate the value(s) of x. 3

105 Grade 12 Core Mathematics

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Financial Maths ⟹ Logarithms: The Log of a number is the index to which a given base must be raised to give the number. e.g. 32 = 9 converts to  2 = log 3 9 and reads : “ the log of 9 to the base 3 is 2 Log form  Exponential form We can always convert the log form to exponential form and vice versa. e.g.

N.B.

log 2 64 = 6  26 = 64 log 3 19 = - 2  3-2 = 19 log 100 = 2  102 = 100 Log a 1 = 0  a0 = 1 92 = 81  log 9 81 = 2 xm = p  logxp = m

Examples: (1) Find log 2 32. (a) Let log 2 32 = x or (b) log 232 = log 225 2x = 32 = 25 =5 (2.)

Find x if log 3 = 4 34 = x (log to exp form) 81 = x

Mathematical definition of a log: z = log a x  az = x a > 0 ; a 1 and x > 0 for z  Laws of logs:

Product Law: Log a xy = log a x + log a y examples: (a) log 3 27 + log 3 3 = log 3 (27 . 3) compare (b) log 3 27 . log 3 3 = log 3 81 =3.1 = log 3 34 =3 =4 c) Expand log10x  log 10 + log x = 1 + log x

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Grade 12 Core Mathematics

Quotient Law: Log a

x y

= log a x â&#x20AC;&#x201C; log a y

Examples: 1. log 2 32 ď&#x20AC; log 2 8 ď&#x20AC;˝ log 2

32 8

ď&#x20AC;˝ log 2 4 ď&#x20AC;˝ 2

ď&#x192;Ś 15.3.4 ď&#x192;ś log15 + log3 â&#x20AC;&#x201C; log 5 + log4 = logď&#x192;§ ď&#x192;ˇ = log 36 ď&#x192;¨ 5 ď&#x192;¸ log 15 + log 3 â&#x20AC;&#x201C; (log 5 + log 4) = log 15 + log 3 â&#x20AC;&#x201C; log 5 â&#x20AC;&#x201C; log 4

Power Law:

log a x m ď&#x20AC;˝ m log a x

Examples: 1. log x 2 = 2 log x 25 2. 2 log 5 â&#x20AC;&#x201C; 3 log 4 = log52 â&#x20AC;&#x201C; log 43 = log 64

Change of base If logs of numbers have different bases, the log laws do not apply . Thus simplification is impossible unless we prove the law which involves a change of base. Law:

loga x ď&#x20AC;˝

logb x logb a

Example.

log 5 2 ď&#x20AC;˝

log 3 2 log 8 2 ď&#x20AC;˝ log 3 5 log 8 5

Inverse functions Exponent Versus Logarithm đ?&#x2019;&#x2021;đ?&#x2019;&#x2013;đ?&#x2019;?đ?&#x2019;&#x201E;đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;?; đ?&#x2019;&#x161; = đ?&#x2019;&#x201A;đ?&#x2019;&#x2122; â&#x2020;&#x2019; đ?&#x2019;&#x2020;đ?&#x2019;&#x2122;đ?&#x2019;&#x2018;đ?&#x2019;?đ?&#x2019;?đ?&#x2019;&#x2020;đ?&#x2019;?đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;&#x201A;đ?&#x2019;? đ?&#x2019;&#x2021;đ?&#x2019;&#x2013;đ?&#x2019;?đ?&#x2019;&#x201E;đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;?. đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;&#x2014;đ?&#x2019;&#x2020;đ?&#x2019;&#x201C;đ?&#x2019;&#x201D;đ?&#x2019;&#x2020; đ?&#x2019;&#x2021;đ?&#x2019;&#x2013;đ?&#x2019;?đ?&#x2019;&#x201E;đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;?: đ?&#x2019;&#x2122; = đ?&#x2019;&#x201A;đ?&#x2019;&#x161; â&#x2020;&#x2019; đ?&#x2019;?đ?&#x2019;?đ?&#x2019;&#x2C6;đ?&#x2019;&#x201A;đ?&#x2019;&#x201C;đ?&#x2019;&#x160;đ?&#x2019;&#x2022;đ?&#x2019;&#x2030;đ?&#x2019;&#x17D;đ?&#x2019;&#x160;đ?&#x2019;&#x201E; đ?&#x2019;&#x2021;đ?&#x2019;&#x2013;đ?&#x2019;?đ?&#x2019;&#x201E;đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;?. đ?&#x2019;&#x2DC;đ?&#x2019;&#x201C;đ?&#x2019;&#x160;đ?&#x2019;&#x2022;đ?&#x2019;&#x2022;đ?&#x2019;&#x2020;đ?&#x2019;? đ?&#x2019;&#x160;đ?&#x2019;? đ?&#x2019;?đ?&#x2019;?đ?&#x2019;&#x2C6; đ?&#x2019;&#x2021;đ?&#x2019;?đ?&#x2019;&#x201C;đ?&#x2019;&#x17D; đ?&#x2019;&#x201A;đ?&#x2019;&#x201D;: đ?&#x2019;&#x161; = đ?&#x2019;?đ?&#x2019;?đ?&#x2019;&#x2C6;đ?&#x2019;&#x201A; đ?&#x2019;&#x2122; Method : Swop the x & y variables and write in the appropriate form. NB If y is an exponent then the correct form is logarithm.

107 GO MATH WORKBOOKS

Grade 12 Core Mathematics Example: đ?&#x2019;&#x2021;đ?&#x2019;&#x2013;đ?&#x2019;?đ?&#x2019;&#x201E;đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;? â&#x2C6;ś đ?&#x2019;&#x161; = đ?&#x;&#x2018;đ?&#x2019;&#x2122;

đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;&#x2014;đ?&#x2019;&#x2020;đ?&#x2019;&#x201C;đ?&#x2019;&#x201D;đ?&#x2019;&#x2020; đ?&#x2019;&#x2021;đ?&#x2019;&#x2013;đ?&#x2019;?đ?&#x2019;&#x201E;đ?&#x2019;&#x2022;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2019;?: đ?&#x2019;&#x2122; = đ?&#x;&#x2018;đ?&#x2019;&#x161; â&#x2020;&#x2019; đ?&#x2019;&#x161; = đ?&#x2019;?đ?&#x2019;?đ?&#x2019;&#x2C6;đ?&#x;&#x2018; đ?&#x2019;&#x2122; Exponential function

hď&#x20AC;¨xď&#x20AC;Š = 3x

tď&#x20AC;¨yď&#x20AC;Š = 3y

Log function: đ?&#x2019;&#x161; = đ?&#x2019;?đ?&#x2019;?đ?&#x2019;&#x2C6;đ?&#x;&#x2018; đ?&#x2019;&#x2122; -5

-2

-4

Logarithms used in Financial Math: Logarithms are used to calculate the time aspect in financial maths. đ??šđ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;˘đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019;

Non â&#x20AC;&#x201C; Annuities:

đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x2018; =

đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D; đ?&#x2018;&#x192;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019; log â Ą(1+đ?&#x2018;&#x2013;)

Annuities: n refers to the number of time periods. ď&#x192;Š Fv (i ) ď&#x192;š log ď&#x192;Ş1 ď&#x20AC;Ť x ď&#x192;şď&#x192;ť ď&#x192;Ť Future value: n ď&#x20AC;˝ log(1 ď&#x20AC;Ť i ) ď&#x192;Š Pv (i) ď&#x192;š log ď&#x192;Ş1 ď&#x20AC; x ď&#x192;şď&#x192;ť ď&#x192;Ť ď&#x20AC;n ď&#x20AC;˝ log(1 ď&#x20AC;Ť i) Present Value:

108 Grade 12 Core Mathematics

Financial Maths

Topics: 1.

Simple Interest

Compound Interest

Simple Decay

Compound Decay

Annuities

5.1

Future Value

5.2

Present Value

5.3

Final Payments

5.4

Deferred Payments

5.5

Fixed Payments

Depreciation: 6.1

Book Value

6.2

Scrap Value

6.3

Sinking Fund

Loans: 7.1

Repayments with Future Value Formula.

7.2

Balance on a loan

7.3

Present Value Formula.

Calculation of Time Period “n”

Microlenders

10.

Pyramid Schemes

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109 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Financial Mathematics: Introduction: No business can exist without the information given by figures. Borrowing, using and making money is the heart of the commercial world thus the principle of interest and interest rate calculations are extremely important. This leads into an examination of the principles involved in assessing the value of money over time and how this Information can be utilized in the evaluation of alternate financial decisions. Remember that the financial decision area is a minefield in the real world, full of tax implications, depreciation allowances, investment and capital allowances etc.

The basic principles in financial decision making are established through the concept of interest and present value: â&#x20AC;&#x201C; Definition of interest:

Interest is the price paid for the use of borrowed money Interest is paid by the user of the money to the supplier of it. It is calculated as a fraction of the amount borrowed or saved over a certain period of time. This fraction is also known as interest rate and is expressed as a percentage per year (per annum).

Present Value of money is the value of the initial investment:

ie PV (Present Value) = P (Principle) PV or P

t = term

r= interest Simple interest:

PV = Present Value or Principle FV = Future Value or Sum Assured

FV or S = P(1 + rt)

110 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Simple interest: is computed on the principle for the entire term of the loan and is thus due at the end of term. Growth Decay A  P(1  ni) A  P(1  ni) S I = Prt I is the interest paid or earned P is the principle or Present value r is the interest rate per annum t is the time or term of loan n is the number of years Compound Interest: Compound interest arises when, in a transaction over an extended period of time, interest due at the end of a payment period is not paid, but added to the principal. Thus interest also earns interest i.e. it is compounded. The amount due at the end of transaction period is referred to as the compounded amount or accrued principal. Interest periods Can vary : daily, monthly, quarterly, half-yearly or yearly. Formula: Compound Growth:

Compound Decay:

Fv  Pv (1  i) n OR

i   Fv  Pv 1   m 

Fv  Pv (1  i) n tm

J   Fv  Pv 1  m  m  

Fv(A)(S) = Amount or Future Value Pv(P) = Principal or Initial value r = rate of interest per annum n (t) = number of years invested m = number of time periods interest is calculated ( annum, quarterly, half yearly, monthly or daily)

(S ) FV  Future Value

Fv(S) = Accrued amount / Future value Pv(P) = Initial principle / present value i = the annual interest rate compounded m = times per year t = the number of years of investment. m = the number of compounded periods per year Jm = the annual interest rate.

111 GO MATH WORKBOOKS

Grade 12 Core Mathematics Further Formulae: 1. Finding Principle: P  A(1  i)  n NB: i 

P  S (1 

j m tm ) m

jm and n  tm m

Jm = nominal interest rate. m = no of interest periods involved. n = tm = total no of time periods. t = no of years invested. S = Accrued amount / Future value P = Initial principle / present value

2. Finding the interest:

I     S  tm  j m  m    1 or J  m m 1  i   1 m  P     Where i = Effective interest rate.



Nominal Interest rates: 1.1.

In cases where interest is calculated more than once a year, the annual rate quoted is the nominal annual rate or nominal rate.

Effective Interest rates: 1.2. If the actual interest earned per year is calculated and expressed as a percentage of the relevant principal , then the so-called effective rate is obtained. This is the equivalent annual rate of interest – that is, the rate of interest earned in one year if compounding is done on a yearly basis.



112 GO MATH WORKBOOKS

Grade 12 Core Mathematics Converting Nominal Rate to Effective Rate:

Take the Nominal Rate and divide by the number of time periods involved and apply this to the formula: Eff Rate = [ 100(1  i) n -100] nominal rate i and n = number of time periods in 1 year. time periods EG The nominal rate of interest is 22% calculated half yearly. What is the corresponding effective rate of interest: 22 2  11 % Thus R100(1.11)  R123.21 effective Interest rate is 23,21% 2

Similarly: J eff

m  J m  m  J  100 1    1 or 1  i   1  m  m m    

J eff  Effective rate

i = effective interest rate

J m  Nominal Rate

m  time periods ( number of time periods per annum)

Annuities: Definition: An annuity is a sequence of equal payments at equal intervals of time.

The payment interval of an annuity is the time between successive Payments while term is the time from the beginning of the first payment interval to the end of the last payment interval.

Future Value Annuities: The formula for the sum of a geometric series is used in financial maths to Calculate values of annuities.

a(r n  1) Sn  r 1 a is the first term or payment made r is the common ratio ( 1  i) n is the number of payments or no of terms in the series.

113 GO MATH WORKBOOKS

Grade 12 Core Mathematics Example:

Twenty equal payments are made into a savings account annually at the beginning of each year. ( effective immediately) Calculate the total accumulated amount at the end of 20 years if an interest rate of 9% compounded annually is applied. 2 8000

8000 17

1 8000

8000 8000

T19 represents the beginning of the 20th year as well as the end of the 19th year. Similarly T1 represents the beginning of the 2nd year and the end of the 1st year. a = R8000(1,09) r =1,0 9% n = 20 a(r n  1) r 1 8000(1,09)[1,09 20  1]  1,09  1  R 446116,24

Sn  S 20 S 20

R700 p.m into a pension scheme for 40 years. Interest rate is 11,4%

a = R700(1,14) r =1,14

n  12  40  480 a (r n  1) r 1 700(1,0095)[1,0095 480  1]  1,0095  1  R6884120,61

Sn  S 20 S 20

114 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Nominal Rate

FORMULAE IN FINANCIAL MATHS: Future Value Formula: x[(1  i) n  1] Fv  i x

x[(1  i) n1  1] Fv  i



Fv (i )

x



(1  i)  1 n



x

No of time periods

Fv (i ) (1  i) n1  1

t = no of years

n = total no of time periods

Present Value Formula: x[1  (1  i)  n ] Pv  i

Jm NB: m n  tm i

Can only use Pv if a Gap in the front

Pv (i ) 1  (1  i )  n

Balance of Loan Formula: Balance of loan = Pv (1  i) n -

OR Pv (1  Jm ) tm m





x (1  i) n  1 i

Jm tm   x (1  )  1 m   Jm m

Sinking Fund Formulae: Scrap Value = A  P(1  i) n New Value = A  P(1  i) n Sinking Fund = New Value – Scrap Value Monthly Payment into Fund :

x

SF (i) (1  i) n  1

SF = Sinking Fund

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Grade 12 Core Mathematics

Effective Rate:

m ď&#x192;Šď&#x192;Ś ď&#x192;š Jm ď&#x192;ś Jeff ď&#x20AC;˝ 100ď&#x192;Şď&#x192;§ 1 ď&#x20AC;Ť ď&#x20AC; 1 ď&#x192;ş ď&#x192;ˇ mď&#x192;¸ ď&#x192;Şď&#x192;Ťď&#x192;¨ ď&#x192;şď&#x192;ť

Jm = Nominal rate Jeff = effective rate

Nominal Rate = Jm ď&#x20AC;˝ 100mď &#x203A;m 1 ď&#x20AC;Ť Jeff ď&#x20AC; 1ď ? Time Periods: Give Present Value and Future Value: A P nď&#x20AC;˝ log(1 ď&#x20AC;Ť i ) log

Give Fv and x ( payment) : Give Pv and x ( payment) ď&#x192;Š Fv (i ) ď&#x192;š log ď&#x192;Ş1 ď&#x20AC;Ť x ď&#x192;şď&#x192;ť ď&#x192;Ť nď&#x20AC;˝ log(1 ď&#x20AC;Ť i )

ď&#x192;Š Pv (i) ď&#x192;š log ď&#x192;Ş1 ď&#x20AC; x ď&#x192;şď&#x192;ť ď&#x192;Ť ď&#x20AC;n ď&#x20AC;˝ log(1 ď&#x20AC;Ť i)

Final Payment: đ??żđ?&#x2018;&#x2019;đ?&#x2018;Ą đ?&#x2018;Ś = đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;Śđ?&#x2018;&#x161;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą

đ?&#x2018;Ś=

đ?&#x2018;Ľ (1â&#x2C6;&#x2019;(1+đ?&#x2018;&#x2013;)â&#x2C6;&#x2019;đ?&#x2018;&#x203A; đ?&#x2018;&#x2013; â&#x2C6;&#x2019;đ?&#x2018;&#x203A; (1+đ?&#x2018;&#x2013;) â&#x2C6;&#x2019;1

đ?&#x2018;&#x192;đ?&#x2018;Łâ&#x2C6;&#x2019;

NB: In final payment calculations any deferment must be taken into account. This affects the Pv value in the formula above. đ?&#x2018;&#x192;đ?&#x2018;Ł = đ?&#x2018;&#x192;đ?&#x2018;Ł(đ?&#x2018;&#x2013;) e.g. a delay of 3months affects the Pv as follows: Recalculated Pv = đ?&#x2018;&#x192;đ?&#x2018;Ł 1 +

đ?&#x2018;&#x2013; đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x2018;

đ?&#x2018;&#x203A;đ?&#x2018;&#x153; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018;&#x201C;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x;đ?&#x2018;&#x161;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ąđ?&#x2018;

116 Grade 12 Core Mathematics

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Deferred Payments: đ?&#x2018;Ľ[ 1+đ?&#x2018;&#x2013; đ?&#x2018;&#x203A; 1 â&#x2C6;&#x2019;1]

đ??šđ?&#x2018;Ł =

Notes:

đ?&#x2018;&#x2013;

Ă&#x2014; (1 + đ?&#x2018;&#x2013;)đ?&#x2018;&#x203A; 2 Ă&#x2014; (1 + đ?&#x2018;&#x2013;)đ?&#x2018;&#x203A; 3

đ?&#x2018;&#x203A;1 = đ??´đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x161;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x2018;: đ?&#x2018;&#x203A;1 â&#x2C6;&#x2019; đ?&#x2018;&#x203A;2 â&#x2C6;&#x2019; (đ?&#x2018;&#x203A;3 ) đ?&#x2018;&#x203A;2 = đ??ˇđ?&#x2018;&#x2019;đ?&#x2018;&#x201C;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x161;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x2018; đ?&#x2018;&#x203A;3 = đ??źđ?&#x2018;&#x203A;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x161;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x2022;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x201C;đ?&#x2018;&#x153;đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x2018; đ?&#x2018;&#x2013;. đ?&#x2018;&#x2019;. đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1 NB: There is a gap at the end. Example:

Formula:

0.16 99 ď&#x192;š ď&#x192;Š 2600ď&#x192;Ş(1 ď&#x20AC;Ť ) ď&#x20AC; 1ď&#x192;ş 12 ď&#x192;Ť ď&#x192;ť ď&#x201A;´ (1 ď&#x20AC;Ť 0.16 ) 9 (1 ď&#x20AC;Ť 0.16 )12 Fv ď&#x20AC;˝ 0.16 12 12 12 Fv ď&#x20AC;˝ R698150,10 Notes:

đ?&#x2018;&#x203A;1 = 120 â&#x2C6;&#x2019; 9 + 12 = 99 đ?&#x2018;&#x203A;2 = 9 đ?&#x2018;&#x203A;3 = 12

117 GO MATH WORKBOOKS

Grade 12 Core Mathematics

FUTURE VALUE ANNUITIES: Investments starting at the beginning of the Month (Immediate) There is a gap at the end. i.e. the investment ends at the second last period. [NB CanT1only use Fv formula here] T0 Tn1

time Tn

Gap at End The investment starts at T0 and ends at Tn -1 . You must add 1 to exponent in the following formula. The formula to us is:

x[(1  i) n  1] Fv  i x

Fv (i ) (1  i)  1 n



x[(1  i) n1  1] Fv  i x



Jm NB: m n  tm i

Fv (i ) (1  i) n1  1

Example: 1.1

An investor decides to save money for ten years in a Unit Trust Fund. He immediately deposits R800 into a savings account. Thereafter , at the end of each month he deposits R800 into the fund and continues to do this for a ten year period. Interest is 15% compounded monthly.

1.2

If the investor leaves his investment in the fund to grow for two years without making further payments of R800. The interest rate changes to 14% p.a. compounded quarterly. Calculate the value of his investment after the two year period. Answer: T0

Tn1

800 1.1

800 Fv 

x[(1  i) n1  1] i

0,15 121 )  1] 12 Fv  0,15 12 Fv = R 223725,81 800[(1 

A  P(1  i ) n

1.2

A  223725,82(1  A  R 272588,19

0.14 8 ) 4

Gap at End

118 Grade 12 Core Mathematics

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Exercise 3.1: 1.

Alida decides to start saving for a car. Starting immediately on her 16th birthday she deposits R5000 into a bank account with an interest rate of 18% p.a. compounded quarterly. She continues to make quarterly payments until the last payment on her 24th birthday. How much money will she then have at her disposal to finance the purchase of a new car? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

John decides to invest money into a share market in order to become a millionaire in ten years time. He believes that he can average a return of 25% p.a. compounded monthly. In one months time he wishes to start making monthly payments into an account. How much must he invest per month in order to obtain his R1 million? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

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Aisha wants to save up R250 000,00 in 5 years time in order to purchase a car. She makes monthly payments into an account paying 13% p.a. compounded monthly, starting immediately. How much will she pay each month? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

R500 is invested each month starting in one months time into an account paying 16% p.a. compounded monthly. How long will it take to accumulate R10 000,00? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

120 Grade 12 Core Mathematics 5.

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R2000 is immediately deposited into a savings account. 6 months later and every 6 months thereafter , R2000 is deposited into the account. The interest rate is 16% p.a. compounded half-yearly. How long will it take to accumulate R100 000,00 ? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

Lebogeng deposits R5000 into an account paying 14% p.a. compounded half-yearly. 6 months later she deposits R400 into the account. 6 months after that she deposits a further R400. She then continues to make half-yearly deposits of R400 for a further 9 years. Calculate the value of her savings at the end of the savings period. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________

121 Grade 12 Core Mathematics 7.

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Mr and Mrs Morogeng are newly â&#x20AC;&#x201C; married and buy a house for R350 000,00. They pay R50 000,00 in cash and take out a home loan for the balance. The interest is calculated at 8,5% p.a. compounded monthly on the home loan. 7.1 Calculate the monthly repayments on the loan if it is repaid over a 30 year period. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 7.2 How much would they need to repay each month if they decided to repay the loan over 20 years? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

122 GO MATH WORKBOOKS

Grade 12 Core Mathematics

7.3 Calculate how much money would be paid in total in each case to repay the loan. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

PRESENT VALUE ANNUITIES: The formula for Pv can only be used if there is a gap at the begining (Delayed payment). Pv  T0

Gap at Begining

x[1  (1  i)  n ] i

800



x

Pv (i ) 1  (1  i )  n Tn

800

Example: How much can be borrowed from a bank if the borrower repays the loan by means of 30 equal monthly repayments of R1250,00 starting in one months time if the interest rate is 14% p.a compounded monthly.

Pv 

x[1  (1  i)  n ] i

0,14 30 )  1] 12 Pv  0,14 12 Pv  R31487,44 1250[1  (1 

123 Grade 12 Core Mathematics

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Exercise 3.2: 1.

How much can be borrowed from a bank if the borrower repays the loan by means of equal quarterly repayments of R2000,00 starting in 3 months time? The interest rate is 18% p.a compounded quarterly and the duration of the loan is 10 years. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

25 semi annual payments are made starting 6 months from now in order to repay a loan of R100 000,00. What is the value of each payment if the interest is 18,6% p.a. compounded semi-annually. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

124 Grade 12 Core Mathematics 3.

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What amount must be invested in order for an investor to receive equal payments of R2000 per month from a bank for 3 years starting in one months time ? Interest is 18% p.a. compounded monthly. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

John inherits R1 000 000,00 from his late father. He invests the money at an interest rate of 14% p.a. compounded monthly. He wishes to earn a monthly salary from the investment for a period of 20 years, starting one month from now. How much will he receive each month? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

125 Grade 12 Core Mathematics 5.

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Mr Govender starts a business and takes out a loan of R500 000,00. He repays the loan by means of equal quarterly payments, starting 3 months after the loan was drawn. The loan is repaid over a ten year period at an interest rate of 8% p.a. compounded quarterly. Calculate his quarterly payments. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

Mr Ndlovu buys a â&#x20AC;&#x17E;bakkieâ&#x20AC;&#x; for his business. The vehicle costs him R120000,00 . He repays 20% in cash and the balance using a bank loan. The interest levied is 11% p.a. compounded monthly. Calculate the monthly repayments if he pays the loan over 4 years. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

126 Grade 12 Core Mathematics 7.

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Mark takes out a retirement annuity that will supplement his pension when he retires thirty years from now. He estimates that he will need R2,5 million in this retirement fund at that stage. The interest rate he earns is 9% p.a. compounded monthly. 7.1

Calculate his monthly payment into the fund if he starts paying immediately and makes his final payment in 30 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

7.2

The retirement fund does not pay out the 2,5 million when Mark retires. Instead he will be paid monthly amounts for a period of 20 years, starting one month after he retires. If the interest rate that he earns over this period is calculated at 7% p.a. compounded monthly, determine the monthly payments he will receive. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

127 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Calculating the balance of a loan:



 

x (1  i) n  1 A  Pv (1  i) i x (1  i) n  1 n OR Balance of Loan = Pv (1  i) i n

and B 



Best to use this formula

Where n = Actual number of payments made. Hence balance of loan = A - B Example: A car loan of R130 000,00 is repaid over 5 years by means of equal monthly payments starting one month after the loan was granted. a)

Calculate the monthly repayments if the interest on the loan is 7% p.a. compounded monthly on a reducing balance.

The client experiences financial difficulties and is unable to pay 18th to 21st payments. Calculate the balance of the loan at the end of the 17th month.

The loan is rescheduled, and payments increased so that the loan is still amortized over the same agreed period. Calculate the increased monthly installments after the default payments , if the interest rate is 7% p.a. compounded monthly.

Pv  130000 T0

Gap at Begining

NB: Pv 

x[1  (1  i)  n ] i

x

n5

Pv (i )

1  (1  i )  n 0,07 130000( ) 12 x 0,07  60 1  (1  ) 12 x  R 2574,16

128 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Balance of loan: Pv (1  i) n 





x (1  i) n  1 (1  i)  1

0.07 17   2574.16(1  )  1 0,07 17 12   Balance  130000(1  )  0 . 07 12 (1  ) 1 12 Balance of loan = R 97 647,50

x

Pv (i ) 1  (1  i )  n  0.07  138924.82   12 

 0.07  1  1   12   x  R 2872,15

 43

Exercise 3.3: 1.

Waydene wants to buy a car costing R192000. She takes out a loan for 5 years with interest at 12% p.a. compounded monthly. 1.1 Calculate the monthly installments she will have to pay on the loan. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

129 Grade 12 Core Mathematics 1.2

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After she has paid 45 installments she decided to settle the balance on the loan. Calculate the lump sum she must pay after the 45th installment. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2.1

Mrs Jones buys a townhouse for R800 000,00 . She pays 15% deposit in cash and takes a home loan for the balance. The interest rate charged is 9,5% p.a. compounded monthly. Calculate her monthly repayments if the loan is amortised over a period of 20 years. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

2.2

How much money does she pay over the 20 years to repay the loan. ______________________________________________ _____________________________________________________ _____________________________________________________

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Calculate the balance on her loan at the end of 7 years. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

Allan takes out a loan to buy a new car valued at R350 000,00. The interest rate charged is 18% p.a. compounded monthly. 3.1

Calculate his repayments if the duration of the loan is 5 years. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

131 GO MATH WORKBOOKS

Grade 12 Core Mathematics 3.2

Allan decides to settle the loan after 3 years. Calculate the balance of the loan. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

Deferred Annuities: If payments are deferred at the beginning of an investment or loan the following formula is used. [A deferred annuity takes place when a payment takes place after the 1st time period.] i.e DELAYED PAYMENTS.

Fv 





x (1  i) n1  1  (1  i) n2 i

n1  (Total number of payments - Deferred Payments) n2  Number of Deferred Payments

Pv (1  i) n2 



x 1  (1  1)  n1 i OR

x

Pv (1  i ) (i ) n2

1  (1  i )  n1



Pv 



x 1  (1  1)  n1 i(1  i ) n2



132 GO MATH WORKBOOKS

Grade 12 Core Mathematics

EXAMPLES: 1. A 20 yr loan of R100 000,00 is repaid by means of equal monthly payments starting 3 months after the granting of the loan. The interest is 18%p.a. compounded monthly. Calculate the monthly repayments.





x 1  (1  1)  n1 Pv (1  i)  i 0.18  238   x 1  (1  )  0.18 2 12  100000(1  )  0.18 12 12 2 100000(1.015) (0.015) OR x 1  (1.015)  238 x  R1591,35 n2

x x

Pv (1  i ) n2 (i ) 1  (1  i )  n1 100000(1.015) 2 (0.015)

1  (1.015)  238 x  R1591,35

2. An investor pays R3000,00 at the end of each month starting in 3 months from now into an account paying 18% p.a. compounded monthly. He pays his final R3000 6 months before the time to withdraw the money. If the investment period starting from now is 8 years calculate the future value of the investment at the end of the 8th year.





x (1  i) n1  1  (1  i) n2 i 0.18 88   3000 (1  )  1 12    (1  0.18 ) 6 Fv  0.18 12 12 Fv  R591969,90

Fv 

n1  88 n2  6

133 Grade 12 Core Mathematics

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Exercise 3.4: 1.

32 semi annual payments of R6000 are made in order to repay a loan. The payments start in 2 years from now. Interest is 18,6% p.a. compounded semi-annually. Find the size of the loan. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

A loan of R120 454,00 is repaid by means of 14 monthly payments, starting 4 years after the granting of the loan. Interest is 15 % p.a. compounded monthly. value of the payments? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

Find the

134 Grade 12 Core Mathematics 3.

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A loan is repaid starting in 5 years time by means of 12 quarterly payments of R7000. What is face value of the loan if interest is 24% p.a. compounded quarterly? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

R2600 is invested monthly into an interest bearing account paying 16% p.a. compounded monthly. The first payment is deferred for 9 months and continues for the duration of the 9 years. The investor requires a lump sum in 10 years from now. What is the future value of the investment at the end of the 10 year? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

135 GO MATH WORKBOOKS

Grade 12 Core Mathematics

FinalPayment on a loan: The final payment on a loan needs to be calculated as it is usually an amount smaller than the normal fixed payment. The number of payments that are required to pay the loan off needs to be calculated. It usually works out to be a fixed number payments and one further payment less than this amount. The final amount(payment) needs to then be calculated using the necessary formula. Example. 1 A loan of R25 000 is to be repaid by means of a number of payments of R500 made at the end of the month, starting in one months time. Interest is 12,3% p.a. compounded monthly. Find: 1.1 the number of payments. 1.2 the final payment.

1.1

Pv  





x 1  (1  i)  n  y(1  i) ( n1) i

 Pv (i )  log 1  x   n  log(1  i )  25000(0.01025)  log 1   800   n  log(1.01025)  n  37,86318614 n  37,86318614 There will be 37 payments of R800 and a final payment less than R800.

1.2

2500 0 T0

800

8000



Pv  x 1  (1  i )  n i y (1  i ) ( n 1)

T37 T38







 800 1  (1.01025) 37  25000    0.01025   y 38 (1.01025) y  R689,73 Final Payment of R689,73

136 GO MATH WORKBOOKS

Grade 12 Core Mathematics

FIXED PAYMENT ANNUITIES. There are occasions when a person wishes to borrow cash from a bank BUT can only afford to pay certain fixed payments per month, which is usually smaller than a normal bank loan monthly repayment. The bank will need to determine the durataion of the loan by finding out the number of fixed payments necessary to amortise the loan. The annuity will then consist of a certain number of negotiated fixed payments and a final payment, which is smaller than the others. The annuity is referred to as a FIXED PAYMENT ANNUITY. Example 1: A loan of R300 000 is to be repaid by means of monthly payments of R5000, starting one month after the granting of the loan. Interest is fixed at 18% p.a. compounded monthly. 1. Determine the number of payments required to amortize the loan. 2. Determine the value of the final payment. 3. Would the bank have granted the loan if he could only afford to pay R1000 p.m. Answers: 1.





5000 1  (1.015)  n 0.015 300000  0.015  1  (1.015)  n 5000 300000  0.015 (1.015)  n  1  5000

300000 

(1.015)  n  0,1 log 0.1 n log 1.015 n  154,6541086

 Pv (i )  log 1  x   n  log(1  i )

 300000(0.015)  log 1   5000   n  log(1.015)  n  154.6541086 n  154.6541086

There are 154 payment of R5000 and a final payment less than R5000.

137 GO MATH WORKBOOKS

Grade 12 Core Mathematics 2.





x 1  (1  i )  n Pv   y (1  i )  ( n1) i 300000 



5000 1  (1.015) 0.015

n

  y(1.015)

y





155

326,2115  y (1.015) 155 326,2115 y (1.015) 155 y  R3278,96



x 1  (1  i )  n i  ( n 1) (1  i )

PV 

log( 3.5) [impossible] log(1.105) Answer is NO.



 5000 1  (1.015) 154  300000    0.015   y 155 (1.015) y  R3278,96 Cannot find the logarithm of a negative number ( By Definition)

n

Example 2: How long will it take William to pay off a loan of R100 000,00 by means of monthly payments of R2500 starting 3 months after the granting of the loan? Interest is levied at 15% p.a. compounded monthly. Answer:

10000 0 T0 T1

2500

T3 T4

2500

  Pv (1  i ) n (i )  log 1   x    n log(1  i )  100000(1.0125) 2 (0.0125)  log 1   2500   n log(1.0125)  n  57,84858933 n  57,84858933 Loan will take 56 months to pay off.

2500

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Exercise 3.5: 1

A loan of R4000 is to be repaid by means of a number of payments of R270 made at the end of the month, starting in one months time. Interest is 12,3% p.a. compounded monthly. Find: 1.1 the number of payments. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ 1.2

the final payment. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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2. A loan of R9000,00 is repaid by means of n equal payments of R1200, which are paid on a semi-annual basis. An final payment of y is made , six months after the final payment of R1200. Interest is 16% p.a. compounded semi-annually. Find the number of payments and the value of y , if the first payment is due to start in 6 months time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ 3. A loan of R8000 is repaid by means of equal payments of R650 every quarter, and a final payment, after the last payment of R650. Interest is 15,6% p.a. compounded quarterly. 3.1

Find then number of payments and the value of the final payment if the first payment is made 3 months from granting the loan. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

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Depreciation and Sinking Funds Definitions: 1. Book Value: It is the value of an asset after depreciation has been taken into account. 2. Scap Value: It is the book value of an asset at the end of its useful life. 3. Sinking Fund: A fund that is set up to finance the replacement of an asset after its useful life. Exercise 3.6: 1.

1.1

A printing press is bought for R340 000,00. The cost of a new press is expected to rise by 15% p.a. while the rate of depreciation is 10% p.a. on the reducing balance. The life span of the press is 8 years. Find the scrap value of the old press. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.2

Find the cost of a new press in 8 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

1.3

Find the value of the sinking fund that will be required to purchase the new press in 8 years time , if the proceeds from the sale of the old press at scrap value will be utilized. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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The company sets up a sinking fund to pay for the new press. Payments are to be made into an account paying 12.5% p.a. compounded monthly. Find the monthly payments , if they are to commence one month after the purchase of the old press and cease at the end of the 8 year period. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

A company bought a large generator for R450 000,00. It depreciates at 18% p.a. on a reducing balance. A new machine is expected to appreciate in value at a rate of 12% p.a. A new machine will be purchased in 6 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

2.1

Find the scrap value of the old machine in 6 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

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Find the cost of a new machine in 6 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2.3

The company will use the money received from the sale of the old machine at scrap value as a part payment on a new one. The rest of the money will come from a sinking fund that was set up when the old machine was bought. Monthly payments, which started one month after the purchase of the old machine, have been paid into a sinking fund account paying 9,5% p.a. compounded monthly. The payments will finish 6 months before the purchase of the new machine. Calculate the monthly payments into the sinking fund that will provide the required capital to purchase the new machine. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

A vehicle is purchased for R300 000,00. The cost of a new vehicle is expected to rise by 12% p.a., while depreciation is 15% on the reducing balance. The lifespan of the vehicle is 5 years. 3.1 Find the scrap value of the old vehicle. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

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Find the cost of a new vehicle in 5 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.3

Find the value of the sinking fund required to purchase the new vehicle in 5 years time, if the old vehicle is sold and the proceeds used towards the new one. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.4

The company sets up a sinking fund to pay for this new vehicle. Payments are to be made into the account returning 12,5% p.a. compounded monthly. Find the value of the monthly payments. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

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A vehicle is purchased for R450 000,00. The cost of a new vehicle is expected to rise by 12% p.a., while depreciation is 10% on the reducing balance. The lifespan of the vehicle is 6 years. 4.1 Find the scrap value of the old vehicle. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 4.2

Find the cost of a new vehicle in 6 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

4.3

Find then value of the sinking fund required to purchase the new vehicle in 6 years time, if the old vehicle is sold and the proceeds used towards the new one. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

4.4

The company sets up a sinking fund to pay for this new vehicle. Payments are to be made into the account returning 14,5% p.a. compounded monthly. Find the value of the monthly payments. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

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A farmer purchases a new combine harvester for R 2 500 000,00. The life span of the harvester is 8 years and depreciates at 10 % p.a. The future price of a combine harvester increases by 12% p.a. The farmer decided to set up a sinking fund with a return of 15,5% p.a. compounded monthly to cover the cost of a new machine in 6 years time.

5.1

Calculate the scrap value of the harvester in 6 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

5.2

Find the cost of a new harvester in 6 years time. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

5.3

Find the value of capital required to purchase a new harvester in 6 years time if the proceeds of the old machine are used towards the purchase. _____________________________________________________ _____________________________________________________ _____________________________________________________

5.4

The farmer sets up a sinking fund to pay for a new harvester in 6 years time. Calculate the monthly payments required into the account. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

146 GO MATH WORKBOOKS

Grade 12 Core Mathematics

4. Functions & Graphs Quadratic Function: Exercise 4.1: Sketch the graphs of the functions below: 1. y  x 2  3x  4 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ _______________________________________________________________ ______________________________________________________________ 6

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Grade 12 Core Mathematics y  x 2  4x  5

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3. y  x2  x  6 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________

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4. y  x 2  3x  10 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 10

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5. y  x 2  2x  8 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics y  x 2  4 x  12

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Grade 12 Core Mathematics

7. y  2x 2  7 x  6 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 12

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Grade 12 Core Mathematics

8. y  2 x 2  5x  3 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 12

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9. y  x 2  6x  7 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 12

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Exercise 4.2: a) Write the following equations in the form: y  a( x  p) 2  q i.e in the completed square form of the equation. b) Write down the coordinates of the turning point: c) Solve the equation and write down the x and y intercepts. d) Sketch the graphs of the equations. 1.

y  x 2  6x  9 .

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2. y  (2 x  1)( x  1) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 12

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3. y  x 2  2 x  3 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 12

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4. y  2 x 2  4 x  6 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ 12

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5. y   x 2  2 x  3 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ 12

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Grade 12 Core Mathematics 10.

y  3x 2  2 x  1

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Grade 12 Core Mathematics Shifting parabolas: Horizontal ( Left or Right)

In the completed square form of the equation simply change the “p” value and multiply the equation out for the ax 2  bx  c form ( if required) Example:

2 x 2  12 x  10  0 2[ x 2  6 x  5]  0 2[( x  3) 2  4]  0 2( x  3) 2  8  0 pq(3;8)

Instruction: Shift the graph of, 2 x 2  12 x  10  0 , 5 units to the left. p3

p1  3  5 p1  2

New equation

y  2( x  2) 2  8 y  2 x 2  8x

NB. Don‟t forget to change the sign when substituting back into y  a( x  p) 2  q Exercise 4.3 : 1. Shift questions 1 to 5 in exercise 2 by 4 moves to the left. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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2 x 2  12 x  10  0 2[ x 2  6 x  5]  0 2[( x  3) 2  4]  0 2( x  3) 2  8  0 pq(3;8)

Instruction: Shift the graph of, 2 x 2  12 x  10  0 , 5 units upwards. q  8

q1  8  5 q1  3

y  2( x  3) 2  3

y  2 x 2  12 x  15

New equation

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Exercise 4.4: 1.

Shift questions 1-5 in Ex 2 by 3 moves upwards. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

164 Grade 12 Core Mathematics

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Finding Equations of Parabolas: A. Given the x- intercepts and one other point. Method: Use the general form of the equation : y  a( x  r1 )( x  r2 ) and substitute the roots(x-values) and the x & y values of the given point into this formula and solve for a. e.g. Find the equation of the parabola that has x-intercepts –3 and 4 which passes through point(1 ; -24) y  a ( x  r1 )( x  r2 )

 24  a (1  3)(1  4)  24  12a a2 y  2( x  3)( x  4) y  2 x 2  2 x  24

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B. Given the turning point and one other point. Method: use the general form : y  a( x  p) 2  q i.e the completed square form of the general equation. Substitute the TP and the other point into this form to solve for a. e.g. Find the equation of a parabola that has a turning point (2 ; 3) And passes through point (1 ; 2)

y  a( x  p) 2  q y  a ( x  2) 2  3 2  a (1  2) 2  3 2  a3 a  1 y  1( x  2) 2  3 y  x 2  4x  1 C: Given a sketch : Use the information supplied on the sketch to find the equation of the parabola and straight line:

y=x + 1

C(-3 ; 0)

A (0 ; -3) Method: A:

y=ax2 +bx +c

Straight line : y  mx  c

y0 x 1 0 x  1 B (1;0)

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Parabola: y  a( x  r1 )( x  r2 ) [Use this form as roots are known] y  a ( x  r1 )( x  r2 ) Substitute the roots into the y  a ( x  1)( x  3) equation and one other point:  3  a (0  1)(0  3) Then solve for „a‟  3  3a a  1 y  1( x  1)( x  3)

y   x 2  4x  3

Exercise 4.5: Find the equations of the following given: 1.

Turning Point ( (2;10) passing through (0 ;2) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Turning point ( -1;5) passing through (1;13) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Turning point ( -4;-1) passing through (-3;2) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

x â&#x20AC;&#x201C; intercepts (1;0) and (-3;0) passing through (-1;-4 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

x â&#x20AC;&#x201C; intercepts (2;0) and (-4;0) passing through (3;-14) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 6.

x – intercepts (1;0) and 5;0) and y –intercept (0;-5) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Exercise 4.6: Quadratic Function: Parabolas. 1.1

Sketch the graphs of y   x 2  x  12 and y  3x  12 on the same system of axes.

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1.2 Write down the co-ordinates of the points of intersection.

1.3 Calculate the distance between the two graphs at x = -2 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.4

Write y  x 2  2 x  8 in the form of y  a( x  p) 2  q . ________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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1.5 Write down the co-ordinates of the turning point.

1.6 Write down the roots ( x-intercepts ) of the graph.

171 GO MATH WORKBOOKS

Grade 12 Core Mathematics 1.7 Write down the co-ordinates of the y – intercept.

______________________________________________________________ ______________________________________________________________

1.8 Sketch the graph. 4

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1.9 Find the new equation if y  x 2  2 x  8 is moved 5 units to the left. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

172 GO MATH WORKBOOKS

Grade 12 Core Mathematics 3.

Write y   x 2  4 x  5 in the form y  a( x  p) 2  q ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.2

Sketch the graph of

y  x 2  4x  5 .

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Sketch the graph of y  x  5 on the same system of axes.

3.1 3.3.1

Write down the co-ordinates of the points of intersection of the two graphs.

Write down the new equation in form y  ax 2  bx  c if the y  2 x 2  8x  10 is moved 3 units to the left. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Write down the new equation in form y  ax 2  bx  c if the y  2 x 2  8x  10 is moved 6 units upwards. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.2

Write down the new equation in form y  ax 2  bx  c if y  2 x 2  8x  10 moved 5 to the right and 4 moves downwards. ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

y D E

C( 6 ; 32)

B 0

The graphs above are of f: y   x 2  8x  20 and g: y  mx  c . 5.1 Find the co-ordinates of A ; B ; H and D. ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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5.2 Write down the lengths of ; AO ; OH ; OB ; AB and SD ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.3 Calculate the length of AC.

5.4

Write down the equation of g. ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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EF = 16 units in length. Calculate the length of OG. ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________

178 GO MATH WORKBOOKS

Grade 12 Core Mathematics Exponential Graphs: y = ax

General formula :

a>0,a1,x  y>0

A. To sketch the graph y = ax (a) x  and y > 0 y = ax lies above the x – axis in quadrants 1 and 2. b) (i) If a > 1 , then as x increases , y increases. e.g. y = 2 x (ii) If 0 < a < 1 , then as x increases , y decreases. e.g. y = ( 12 )x. (i)

y = 2x

x y

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-1

1 4

1 2

(ii)

y = ( 12 )x.

x y

-2 4

-1 2

0 1

1 2

2 4

0 1

1 2

1 4

1 y    or  2

y = 0,5x y = 2x y

1 x

These 2 graphs are mirror images of each other, the axis of symmetry being the y- axes (x = 0 ). The x- axes is a horizontal asymptote, as y will never equal zero.

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Grade 12 Core Mathematics Log Graphs: General formula : y  log a x

y  log a x

To sketch the graph of

a) x>0 and y   thus y  log a x lies on the right of the y- axes in quadrants 1 and 4. b) If a > 1, then as x increases so y- increases c) If o < a < 1 , then as x increases so y decreases. (1)

y  log 2 x ( x  2 y ) x y

(2)

1 4

1 2

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-1

y  log 1 x ( x   12  ) y

1 4

1 2

-1

-2

y y = log 2x

1 1

y = logo,5x The graph y   log 2 x is exactly the same as y  log 1 x 2

These two graphs are mirror images of each other, the axix of symmetry being the x – axes. The y- axes is a vertical asymptote, as x will never equal zero.

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Shifting the exponential graph: 1. Vertical Shifts: y  a x or y  a  x 1. y  2 x Original graph 2. y  2 x  2 Graph shifted 2 units upwards. 3. y  2 x  4 Graph shifted 4 units downwards. NB: the values are added after the base (2x) for vertical movement.

g x  = 2 x +2 2

f x  =

new horizontal asymptote for g(x) =2x + 2

2. -5

-2

h x  = 2 x -4

3. -4

new horizontal asymptote for h(x) =2x - 4

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Grade 12 Core Mathematics Exercise 4.7: Sketch the graph of y  3 x on a Cartesian plane.

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Shift y  3 x by 2 units upwards and sketch this graph on the same system of axes.

Shift y  3 x by 4 units downwards and sketch this graph on the same system of axes. 2.

Write down the equations of the asymptotes after the shifts in question 1.

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Grade 12 Core Mathematics Sketch the graph of y  4  x on a Cartesian plane.

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Shift y  4  x by 3 units upwards and sketch this graph on the same system of axes, Shift y  4  x by 4 units downwards and sketch this graph on the same system of axes. Write down the equations of the asymptotes after the shifts in 3.

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Grade 12 Core Mathematics

Horizontal Shifts: y  a x or y  a  x y  2 x Original graph 2.1 y  2 x  4 Graph shifted 4 units to the right. 2.2 y  2 x  4 Graph shifted 4 units to the left. NB: the movement is added or subtracted in the exponent for lateral shifts. 2.

s x  =

r x  = 2 x-4

2 x+4

f x  = 2 x 4

3. New vertical Asymptote x = -1

New vertical Asymptote x= 7

2. -5

-2

3. -4

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Grade 12 Core Mathematics Exercise 4.8:

1. Sketch the graph of y  2 x on a Cartesian plane. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 6

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-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.1 Shift y  2 x by 4 units to the left and sketch this graph. 1.2 Shift y  2 x by 4 units to the right and sketch this graph.

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Write down the equations of the asymptotes after the shifts in question 1. ___________________________________________________________________ __________________________________________________________________

Sketch the graph of y  2  x on a Cartesian plane, ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

3.1 3.2

Shift y  4  x by 3 units to the left and sketch this graph on the same system of axes. Shift y  4  x by 4 units to the right and sketch this graph on the same system of axes.

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Grade 12 Core Mathematics 4

Write down the equations of the asymptotes after the shifts in 3. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Sketch the graph of y  3 x ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

5.1

Shift the graph 4 to the right and 3 upwards and sketch the new position. i.e. the graph of y  2 x 4  3 , on the same system of axes.

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Grade 12 Core Mathematics The Hyperbola Graph. k k or y  x x

General equation: y  Sketching the graphs:

Method 1. Table Method 4 1. y x X -4 -2 Y -1 -2

-1 -4

1 4

f x  =

2 2

4 1

2 -2

4 -1

4 x

-5

-2

-4

Sketch of y 

2. X Y

4 x

-4 1

-2 2

-1 4

1 -4

f x  =

-4 x

-5

-2

-4

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Grade 12 Core Mathematics Shifting the hyperbola graph: 1.

If a constant is added to the equation after

k then this will cause a vertical shift: e.g. x

4 4 is shifted upwards by 3 units.  3 : the graph of y  x x 4 Asymptotes of y  are y  0 ( x – axis) and x  0 ( y- axis). x y

This shift will change the horizontal asymptote BUT not the vertical. Asymptotes are: x  0 ( y- axis). And y  3 (New horizontal asymptote)

(1;7) 6

(2;5) 4

h x  =

(4;4)

(1;4)

4 x

horizontal asymptote is y = 3 (-4;2) 2

(2;2)

(-2;-1)

(4;1)

f x  =

4 x

horizontal asymptote is y = 0 -10

-5

(-4;-1)

(-1;-1) (-2;-2)

(-1;-4)

-2

-4

-6

Horizontal shifts are caused when a constant value is added to the x- value in the denominator of the hyperbola equation.

The sift is in the opposite direction of the integer in the equation. i.e if the integer is +ve then the shift is to the left and if the integer is – ve then gthe shift is to the right.

Shift the graph of y 

4 4 as follows: y  x3 x

This shift is 3 units to the right. The horizontal asymptote stays the same BUT the vertical asymptote changes to the line x  3

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Grade 12 Core Mathematics

vertical asymptote x = 0 vertical asymptote x = -3

(-2;4)

(-1;2)

(1;4)

(2;2) (4;1) (1;1)

f x  =

4 x

horizontal asymptote is y = 0 -10

-5

(-1;-1)

(-4;-1) (-7;-1) (-5;-2)

(-4;-4)

(-2;-2)

(-1;-4)

-2

-4

-6

h x  =

4 x+3

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Grade 12 Core Mathematics Inverse Functions: Definition: fg ( x)  gf ( x)  x „g‟ is the inverse of „f‟ and similarly „f‟ is the inverse of „g‟. Example:

f ( x)  2 x  3 the inverse g ( x) 

g ( fx )  g (2 x  3) 

g ( x)  5 x  2

2x  3  3 x 2

f ( x) 

fg ( x)  5 

x3 2

x2 5

x2 2  x 22  x 5

Functions:

y  mx  c y  ax  bx  c 2

y   r 2  x2 y  ax

Straight line 1 – 1 Function Parabola 1 – M Function Circle

M – M Non –Function ( Relation) Exponetial 1 – 1 Function

Use a vertical line test to ascertain whether a function or non-function Horizontal line test will give the mapping ( 1 -1 etc) 1  1  function m  1  function 1  m  non  function m  m  non  function

Many functions have inverses that are not functions: Parabolas in the form of y  ax 2  bx  c have inverses xare not functions; Examples: 1. or

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Grade 12 Core Mathematics 2.

where

fx = x2-2x-3 4

-5

gy = y2-2y-3 -2

-4

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Grade 12 Core Mathematics 4.

hx = 3x 4

Exponential function

-5

ty = 3y

Log function

-2

-4

Exercise 4.9: 1. A. Find the inverse functions of the following , all answers in the y-form: B. Draw a neat sketch of the original and its inverse. 1.1 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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1.2 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.3 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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1.4 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.5 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

1.6 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2. Which of the following graphs of functions have inverses that are functions. Justify your answers. 2.1 fx = x+1 2

-5

-2

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Grade 12 Core Mathematics 2.2

gx = -x2+2

-5

-2

2.3 hx = 2x 2

-5

-2

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Grade 12 Core Mathematics Trigonometric Functions: Method:

Sketch the graphs using the values of the special angles. Viz using 0º ; 90º ; 180º ; 270º and 360º (You do not have to set up a table) Example 1: Sketch the graph of f ( x)  sin x where x  [360 ;360 ]

f x  = sin x 

-360

-270

-180

-90

90

180

360

270

-1

Example 2:

Sketch the graph of f ( x)  cos x where x  [360 ;360 ]

g x  = cos x  -270

-360

-180

-90

90

180

270

360

-1

Example 4:

Sketch the graph of f ( x)  tan x where x  [360 ;360 ]

h x  = tan x  1

-360

-270

-180

-90

-1

90

180

270

360

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Grade 12 Core Mathematics

Exercise 4.10: 1. Sketch the graphs of y  sin x and y  cos x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

180

270

-1

-2

From the sketch find the following: 1.1

the period of y  sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.2

the range of y  cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.3

the amplitude of y  sin x ______________________________________________________________

360

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Grade 12 Core Mathematics 1.4

the value for x for sin x  cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2. Sketch the graphs of y  2 sin x and y  cos x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

180

270

360

-1

-2

2.1

From the sketch find the following: the period of y  2 sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.2

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Grade 12 Core Mathematics 2.3

the amplitude of y  2 sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.4

3. Sketch the graphs of y  sin 2 x and y  cos x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

180

270

360

-1

-2

From the sketch find the following: 3.1

the period of y  sin 2 x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 3.2

3.3

the amplitude of y  sin 2 x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.4

the value for x for sin 2 x  cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4. Sketch the graphs of y   sin x and y  cos 2 x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

-1

-2

180

270

360

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From the sketch find the following: 4.1

the period of y   sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.2

the range of y  cos 2 x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.3

the amplitude of y   sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.4

the value for x for  sin x  cos 2 x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

5. Sketch the graphs of y  sin x  1 and y  cos x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

180

270

360

-1

-2

From the sketch find the following: 5.1

the period of y  sin x  1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.2

5.3

the amplitude of y  sin x  1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 5.4

the value(s) for x for sin x  1  cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6. Sketch the graphs of y  2 sin x and y  cos x  1 on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

180

270

360

-1

-2

From the sketch find the following: 6.1

the period of y  2 sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 6.2

the range of y  cos x  1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.3

6.4

the value(s) for x for sin x  cos x  1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

7. Sketch the graphs of y  sin( x  30  ) and y  cos x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

-1

-2

180

270

360

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From the sketch find the following: 7.1 the period of y  sin( x  30  ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 7.2

7.3

the amplitude of y  sin( x  30  ) ______________________________________________________________ ______________________________________________________________

7.4

the value for x for sin( x  30 )  cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

8. Sketch the graphs of y   sin( x  30  ) and y  cos 2 x on the same set of axes for the interval x  [0  ;360  ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  y 2

x 0 90

180

270

360

-1

-2

From the sketch find the following: 8.1

the period of y   sin( x  30  ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

8.2

the range of y  cos 2 x ______________________________________________________________ ______________________________________________________________

8.3

the amplitude of y   sin( x  30  ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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the value for x for  sin( x  30  )  cos 2 x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Linear Programming:

Linear programming is a method of approaching a given problem, which allows you to reach certain conclusions using the constraint supplied. The constraints are represented by one or more linear inequalities which when applied graphically allow one to reach the required conclusion(s).The linear inequalities are either given or formulated. Constraints: These are the limitations that are provided in word or equation form in the problem. These are converted into the necessary linear inequalities. Implicit constraints: These are the conditions that arise naturally in the given problem. i.e. x and y are necessarily whole numbers greater than or equal to zero and can represent people, animals or products. Feasible region: Once the constraints are sketched onto a system of axes, the feasible region can me marked off by shading etc. Any co-ordinate within this region represents a solution to the given problem.

Objective function: This function allows us to either maximise or minimise a situation as is required. This is also referred to as Optimisation. The gradient of the equation that represents this optimisation is very important in pinpointing the maximum or minimum production etc.

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Grade 12 Core Mathematics Methods of approach: 1.

If inequalities are supplied simply sketch them onto a graph and find the feasible region .

If a graph is supplied with the inequalities drawn work out what each inequality should be from the given information.

If information is given in word form work out what the inequalities representing the given information should be.

Examples: 1.

Point B(5;30) Maximum optimisation

y 80

Optimisation: 3y +2x=60 m

2 3

Feasible Region

20 Point A(4;19) Minimum optimisation 0

1.1 Write down the linear inequalities represented on the graph. Answers: 1.1.1 4 x  y  80 1.1.2 x  3 y  60 1.1.3 x  4 1.1.4 y  30 1.2 Given that the equation representing optimisation is 3y +2x=60, write down the co-ordinates that show the points of optimisation. Answers:Maximum optimisation = B (5; 30) Minimum optimisation = A(4 ; 19)

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Grade 12 Core Mathematics

2. The following inequalities are supplied place them on a graph and shade in the feasible region: x <20; y > 0 ; 8x +3y ď&#x201A;ł 240; 3x + 4y ď&#x201A;Ł 240 Answer:

y 80

Feasible region

Kevin Stevenson, the Sharks fitness coach, gave a Craig Davidson the following advice. He should be supplementing his daily vitamin intake with at least 6000 units of Vitamin A, at least 200 mg Vitamin C and at least 600 vitamin D. Craig finds out that Broadway Pharmacy sells blue pills at 50 cents each and red pills at 40 cents each. Each blue pill contains 3000 mg Vitamin A, 50 gm Vitamin C and 75 units of Vitamin D. Each red pill contains 1000 units Vitamin A, 50 gm vitamin C and 200 units Vitamin D. The table below sets out the required constraints, use this table to help get the necessary inequalities.

Blue Pills

Minimum Dose.

Red Pills

Vitamin A

3000

1000

6000

Vitamin C

200

Vitamin D

200

600

3.1 Let the number of blue pills be x and the number of red pills be y. Write down the set of three linear inequalities that satisfy the constraints mentioned above.

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Grade 12 Core Mathematics

3.2 Write down the equation of the objective function that is used to determine the minimum cost of the pills.

3.3 Use a graph to illustrate this optimization graphically. Clearly showing the feasible region.

3.4 Determine what combination of pills the sportsman should buy at the minimum cost.

3.5 When Craig went to get more supplies he found out that a discount of 40% was being offered on blue pills and the red pills had increased by 30 cents each. Determine the combination of pills that Craig should now buy at the minimum cost.

Memo: 3.1

Linear inequalities:

3000 x  1000 y  6000 3 x  y  6..................(1) 50 x  50 y  200 x  y  4....................(2) 75 x  200 y  600 3 x  8 y  24...............(3)

Also x; y  0

y 3.2

C  50 x  40 y

3x + y = 6

5 C y  x 4 40

3.3 As per graph

Ist Min profit line

4 RED PILLS

2nd Min profit line

3x + 8y = 24

3.4 At pt (1;3) 1 blue to 3 red pills

x+y=4 4 BLUE PILLS

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Grade 12 Core Mathematics 3.5 New price: Blue 0,6 x 50c = 30c Red 30c + 40c = 70c Objective function: C  30 x  70 y

30 C x 70 70 3 M  7 Buy 3 Blue and 2 Red at new rate. x; y  N NB: We use natural numbers in linear programming and as the minimum is at pt(2,4 ; 1,6) we take the nearest combination. i.e (3;2) C

Exercise 5.1: 1. Sketch the following inequalities on the grids supplied below. Use arrows to show which side of the line the solution should lie. 1.1

x y  6

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Grade 12 Core Mathematics 1.2 3 y  2 x  24

1.3 x  2 y  0

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Grade 12 Core Mathematics 1.4 4  x  8

1 y  5

1.5

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Grade 12 Core Mathematics

1.6 Sketch the five inequalities above on a set of axes and shade in the feasible region

y _

90 _

B(0;40)

45 _ 40 _ C(10;35) _

D(30;15) _

A(0;0) _

_40 _45

_80

x _

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Grade 12 Core Mathematics

2.1 Write down the inequalities represented in the graph above.

2.2 Shade in the feasible region. 2.3 Given that the maximisation equation is M  6 x  4 y , Sketch the optimisation line on the graph above and ascertain which point (A;B;C or D) will maximise the objective function.

3.6 Represent the following inequalities on a set of axes:

y  2 x  16 ; 7 y  6 x  42 ; x  2 and y  3 . 3.2 Shade in the feasible region.

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Grade 12 Core Mathematics 4. y

Line 1

Line 2

30 Line 5 Line 6 T y 20 p e 15

Line 3

B 10 Line 4

Type A

4. A shopkeeper sells 2 types of products, A & B. The shaded area on the graph above represents the feasible region if he sells x- articles of A and y- articles of B subject to certain constraints. 4.1

Write down these constraints as inequalities. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.2

If the shopkeeper makes R2 profit on every article of A and R5 on every article of B, write down an equation in terms of x and y which will represent the profit (P) that the shopkeeper makes. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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If the shopkeeper desires to maximise his profit, how many articles of each type should he sell. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.4 What is his maximum profit? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.5 What is his minimum profit subject to the same constraints and how many articles of each are sold. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.6 If the shopkeeper increases his prices and now makes R3 on each article of A and R6 on each of B. 4.6.1 How many of each type must he sell? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.6.2 What is his new profit? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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5. A bus company assembles two models of minibuses , a 12 seater and a 16 seater. The minibuses must go through 2 processes are, bodywork and engine work.  The factory cannot operate for less than 360 hours on engines. The factory has a maximum capacity of 480 hours for bodywork. 

1 1 hour of engine work and hour of body work is required to produce a 16 seater bus. 2 2



1 1 hour of engine work and hour of bodywork is required to produce a 12 seater bus 3 5

The ratio of 12 seater busses to 16 seater busses produced per week must be 3 : 2 A minimum of 200 12 seater busses must be produced per week. Let the number of 12 seaters be x Let the number of 16 seaters be y. If two of the constraints are x  200 and 2 x  3 y  2160

5.1

Write down the remaining constraints in terms of x and y, from the information above. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.2

Use the graph below to represent the constraints graphically.

5.3

Clearly indicate the feasible region by shading it.

5.4

If the profit on a 12 seater is R4000 and 16 seater is R12000 write down an equation that will represent the profit on minibuses.

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Grade 12 Core Mathematics 5.5 5.5.1

If the aim is to maximise the profit : How many of each type of buss is sold? and,

5.5.2

What is the maximum profit?

[Hint let each square represent 100 units] 16

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Grade 12 Core Mathematics Exercise 5. 2: 1.

1.1

A toy factory produces wooden box-carts and rocking-horses. The profit on a box-cart is R12,00 and on a rocking-horse is R8,00. The owner wants to maximize his profit but there are certain constraints that he must consider. Under contract he has to deliver at least 50 rocking-horses to the wholesaler. Market research has shown that he will not sell more than 120 box-carts per week. He must not produce more than 2 rocking-horses for every box-cart to ensure that he does not waste wood. Let x represent box-carts and y the rocking-horses. If the following constraints in the form of inequalities are given y  50 ; x  120 ; x  0 and y  0 where x, y  Z . Write down the remaining constraint as an inequality ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.2

Draw the graphs and determine the feasible region.

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How many box-carts and how many rocking-horses must he produce in order to maximize his profit. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.4

Determine the maximum profit. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Your mother needs to bake cakes for a fete. She decides to bake fruit cakes and sponge cakes. She has only 2kg flour and 1,2kg sugar. She has an abundance of all the other ingredients. For a fruit cake she needs 500gms flour and 100gms sugar. For a sponge cake she will use 200gms flour and 200gms sugar. She would like to bake at least 5 cakes. Let x be the fruit cakes and y the sponge cakes 2.1

If the following constraints in the form of inequalities are given 100 x  200 y  1200 ; 500 x  200 y  2000 ; x  0 and y  0 where x, y  Z . Write down the remaining constraint as an inequality. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 2.2

Draw the graphs and determine the feasible region.

2.3

Using the graphs, make a list of all possible combinations she could bake. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.4

The profit on a fruit cake is R8 and on a sponge cake is R4. How many of each must she bake to ensure a maximum profit? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 3.

In the sketch , the shaded area , including the boundary lines, represents the feasible region of a set of inequalities. The equation of the objective function is given by P ď&#x20AC;˝ mx ď&#x20AC;Ť y . B

A 8

D C

3.1

Write down all the inequalities. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.2

Determine the minimum value of P, if m = 0. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.3

Determine the possible value(s) of m if P is minimized at D. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

225 GO MATH WORKBOOKS

Grade 12 Core Mathematics

4. A Certain industrial problem can be reduced to the following set of inequalities: x  10 ; y  8 ; 2 x  y  12 ; x  2 y  12 where x, y   4.1 Sketch the set of inequalities graphically in order to determine the feasible region 16

4.2 The objective function K  5x  2 y must be minimized. 4.2.1 Draw a line representing the objective function if K = 10. 4.2.2 Explain how you would minimize K. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.2.3

Write down the minimum value of K.

226 GO MATH WORKBOOKS

Grade 12 Core Mathematics 5

A factory has a contract to deliver at least 90 units of a certain piece of furniture per week. There may not be more than 18 employees. An artisan , who earns R600 per week, can produce 7 units per week. While an apprentice, who earns R300 per week, can only produce 4 units per week. The labour laws specify that at least one apprentice should be employed for every 5 artisans. The labour union , however , insists that the ratio of apprentices to artisans should not exceed 1: 2. 5.1 Let the number of artisans be x and the number of apprentices be y. 4 of the constraints are given by the following inequalities: x  0 ; y  0 ; 5 y  x and 7 x  4 y  90 . Write down two more constraints in terms of x and y. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.2

Represent all the constraints graphically and indicate the feasible region.

227 Grade 12 Core Mathematics 5.3

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Express the amount (L) representing weekly wages in terms x and y.

Show the optimal position of the curve of the objective function which will minimize the wage bill.

How many artisans and apprentices should be employed so that the wage bill is kept to a minimum but the largest number of units are delivered.

6. 12

228 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Feasible Region

i n i B u s

Passengers Transported by Bus The graph above represents data from a company using buses and minibuses to transport a minimum of 800 and a maximum of 1200 passengers per day. The number of passengers, x , transported by bus, must be a minimum of 400 per day, but cannot be more than 3 times the number of passengers transported daily by minibus. ( x, y ď&#x192;&#x17D; N ) 6.1

Extrapolate the set of inequalities that represent the above constraints. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.2

If the daily profit per passenger traveling by bus is R1 and the daily profit per passenger per minibus is 80 cents . Use the graph to determine the values of x and y which will give a maximum profit. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Hence determine the maximum daily profit. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

Probability Theory:

Counting Different Options: Example: A diner has the following options available to customers:

Main Course Curry and Rice Pap and Beans Boerewors Roll Chicken Veg Salad

Dessert Ice Cream Fruit Salad Pumpkin Fritters Pancakes

Total Number of Options available

Drinks Orange Juice Milk Tea Coffee Lemonade Cola 6

= 5x4x6 = 120

2. Different Standard Number Plates available if format used is 3 alphabetical characters (excluding Vowels and Q) followed by 3 numeric digits. There are: 20 alphabet characters available and 10 digits Answer: 20 x 20 x 20 x 10 x 10 x 10 = 8 000 000 combinations 3. A maze is constructed so that at 8 places a choice is made between turning left or right. Number of routes in the maze is thus:

M 1  M 2  M 3  ........M n  Mn The above are based on the “FUNDAMENTAL COUNTING PRINCIPLE” Factorial Notation: 1.

In how many different ways can 3 stamps be placed in a row in the top right hand corner of an envelope. Any of the three can be placed 1st then only 2 left to choose from then only 1 left 3X2X1=6 This kind of product is called 3 factorial and is written as 3! 5 x 4 x 3 x 2 x 1= 5! Called 5 factorial. The number of ways that m different terms can be arranged is : m(m  1)(m  2) and is written as m!. and is called m factorial.

231 GO MATH WORKBOOKS

Grade 12 Core Mathematics 2.

The number of ways that the 1st three places can be recorded in a 10 lane pool is 10 ( Any one of the 10 swimmers can come 1st ) multi[plied by 9( any of the remaining 9 swimmers can come 2nd ) multiplied by 8 (any of the remaining 8 swimmers can come 3rd .) Answer is : 10 x 9 x 8 10  9  8 

10  9  8  7  6  5  4  3  2  1 10! = 720  7  6  5  4  3  2 1 7!

Can use the function on the calculator :

n Pr

P is number of permutations r is number of successive digits n is largest of the digits. 10 P3

= 720

m different terms in r number of arrangements : Total number of arrangements will be m Pr Example: 20 permutations with 5 successive digits will produce a total number of permutations: 20 P5  1860480 i.e m different items will produce of arrangements r will produce a total number of m! arrangements : or m Pr (m  r )! Exercise 6.1: 1. Use your calculator to calculate the following: 1.1 8 ______________________________________________________________ ______________________________________________________________ 1.2 18 ______________________________________________________________ ______________________________________________________________ 1.3 24 ______________________________________________________________ ______________________________________________________________

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2. Write 15 in expanded mode instead of scientific notation as the calculators answer. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3. Use the n Pr key on the calculator to calculate the product: 3.1 60 × 59 × 58 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.2

18 × 17 × 16 × 15 × 14 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4. Check if the answers in 3 are the same as follows: 60! 18! 4.1 4.2 (60−3)! (18−5)! ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5. State whether the following statements are true or false, without using a calculator: 5.1 10 × 9! = 9! ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.2

20! 19!

= 20

______________________________________________________________ ______________________________________________________________ ______________________________________________________________

233 Grade 12 Core Mathematics 15!

5.3

4!×3!

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= 1! = 1

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 10!

5.4

6!×4!

= 210

5.5

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ Example: There are flags of 16 different countries competing in the African Cup of Nations. a) In how many ways could you display the flags in a single row? The total number is: 16 × 15 × 14 × 13 × … .× 3 × 2 × 1 = 16! = 2.092278989 × 1013 . b)

In how many different ways can you determine the winner and runner-up, assuming that every nation could win and every other nation could be runner-up. 16 X 15 = 240 Or

16!  240 14!

or 16 P 2 

240

i.e.

m! 16!   240 (m  2)! (16  2)!

234 GO MATH WORKBOOKS

Grade 12 Core Mathematics Example:

How many code numbers of 3 digits can be made using digits 1; 2; 3; 4 & 5 if the order of the digits are important and: a) repetition is not permitted b) repetition is permitted

Choice: a) Using fundamental counting principle:

5(4)(3) = 60 or 5 P 3 ď&#x20AC;˝ 60

b) Using fundamental counting principle:

(5)(5)(5) = 53 ď&#x20AC;˝ 125

Example: Write down the different four letter arrangements or â&#x20AC;&#x17E;wordsâ&#x20AC;&#x; that do not have to have any meaning, that can be formed using the following letters: a) LIKE b) LEEK c) LULL d) LULU Solution: a) b)

4 x 3 x 2 x 1 = 24 i.e 4! = 24 4! = 12 (NB 2 in the denominator is due to the fact that a letter us repeated twice.) 2 4! ď&#x20AC;˝ 4 (NB The number of permutations of n items where r are identical ( and the 3! đ?&#x2018;&#x203A;!

remaining (n â&#x20AC;&#x201C; r) are all different is đ?&#x2018;&#x;! .) d) e)

4! ď&#x20AC;˝ 6 arrangements. 2!ď&#x201A;´2! 5! Using NANNA ď&#x20AC;˝ 10 arrangements. 3!ď&#x201A;´2!

BANAN;

5! ď&#x20AC;˝ 30 2!ď&#x201A;´2!

arrangements.

235 GO MATH WORKBOOKS

Grade 12 Core Mathematics Example: NDUNDULU: a) how many different „words‟ can be formed. How many „words‟ in (a) Start with N

in (a) Start and end with N There are 8 letters: 3 U‟s ; 2N‟s and 2 D‟s. Hence the number of different words:

8!  1680 Words 3!2!2!

b) 1)

If we take one of the N‟s for first letter, there are seven letters left, of which 3 are U‟s and 2

are D‟s:

7!  420 3!2!

2) If we take both N‟s for first and last letters, there are 6 letters left, of which 3 are U.s are D‟s

and 2

6!  60 3!2!

c) How many code numbers of 3 digits can be made using digits 1;2;3;4 and 5 if the order of the digits is important and: 1. repetition is not permitted. 2. repetition is permitted Solutions: 1. There is a choice of five digits to choose from for the 1st digit of the code :1; 2; 3; 4; 5. Once the 1st digit is chosen there are four left from which to choose the second digit. Etc By the fundamental counting principle: The tota; number is 5 x 4 x 3 = 60 2. When repetition is allowed any of the five digits can be chosen for each of the three digits in the code: Using the fundamental counting principle: 5 × 5 × 5 = 53 = 125 Exercise 6.2: 1. How many four digit numbers can be made from digits 1 to 6 if: 1.1 no digit may be repeated. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________

236 Grade 12 Core Mathematics 1.2

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repetition is allowed.

2. How many ways can a captain and then a vice-captain be chosen from a rugby team of 15 members? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3. Assuming any combination of letters form a word. How many different words can be formed using the following letters: 3.1 RAT. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.2

NAIL.

237 GO MATH WORKBOOKS

Grade 12 Core Mathematics 3.3

TIMBER

n(S )  4! To calculate the number of ways the event E can occur, we need to remember that the „G‟ and „L‟ are fixed . THUS n( E )  2! Then the probability that a randomly generated word will star with a „G‟ and end in a „L‟ is:

P( E ) 

n( E ) 2! 1    0,0833.... n( S ) 4! 12

238 GO MATH WORKBOOKS

Grade 12 Core Mathematics Tree Diagram illustrates:

1 2

2 3

not L

1 2

not L

G 1 3

1 4

P(G __ L) 

3 4

1 2 1 1   1  4 3 2 2

not G

Example 2: Suppose that a number plate is formed using three letters of the alphabet , excluding the vowels, and any three digits. Calculate the probability that a number plate , chosenat random: a) starts with a‟B‟ and ends in a „5‟ b) has exactly one „B‟ c) has at least one „5‟. SOLUTION: a) Let E be the event that the number plate starts with a „B‟ and ends in a „5‟. We are looking ( eg BRR615) 20 letters and 10 digits can be used. n( S )  20  20  20  10  10  10  8000000

n( E )  1  20  20  10  10  1  40000  P( E ) 

n( E ) 40000 1    0,005 n( S ) 8000000 200

239 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Example: If the letters of the word STATISTICS are randomly arranged, what is the probability that the word will start and end with the same letter ? Solution: STATISTICS has 10 letters. It consists of three S‟s, three T‟s and 2 I‟s. To find the possible arrangements use the formula: n( S ) 

10!  50400 [10 letters of which 3 S’s, 3 T’s 2 I’s] 3!3!2!

If E is the event of the word starting and ending with the same letter, then we are looking for S_ _ _ _ _ _ _ _ S or T _ _ _ _ _ _ _ _ T Or I_ _ _ _ _ _ _ _ I Then n(E) =number of words using letters TATISTIC + the number of words using SATISICS + the number of words using letters STATSTCS 8! 8! 8!    7840 3!2! 3!2! 3!3! 8! 8! 8! e.g. n( E )  (3T‟s & 2I‟s) + (3S‟s & 2I‟s) + (3S‟s & 3T‟s) 3!2! 3!2! 3!3! n( E ) 

Probability is P( E ) 

7840  0,15 50400

ALTERNATIVE METHOD: P(choosing an „S‟) =

3 ; because 3 of the 10 letters are „S‟ 10

The probability of choosing a second “S” is not independent of whether or not an “S” is chosen in the Ist Draw> P(choosing a second “S”) =

2 (once it is known that the Ist letter drawn is an 9

“S”. Similarly the probabilities of the successive events, of drawing ane and another “T” and an “I” can be calculated. P( E ) 

3 2 3 2 2 1       0,15 10 9 10 9 10 9

240 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Exercise 6.3: 1. The Matric Dance Committee has decided on the menu below for the 2008 Matric Dance. A person attending the dance must choose only ONE item from each category, that is starters, main course and dessert.

STARTERS Crumbed Mushroom Garlic Bread Fish 1.1.1

MAIN COURSE Fried Chicken Beef Bolognaise Chicken Curry Vegetable Curry

DESSERT Ice-Cream Mulva Pudding

How many different meal combinations can be chosen? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________

1.1.2 A particular person wishes to have chicken as his main course. How many different meal combinations does he have? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ 1.2

A photographer has placed six chairs in the front row of a studio. Three boys and three girls are to be seated in these chairs. In how many different ways can they be seated if: ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

241 Grade 12 Core Mathematics 1.2.1

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Any learner may be seated in any chair ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________

1.2.2

Two particular learners wish to be seated next to each other ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________

2. A smoke detector system in a large warehouse uses two devices, A and B. If smoke is present, the probability that it will be detected by device A is 0,95. The probability that it will be detected by device B is 0,98 and the probability that it will be detected by both devices simultaneously is 0,94. 2.1

If smoke is present, what is the probability that it will be detected by device A or device B or both devices? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

242 Grade 12 Core Mathematics 2.2

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What is the probability that the smoke will not be detected? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Compiled by Chez Nell

2 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Forward: Welcome to “GO MATH WORKBOOKS”. This workbook is designed to be a text book and class work book in one. There are sufficient exercises to ensure that learners get the required practice. A detailed memorandum booklet is available for each workbook. The statement “You get out what you put in.” is very apt where maths is concerned. To succeed in mathematics one must be prepared to invest the time and effort to achieve that success. The partnership that you as a learner and this GO MATH WORKBOOK develop will be profitable if you allow it to be. Chez Nell: Mathematics Educator: Northwood School  Norma Nell 2011

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Grade 12 Core Mathematics

GRADE 12 CORE MATHEMATICS Contents: Paper Two: Topic:

Pages:

Analytical Geometry

( 4 – 15)

Transformation Geometry

Trigonometry

(20 - 45)

Data Handling

(46 – 69)

Circle Geometry

(70 – 85)

(16 – 19)

4 GO MATH WORKBOOKS

Grade 12 Core Mathematics

Paper Two: 1.

ANALYTICAL GEOMETRY

Analytical geometry - Studies the properties of geometric figures Algebraically. This is pursued by the means of examining significant points (co-ordinates) of these figures in a Cartesian Plane. Hence also referred to as Co-ordinate Geometry. Formulae: 1.

Length of a line:

A(2 ; 5)

B(-4 ; -3)

Length of AB = (x1  x 2 ) 2  (y 1  y 2 ) 2 = (6) 2  (8) 2 = 100 = 10 2.

Mid – Point of a line:

 (x 1  x 2 ) (y 1  y 2 )  ;   2 2  Mid – point =  A(2 ; 5)

 C (x ; y )

B(-4 ; -3)

Mid – Point AB = C (-1; 1) 3.

Gradient of Straight Line: Gradient is represented using the symbol „m‟ [from y= mx+c] M=

y [ i.e the difference in y divided by the difference in x] x

Equation of a Straight Line: y  mx  c or y  y1  m( x  x1 )

Equation of a circle centre origin on a cartesian plane. x2 + y2 = r2

5 GO MATH WORKBOOKS

Grade 12 Core Mathematics 5.

The equation of a circle centre not the origin

(x  a) 2  (y  b) 2  r 2 Where (a ; b ) represents the co-ordinates of the centre and (x ; y ) a point on the circumference.

Examples: 1. Determine the equation of a circle centre (2 ; 5) and radius 4 units (x  a) 2  (y  b) 2  r 2 (x -2)2 + (y – 5)2 = 42 (x -2)2 + (y – 5)2 = 16 OR

x2 – 4x +4 + y2 – 5y + 9 = 0

2. Determine the centre and radius of x2 + 6x + y2 – 4y - 12 = 0. One must use the method of “completing the square” to calculate the centre and radius. The given equation must be converted to the correct form i.e. (x  a) 2  (y  b) 2  r 2 x2 + 6x + y2 – 4y - 12 = 0. x2 + 6x +( 3 )2+ y2 – 4y + (2)2 = 12 + 9 + 4 (x + 3)2 + (y – 2)2 = 25 Centre ( -3 ; 2) and radius = 5

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Grade 12 Core Mathematics Tangents to Circles. Remember: Tangents are perpendicular to the radius of any circle.

Essential to know the centre of the circle in order to work out the gradient of the radius. Examples. 1. Determine the equation of the tangent to the circle x - 2x + y + 4y = 5 at the point (-2;-1) Use completing the square to form two binomials: To find the centre x2 - 2x + (-1)2 + y2 + 4y + (2)2 = 5 + (-1)2 + ( 2)2 (x-1)2 + (y+2)2 - 1 - 4 = 5 (x-1)2 + ( y +2)2 = 10 centre is (1;-2) gradient of radius =

 2 1 1 2 1 3

Gradient of tangent = 3

(radius perpendicular to tangent)

Equation of tangent is y = 3x + c Substitute (-2;-1) thus -1 = 3(-2) + 6 c=5 equation of tangent is y = 3x + 5

7 Grade 12 Core Mathematics Exercis 1.1: A:

GO MATH WORKBOOKS

Analytical Geometry. Equations of circles of the form: (x  a) 2  (y  b) 2  r 2

1. Determine the equation of the circle with: 1.1 centre (2 ; 3) and radius 5 units.

1.2

Centre (4; -5) and radius 10 units

1.3

Centre (-2 ;-3) passing through (-2; 4)

8 Grade 12 Core Mathematics

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2. Determine the centre and radius of each of the following circles: 2.1

(đ?&#x2018;Ľ â&#x2C6;&#x2019; 4)2 + (đ?&#x2018;Ľ + 7)2 = 64

2.2

đ?&#x2018;Ľ 2 + đ?&#x2018;Ś 2 â&#x2C6;&#x2019; 6đ?&#x2018;Ś â&#x2C6;&#x2019; 27 = 0

2.3

đ?&#x2018;Ľ 2 + 4đ?&#x2018;Ľ + đ?&#x2018;Ś 2 â&#x2C6;&#x2019; 5 = 0

2.4

đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 4đ?&#x2018;Ľ + đ?&#x2018;Ś 2 + 2đ?&#x2018;Ś â&#x2C6;&#x2019; 20 = 0

9 Grade 12 Core Mathematics 2.5

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đ?&#x2018;Ľ 2 + 6đ?&#x2018;Ľ + đ?&#x2018;Ś 2 â&#x2C6;&#x2019; 4đ?&#x2018;Ś â&#x2C6;&#x2019; 12 = 0

2.6

đ?&#x2018;Ľ 2 + đ?&#x2018;Ś 2 + 4đ?&#x2018;Ľ + 6đ?&#x2018;Ś â&#x2C6;&#x2019; 3 = 0

Find the equations of the tangents in the following: 3.1

to circle đ?&#x2018;Ľ 2 + đ?&#x2018;Ś 2 = 5 at the point (3 ; 4).

3.2

to circle đ?&#x2018;Ľ 2 + đ?&#x2018;Ś 2 = 36 at the point (â&#x2C6;&#x2019;2 ; 3).

10 Grade 12 Core Mathematics 3.3

GO MATH WORKBOOKS

To circle (đ?&#x2018;Ľ + 1)2 + đ?&#x2018;Ś 2 = 20 which is parallel to 2đ?&#x2018;Ś â&#x2C6;&#x2019; đ?&#x2018;Ľ = 0

3.4

to (đ?&#x2018;Ľ â&#x2C6;&#x2019; 2)2 + (đ?&#x2018;Ś + 3)2 = 16 which is parallel to the y â&#x20AC;&#x201C; axis.

11 GO MATH WORKBOOKS

Grade 12 Core Mathematics B. 1.

A C 0

The Points A(-4 ;3) ; B(-4 ; -4) ; C(6 ; 1) and D(6 ; 8) lie on a cartesian plane. Determine: 1.1

the length of AD.

1.2

the mid-point of DC ______________________________________________________________ ____________________________________________________________

1.3

The gradient of BC ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

12 Grade 12 Core Mathematics 1.4

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the inclination of BC ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.5

the equation of BC ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.6

Prove that ABCD is a parallelogram. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Points A( 2 ; - 3 ) ; B( - 1 ; p ) and C ( 4 ; 3 ) are co-linear. Calculate the value of p. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

13 GO MATH WORKBOOKS

Grade 12 Core Mathematics 3 y

A(4 ; 12)

D ( 8; 4)

B(1 ; 3) C4;2) 0

In the figure ABCD is a quadrilateral with, A(4 ; 12 ); B(1 ; 3); C(4 ; 2) and D(8 ;4). 3.1

Find the length of AB. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.2

Find the co-ordinates of the mid-point of AC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.3

If E(p ; -3) is co-linear with A and B, find the value of p. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

14 Grade 12 Core Mathematics 3.4

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Determine the angle of inclination of the line through BC with the x- axis. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.5 Determine the size of BCˆ D .

______________________________________________________________ ______________________________________________________________ 3.6

Show that AB  BC.

15 GO MATH WORKBOOKS

Grade 12 Core Mathematics 5.

Locus of a Point ( Not in syllabus)

The locus of a point P is the set of all possible positions/locations of P under given conditions. The locus is expressed in terms of x and y in equation form, since the point P can vary as long as the conditions are upheld. Example: NB a sketch is extremely useful when finding loci 1.Find the relationship between x and y if the distance of P(x;y) from the origin must remain equal to its perpendicular distance from the line x = 2. P(x ; x=2 0;0

Note: The points P form a parabola thus the locus of P will be a Parabola P is tyhe Pt (x;y) M is the point( 2;y) Condition on P is that PM = OP PM = OP ( y - y ) + (2 - x) = y - 0) + ( x - 0) 4 - 4x + x = y + x y = -4x + 4 or x = y + 1 (it is the inverse of y = - x + 1 This is the relationship between the co-ord's of P that satisfy the given condition ( PM + OP always) and this is the locus of P. P (x ; y) 5

2. P is 5 units from B(2;3).

B(2 ; 3)

P is the point(x ;y) BP = 5 thus BP2 = 25 (y - 3)2 + ( x - 2 )2 = 25 (square to get root of root sign.) y2 -6y + 9 + x2 - 4x + 4 = 25 thus y2 - 6y + x2 - 4x = 12 is the equation of the locus required.

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Grade 12 Core Mathematics

Transformation Geometry

Rotation of an angle on a cartesian plane. Example 1. Rotate a point P(2;4) about the origin through an angle of 30º.

P' ( x cos   y sin  ; y cos   x sin  ) P' ( x cos 30   y sin 30  ; y cos 30   x sin 30  ) 2 3 4 1 4 3 2 1 P'     ;     1 2 1 2 1 2 1 2 





P' 3  2;2 3  1 Example 2:

Rotate a point P(2;4) about the origin through -135º. P' ( x cos   y sin  ; y cos   x sin  )

P'[ x cos(135  )  y sin(135  ); y cos(135  )  x sin( 135  )] 2  2 4 2 4  2 2 2  P'     ;     1 2 1 2 1 2 1 2  



P'  2  2 2 ;  2 2  2



P ' ( 2 ; 3 2 ) Example 3: Rotate a point P(4 ; 2) about the origin through 90º in an anticlockwise direction and then rotate P( x; y ) a further 120º anticlockwise and find the coordinates of P( x; y )

Px cos   y sin  ; y cos   x sin    1 3 1 3  P  (4   )  (2  ); (2   )  (4  )  2 2 2 2   P  2  3;1  2 3





17 Grade 12 Core Mathematics

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Exercise 2.1: Calculate the co-ordinates of Pˊ of the point ( 2 ; 4) after rotation about the origin, through an angle of: 1.

60° ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

150° ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

210° ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

−30° ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

−225° ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

The point P ( 2 ;4) is rotated about the origin through an angle of 75 in an anticlockwise direction. Without the use of a calculator , determine the x – coordinate of the image Pˊ of P. Simplify your answer. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

The point (3 ; 1) is rotated in an anticlockwise direction about the origin through an angle . If the image is

3− 3 1+3 3 2

;

. Calculate .

19 Grade 12 Core Mathematics

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NB: You may not use a calculator for this question:

The point P( 3 ; 2) is rotated about the origin through an angle of 120 in an anticlockwise direction. Determine xˊ and yˊ , the coordinates of Pˊ. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _____________________________________________________________

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Grade 12 Core Mathematics

TRIGONOMTRY FORMULA SHEET Compound Formula: cos(a-b) = cosacosb + sina sinb

REDUCTION FUNCTION

cos(a+b) = cosa cosb - sina sinb 90°

Quadrant 2

sin(a-b) = sinacosb - sinb cosa



180  

sin(a+b) = sina cosb + sinb cosa 180°

Double Angles

Quadrant 1  90  

 90  

S A   180   T C 360  

cos2x  cos 2 x  sin 2 x/2cos 2 x  1/1  2sin 2 x Quadrant 3 sin2x  2sinxcosx

0 360

Quadrant 3 270°

Complimentary Functions: sinx  cos(90   x) x =0 y=1 r=1 90°° Quadrant 2

SPECIAL ANGLES Quadrant 1

x = -1 y = 0 180° r=1

S A T C Quadrant 3

0° 360°

x =1 y=0 r=1

Quadrant 4

270° x =0 y = -1 r=1

Basic ratios:

sin x 

y r

cos x 

x r

tan x 

y x

FUNDAMENTAL IDENTITIES ` Quotient Identities: Squared Identities: cos 2 x  sin 2 x  1 sin x tan x  cos x

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Grade 12 Core Mathematics Solution of Triangles: sine rule :

a sin Aˆ



b sin Bˆ



c sin Cˆ

Cosine Rule: 1. a 2  b 2  c 2  2bc cos Aˆ

b2  c2  a2 OR cos Aˆ  2bc

a2  c2  b2 2. b 2  a 2  c 2  2ac cos Bˆ OR cos Bˆ  2ac 3. c 2  a 2  b 2  2ab cos Cˆ

OR cos Cˆ 

a2  b2  c2 2ab

Area rule : AreaABC  12 ac sin Bˆ ; 12 ab sin Cˆ ; 12 bc sin Aˆ

GRAPHS: y  asinb(x  c)  d ; ETC. a  affects the amplitude of the graph and also inverts the graph when negative.

b  affects the period or frequency c  shifts the graph horizontall

d  shifts the graph vertically

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Grade 12 Core Mathematics

3. Trigonometry: Compound Angles.

cos( A  B)  cos A cos B  sin A sin B cos( A  B)  cos A cos B  sin A sin B

Examples: 1. cos 2 x cos x  sin 2 x sin x

Reduce by using compound angle Formula cos(A+B)=cosAcosB-sinAsinB

 cos(2 x  x)  cos 3x 2. cos(3x  45 )  cos 3x.cos 45  sin 3x.sin 45

Use reduction function to reduce to acute angles

3. cos 290  . cos 320   sin110  . sin140   cos 70  . cos 40   sin 70  . sin40   cos(70  40 ) 



 cos 30  1  2

sin( x  y)  sin x cos y  sin y cos x sin( x  y)  sin x cos y  sin y cos x

1. sin 2 x cos x  sin x cos 2 x

 sin(2 x  x)  sin 3x 2. sin(x  90)  sin x cos 90  sin90 cos x

 Sinx(0)  1(cos x )  cos x

Use special angles to obtain answer

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Grade 12 Core Mathematics 3. sin 50 cos10  sin 10 cos 50

 sin(50  10)  sin 60 

3 2

4. Using compound angles to prove a statement.

3 sin( x  60)  sin( x  30)  cos x LHS  3[sin x cos 60  sin 60 cos x]  [sin x cos 30  sin 30 cos x] LHS  3[ 12 sin x  LHS 

3 2

cos x] 

sin x  32 cos x 

3 2

Use Expansion for sine Compound formula

sin x  12 cos x

LHS  cos x LHS  RHS . Exercise 3.1: 1. Use the compound angle formulae to simplify each expression to one term only: cos 3x cos 2x  sin 3x.sin 2x . 1.1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.2

sin 3x cos 2x  cos 3x.sin 2x . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.3

cos 5 x. cos 2 x  sin5 x sin2 x . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

24 Grade 12 Core Mathematics 1.4

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sin 3x.sin 2x  cos 3x. cos 2x . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.5

sin 2 x.sin x  cos 2x. cos x . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.6

sin 2 x. cos x  cos 2x. sin x . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.7

sin50 . cos 10  cos 50 . sin10 . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.8

sin81 . cos 23  sin23 . cos 81 . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

25 Grade 12 Core Mathematics 1.9

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cos 18 . sin31  sin18 cos 31 . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2. Expand each of the following using compound angle formulae. 2.1 sin(x  20 ) . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 2.2

cos(2 x  10 ) . ______________________________________________________________

2.3

sin(a  2b) . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

26 Grade 12 Core Mathematics 2.4

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cos(a  2b) . ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.5

sin(2a  20 ) . ______________________________________________________________

cos(a  30 ) . ______________________________________________________________

3. Evaluate the following without a calculator: sin40 . cos 20  cos 40 . sin20 3.1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

27 Grade 12 Core Mathematics 3.2

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cos 40 . cos 20  sin40 . sin20 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.3

cos 100 cos 280  sin100 . sin280 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.4

sin80 . sin40  sin10 . sin50 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4. Prove the following using compound angle formula: 4.1 sin(180   )   sin ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

28 Grade 12 Core Mathematics 4.2

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cos(360   )  cos ______________________________________________________________

cos(a  60 )  cos(a  60)  cos a ______________________________________________________________

5.3

sin(a  30 )  sin(a  30 )  cos a ______________________________________________________________

29 Grade 12 Core Mathematics 5.4

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cos( P  Q)  cos( P  Q)  2 cos P cos Q ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.5

sin5 A  sin3 A  2 sin4 A. cos A ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6. Prove that: sin75  cos 105  sin15  cos 15  0 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

30 Grade 12 Core Mathematics Double Angle Formulae. 1.

sin 2 x  2 sin x cos x i.e.

sin( x  x)  sin x cos x  sin x cos x  2 sin x cos x

cos 2 x  cos 2 x  sin 2 x OR  2 cos 2 x  1

OR  1  2 sin 2 x

cos 2 x  cos( x  x)  cos x cos x  sin x sin x  cos 2 x  sin 2 x OR i.e

 cos 2 x  (1  cos 2 x)  2 cos 2 x  1 OR  1  sin 2 x  sin 2 x  1  2 sin 2 x

tan 2 x 

2 tan x 1  tan 2 x

tan 2 x  tan( x  x) i.e

tan x  tan x 1  tan x tan x 2 tan x  1  tan 2 x 

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Grade 12 Core Mathematics

Proving Identities which include Double Angles. EXAMPLES: 1. Expand using appropriate sinx  sin2x  tanx formula 1  cosx  cos2x sin x  2 sin x cos x LHS  1  cos x  2 cos 2 x  1 sin x (1  2 cos x ) LHS  Factorise and cos x (1  2 cos x ) simplify sin x LHS  cos x LHS  tan x Simplify using LHS  RHS basic Identities 2. 8sinAcosA. cos2Acos4A  sin8A

RHS  sin 8 A

continue expanding using double angle formulae

RHS  2 sin 4 A cos 4 A RHS  4 sin 2 A cos 2 A cos 4 A RHS  8 sin A cos A cos 2 A cos 4 A RHS  LHS. 2. cos4A  cos2A 1  sin4A  sin2A tanA 2 cos 2 2 A  cos 2 A  1 look at the numerator and denomenato r searately 2 sin 2 A cos 2 A  sin 2 A (2 cos 2 A  1)(cos 2 A  1) LHS  and use the specific expansion needed. sin 2 A(2 cos 2 A  1) cos 2 A  1 LHS  If cos functions needed use only cos double L expansion etc. sin 2 A LHS 

2 cos 2 A 2 sin A cos A cos A LHS  sin A 1 LHS  tan A LHS  RHS LHS 

NB constants can be written in form of cos 2 x  sin 2 x 2  2sin 2 x  2cos 2 x etc.

32 Grade 12 Core Mathematics 3. cos3x  4cos 3 x  3cosx

LHS  cos(2 x  x) LHS  cos 2 x cos x  sin 2 x sin x LHS  cos x(2 cos 2 x  1)  2 sin x cos x sin x LHS  2 cos 3 x  cos x  2 sin 2 x cos x LHS  2 cos 3 x  cos x  2(1  cos 2 x) cos x LHS  2 cos 3 x  cos x  2 cos x  2 cos 3 x LHS  4 cos 3 x  3 cos x LHS  RHS 4. cos3x  1  4sin 2 x cosx cos(2 x  x) LHS  cos x cos 2 x cos x  sin 2 x sin x LHS  cos x cos 2 x cos x  2 sin 2 x cos x LHS  cos x LHS  cos 2 x  2 sin 2 x

LHS  1  2 sin 2 x  2 sin 2 x LHS  1  4 sin 2 x LHS  RHS .

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Grade 12 Core Mathematics 5. 1 sina  cosA  tan2A  cos2A cosA  sinA 1 sin 2 A LHS   cos 2 A cos 2 A 1  sin 2 A LHS  cos 2 A sin 2 A  2 sin A cos A  cos 2 A LHS  cos 2 A  sin 2 A (sin A  cos A)(sin A  cos A) LHS  (cos A  sin A)(cos A  sin A) sin a  cos A LHS  cos A  sin A LHS  RHS .

Exercise 3.2:

Rewrite 1 as sin2x + cos2x

Proving Identities.

2 cos x sin x  sin 2 x 1 sin 2 x

34 Grade 12 Core Mathematics 2.

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1 sin 2 x sin x  cos x   cos 2 x cos 2 x cos x  sin x

sin x  sin 2 x  tan x 1  cos x  cos 2 x

35 Grade 12 Core Mathematics 4.

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sin 3x  sin x  2 sin x 1  cos 2 x

sin 2 x  tan x 1  cos 2 x

36 Grade 12 Core Mathematics 6.

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sin 4 x  sin 2 x. cos 2 x  1  cos x 1  cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

sin 2 x  tan x  tan x cos 2 x

37 Grade 12 Core Mathematics 8.

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sin 2 x  cos x cos x  sin x  cos 2 x sin x  1

1  cos 2 x tan 2 x  cos 2 x tan x

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Grade 12 Core Mathematics General Solution of Trigonometric Equations:

If an equation is solved in general terms specific answers are not given unless asked for: The variable “k” is used to refer to the number of revolutions that are utilized to find the specific solution: Examples: 1.

Find the general solution for x in the following:

2 sin 2 x  1,630 sin 2 x  0,815 KeyL (2 x)  54,6  In Q1: 2x = 54,6 + k .360º x = 27,3º + k .180º In Q2 2x = (180º - 54,6º) + k.360º 2x = 125,4º + k.360º x = 62,7º + k.180º

General solutions:

2. If the specific solution is required then the period (Domain)must be stated: Example: Now find the value(s) for x when x  [0  ;360  ] If k = 0 then x = 27,3º or 62,7º k = 1 then x = 207,3º or 242,7º

3. (basic) cos x  0,5

x  {-180  ;180  }

key  30  x  150   360k or Or x  30  180k

x  210   360k

If k  1 then x  150  If k  0 then x  150 

39 Grade 12 Core Mathematics 4.

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(Difficult)

2 cos 2 x ď&#x20AC;Ť cos x ď&#x20AC;Ť 2 ď&#x20AC;˝ 0 4 cos 2 x ď&#x20AC; 2 ď&#x20AC;Ť cos x ď&#x20AC;Ť 2 ď&#x20AC;˝ 0 cos x(4 cos x ď&#x20AC;Ť 1) ď&#x20AC;˝ 0

cos x ď&#x20AC;˝ 0 x ď&#x20AC;˝ 90 ď&#x20AC;Ť 180k ď Ż

4 cos x ď&#x20AC;˝ ď&#x20AC;1 cos x ď&#x20AC;˝ ď&#x20AC;0,25 keyď&#x192;? ď&#x20AC;˝ 75,5ď Ż x ď&#x20AC;˝ ď&#x201A;ą104,5ď Ż ď&#x20AC;Ť 360k

TRIG EQUATIONS INVOLVING COMPLIMENTARY ANGLES. Examples: 1. Cos x = sin40 = cos[90-40] convert sine to cosine by using a complimentary function) = cos 50 The Key Angle is now 50 and answers lie in quadrants 1 and 4 x = 50 + 360k OR x = 310 + 360k 2.

sin(2 x ď&#x20AC;Ť 5ď Ż ) ď&#x20AC;˝ cos(ď&#x20AC;20ď Ż ď&#x20AC;Ť x) = sin[90-(-20 + x)] Use the complimentary angle = sin (110 - x) {1/2}

Use (110 - x) for the KEY ANGLE

2 x ď&#x20AC;Ť 5ď Ż ď&#x20AC;˝ 110 ď&#x20AC; x ď&#x20AC;Ť 360k 3x ď&#x20AC;˝ 105 ď&#x20AC;Ť 360k x = 35 + k.120 OR

2 x ď&#x20AC;Ť 5ď Ż ď&#x20AC;˝ 180 ď&#x20AC; (110 ď&#x20AC; x) ď&#x20AC;Ť 360k

2 x ď&#x20AC;Ť 5ď Ż = 70Â° + x + 360k đ?&#x2019;&#x2122; = 65Â° + 360k

40 Grade 12 Core Mathematics

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Exercise 3.3: A. Find the general solution for the following: 1.

sin x  0,235 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3 cos x  1,2066 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

tan 2 x  4,302 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2 tan 3x  2,3648 3 ______________________________________________________________

41 Grade 12 Core Mathematics

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B: Find the specific solutions of the above equations if x  [360 ;360  ] ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics Exercise 3.4: Example: 1. Solve for 2 sin x  3 cos x  0 , where x  0 ;360





2 sin x  3 cos x 2 sin x 3 cos x  cos x cos x 2 tan x  3

1 to convert to cos x a tan function

multiply by

3 2 Key L = 56,3 tan x  

x  123,7  / 303,7 

Find the general solution of : 4sin2x – 3 = 0 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Solve for x for x[-180;90] when cos2x – 7cosxtanx = 4 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

43 Grade 12 Core Mathematics 3.

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Solve for x if 4sinxcosx = 1 and x[0;360] ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Find the general solution of 4cos2x – 6cosx + 5 = 0 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 5.

2 Solve for : 2 sin x ď&#x20AC; sin x ď&#x20AC; 1 ď&#x20AC;˝ 0

ď &#x203A;

where x ď&#x192;&#x17D; 0ď Ż ;360ď Ż

ď ?

2 Solve for: 2 sin x ď&#x20AC;Ť 5 cos x ď&#x20AC;˝ 4

ď &#x203A;

where x ď&#x192;&#x17D; 0ď Ż ;360ď Ż

ď ?

Find the general solution of đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Ľ đ?&#x2018;Ľ + 20Â° = đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;Ľ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

45 Grade 12 Core Mathematics 8.

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Find the general solution of cos 2đ?&#x2018;Ľ + 23Â° = sinâ Ą (4đ?&#x2018;Ľ + 25Â°) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _____________________________________________________________

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Grade 12 Core Mathematics

Data Handling

MODIFYING BOX AND WHISKER DIAGRAMS. Box and Whisker diagrams: The inter-quartile range gives the spread of the middle 50% of data values and is not affected by the extremes. Outliers are values that stand apart from the rest of the values. An outlier is a value that is more than 1,5 times the interquartile range from the nearest quartile. E.G. If Q1 = 46 and Q3 = 60 then the IQR = 14. IQR x 1,5 = 14 x 1,5 = 21 Q1 – 21 =46 – 21 = 25. : Any value less than 25 will be an outlier Q3 + 21 = 60 + 21 = 81 : Any value greater than 81 will be an outlier. Outliers greatly affect the mean but have no more affect on the median or mode than any other value. THE BOX AND WHISKER DIAGRAM CAN BE MODIFIED TO MAKE IT MORE DESCRIPTIVE BY EXCLUDING OUTLIERS. If Q1 = 20 and Q3 = 34 then the IQR = 14. IQR x 1,5 = 14 x 1,5 = 21 Q1 – 21 =20 – 21 = -1. : Any value less than -1 will be an outlier Q3 + 21 = 34+ 21 = 55 : Any value greater than 55 will be an outlier.

Original Box & Whisker

80 20

Modified Box & Whisker 12

55 20

34 30

100

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Grade 12 Core Mathematics STANDARD DEVIATION: From a given table of data: 67; 70; 71; 71; 73; 74; 75; 75; 75; 77; 78; 78; 78; 78; 79; 80; 81; 82; 82; 83; 86; 86; 87; 91

67 70 71 73 74 75 77 78 79 80 81 82 83 86 87 91 x  78,2

1 1 2 1 1 3 1 4 1 1 1 2 1 2 1 1

n  24

xx 11,2 8,2 7,2 5,2 4,2 3,2 1,2 O,2 -0,8 -1,8 -2,8 -3,8 -4,8 -7,8 -8,8 -12,8

F x1 67 70 142 73 74 225 77 312 79 80 81 164 83 172 87 91

( x  x )2 125,44 67,24 51,84 27,04 17,64 10,24 1,44 0,04 0,64 3,24 7,84 14.44 23,04 60,84 77,44 163,84

 (x  x

STD DEV

s

 ( x  x1 ) 2 n 1

 

 (x  x ) 1

= 652,24

 5,8

s  5.9

Standard Deviation is a Measure of Dispersion about the mean: It measures how far each data item is from the mean and takes into account all data items. If the differences of the scores above the mean are added to the differences below the mean the answer will be zero. The differences are recorded as positives whether the score is above or below the mean.

Variance is defined as OR

 (x  x ) 1

n 1

, when working with a sample of a population.

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Grade 12 Core Mathematics

A small standard deviation indicates that the data items are clustered around the mean. While a large standard deviation indicates that the items are more spread out. STATISTICAL DATA & THE CALCULATOR: Mean & Std Deviation using a calculator The Casio f(x) 82ES and STD Dev Key MODE  2: STAT THEN 1 –VAR To enter Data into table:

key (data list) into each row.

To calculate the MEAN : key SHIFT -1 THEN No 5 : VAR : 2: x To calculate Std Dev: key SHIFT : 1 then 5 : VAR then 3 xn [OR 4 : xn  1] Mean & Standard Deviation on a Frequency Table using a Casio: Key MODE 2 : STAT then 1: 1 – VAR. Set a table up: Key SHIFT SETUP Scroll down to new screen: then key 3:STAT ; 1 : ON. Enter Data into column 1 and frequency into column 2 When complete key [AC] For the MEAN: key Shift 1 then 5:VAR then 2 : x . For Std Dev: key Shift 1 then 5 : VAR 3: xn [OR 4 : xn  1] Example: Speed in Kph 5060708090100110120-

Midpoint of Interval X 55 65 75 85 95 105 115 125

No of cars „f‟ 20 27 25 54 21 15 8 5

mean  82,48 StdDEv  17,5 n  175 175 Median lies in interval 80 – 89 thus Median = 84,5  87,5 2

Total No fX 1100 1755 1875 4590 1995 1575 920 625

49 GO MATH WORKBOOKS

Grade 12 Core Mathematics SYMMETRIC & SKEWED DATA:

a) Equal spread either side of the median in a box & Whisker diagram portrays a symmetrical spread. b) If data values are spread out more on one side than the other of the median then the data is said to be skewed.

xM 0

SYMMETRICAL

NORMAL DISTRIBURION

Q2 DIAGRAM SKEWED TO LEFT

X  X M 0

NEGATIVELY SKEWED DISTRIBUTION

DIAGRAM SKEWED TO RIGHT

 X

 X M 0 POSITIVELY SKEWED DATA

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Grade 12 Core Mathematics

Stem & Leaf diagrams; standard deviation an Ogive curves. Example: 1. The following marks were recorded for a math class: 28 53 75 63

45 75 63

36 58 75 63

36 60 78 67

36 60 81 68

38 60 83 68

45 71 84 69

42 71 84 76

45 75 90 79

1.1 Do a stem and leaf diagram for the data 1.2 Find the median, mode and mean for the data 1.3 Find the lower and upper quartile 1.4 Calculate: 1.4.1 the interquartile range 1.4.2 the semi-interquartile range 1.4.3 the range for the class 1.5 Write down the maximum and minimum scores. 1.6 Do a box and whisker diagram using the five-number summary

Answer: Stem 2 3 4 5 6 7 8 9

Leaf 8 6668 2555 38 0003337889 115555689 1344 0

Mode = 75 ; Mean = 62.9 ; Number = 35 Interquartile range = 30 ; Semi- interquartile ; range = 15 Range = 62 Standard Deviation = 16.6 Lowest = 28 ; Q1 = 45 ; Median = 67 ; Q3 = 75 ; Highest = 90

28 45 0

67 50

75 80

100

x  Q2  62.9  67  4.1  0 Data is negatively skewed i.e. skewed to the left. The marks are concentrated to the right of the median and spread out to the left of median.

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Grade 12 Core Mathematics Using Standard Deviation to reach conclusions:

Provided that a sample is reasonably large and the data is not too skewed ( that is , it does not have some very large or very small values), it is possible to make the following approximate statements. 

About 66% of the individual observations will lie within one standard deviation of the

mean. 

For most sets of data, about 95% of the individual observations will lie within 2

standard deviations of the mean. 

Almost all of the data will lie within 3 standard deviations of the mean.

Exercise 4.1: 1.

The marks , out of 150, for 30 learners were as follows:

100

109

122

118

124

127

105

112

128

107

114

115

121

135

111

117

120

130

123

141

107

113

116

119

121

131

129

139

1.1

Organise the marks using a stem & leaf diagram. ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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1.2 Draw a Box & Whisker diagram to illustrate the dispersion of the marks. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.3

Determine the mean for the above data. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.4

Determine the standard deviation, â&#x20AC;&#x2DC;sâ&#x20AC;&#x2122; (correct to 1 decimal place ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.5

What percentage of calls lie within one standard deviation of the mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.6

What can the teacher conclude about these marks. ______________________________________________________________ ______________________________________________________________ _____________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 2.

The following marks were recorded for a maths class:

23 48 30 46 2.1

30 45 25 45

31 50 35 44

52 47 24 53

42 20 30 45

15 43 40 54

45 60 52 35

36 40 75 28

Do a stem and leaf diagram for the data ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.2

Find the median, mode and mean for the data

______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.3

Find the lower and upper quartile ______________________________________________________________ ______________________________________________________________

29 37 34

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Calculate: 2.4.1 the interquartile range. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2,4,2 the semi-interquartile range.

2.4.3

the range for the class.

2.5

Write down the maximum and minimum scores.

55 Grade 12 Core Mathematics

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2.6 Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.7 Standard Deviation. 2.7.1

What % of scores lie within 1 standard deviation from the mean.

2.7.2

What % of scores lie within 2 standard deviations of the mean.

State whether the data is negatively or positively skewed and give a reason for your decision. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 3.

The following marks were recorded for a math class: 80 75 67 80

64 88 75 3.1

75 62 72 65

74 65 74 68

72 55 50 79

66 73 64 89

53 84 75 72

82 90 80 90

Do a stem and leaf diagram for the data

3.2

Find the median, mode and mean for the data

80 78 80

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Find the lower and upper quartile

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.4

Calculate: 3.4.1

the interquartile range.

3.4.2

the semi-interquartile range.

3.4.3

the range for the class.

3.5

Write down the maximum and minimum scores.

______________________________________________________________ ______________________________________________________________

58 Grade 12 Core Mathematics 3.6

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Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H)

3.7

Standard Deviation. 3.7.1 What % of scores lie within 1 standard deviation from the mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.7.2

What % of scores lie within 2 standard deviations of the mean.

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Grade 12 Core Mathematics 4. Girls 64 Boys 75 4.1

The following marks for a class of Girls and Boys were recorded : 90 85 57 72

95 92 62 75

84 90 85 88

82 75 80 79

76 83 64 89

80 64 75 72

88 75 95 80

Do a back to back stem and leaf diagram for the data

4.2

Find the median, mode and mean for both sets of data

70 72 80 81

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Grade 12 Core Mathematics 4.3

Find the lower and upper quartile of each set of data

4.4

Calculate: 4.4.1

the interquartile ranges for: 4.4.1.1

girls

4.4.1.2

boys

4.4.2

the semi-interquartile ranges for: 4.4.2.1 girls

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Grade 12 Core Mathematics 4.4.2.2 boys

4.4.3

the ranges for: 4.4.3.1 girls ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.4.3.2

boys

______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.5

Write down the maximum and minimum scores of each set of data ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.6 Do separate box and whisker diagrams for the girls and the boys ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _____________________________________________________________

62 Grade 12 Core Mathematics 4.7

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Standard Deviation : 4.7.1 the girls. ______________________________________________________________ ______________________________________________________________

4.7.2 the boys ______________________________________________________________ ______________________________________________________________

4.8

What % of scores lie within 1 standard deviation from the mean for: 4.8.1 girls ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.8.2 boys ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.9

What % of scores lie within 2 standard deviations of the mean for: 4.9.1 girls ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.9.2 boys ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

5. The following table represents the maths scores for the entire grade 11 maths group at Northwood School. The data is grouped due to the size of group. Class 0 to 9 10 to 19 20 to 29 30 to 39 40 to 49 50 to 59 60 to 69 70 to 79 80 to 89 90 to 99 100 to 109 Totals

Frequency(f) 15 10 17 40 35 22 20 20 15 5 1 200

Mid-points(X) 4.5 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5

5.1

Complete the last column of the table i.e (fX)

5.2

Find the modal class

fX 67.5

Find the median class ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.4

Find the interval where Q1 and Q3 lie. ______________________________________________________________

5.5

Calculate the estimated mean. NB estimated mean =

ď&#x192;Ľ fX n

64 Grade 12 Core Mathematics 5.6

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Use the grouped data to display the data on a histogram ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.7 Draw the relevant frequency polygon on the histogram.

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Grade 12 Core Mathematics STANDARD DEVIATION: 67; 70; 71; 71; 73; 74; 75; 75; 75; 77; 78; 78; 78; 78; 79; 80; 81; 82; 82; 83; 86; 86; 87; 91 f

x1 67 70 71 73 74 75 77 78 79 80 81 82 83 86 87 91 x  78,2

1 1 2 1 1 3 1 4 1 1 1 2 1 2 1 1

n  24

xx 11,2 8,2 7,2 5,2 4,2 3,2 1,2 O,2 -0,8 -1,8 -2,8 -3,8 -4,8 -7,8 -8,8 -12,8

f x1 67 70 142 73 74 225 77 312 79 80 81 164 83 172 87 91

( x  x )2 125,44 67,24 51,84 27,04 17,64 10,24 1,44 0,04 0,64 3,24 7,84 14.44 23,04 60,84 77,44 163,84

 (x  x

STD DEV

s

 ( x  x1 ) 2 n 1

 

 (x  x ) 1

= 652,24

 5,8

s  5.9

Standard Deviation is a Measure of Dispersion about the mean: It measures how far each data item is from the mean and takes into account all data items. If the differences of the scores above the mean are added to the differences below the mean the answer eill be zero. The differences are recorded as positives whether the score is above or below the mean. Variance is defined as OR

 (x  x ) 1

n 1

, when working with a sample of a population.

when working with a population n Variance is called the standard deviation and is considered the best measure of dispersion. The symbol “  ” is used to denote Standard Deviation when referring to a population AND “s” when referring to a sample of a population. A small standard deviation indicates that the data items are clustered around the mean. While a large standard deviation indicates that the items are more spread out.

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Grade 12 Core Mathematics STATISTICAL DATA & THE CALCULATOR Mean & Std Deviation using a calculator The Casio f(x) 82ES and STD Dev Key MODE  2: STAT THEN 1 –VAR To enter Data into table:

key (data list) into each row.

Midpoint of Interval X 55 65 75 85 95 105 115 125

No of cars „f‟ 20 27 25 54 21 15 8 5

mean  82,48 StdDEv  17,5 n  175 175 Median lies in interval 80 – 89 thus Median = 84,5  87,5 2

Total No fX 1100 1755 1875 4590 1995 1575 920 625

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Grade 12 Core Mathematics Exercise 4.2:

1. Complete the table and calculate the variance and the standard deviation. 40; 50; 65; 65; 70; 75; 75; 75; 75; 78; 78; 78; 78; 78; 79; 80; 81; 81; 82; 82; 82; 86; 88; 90

40 50 65 70 75 78 79 80 81 82 86 88 90 x  75,5

1 1 2 1 4 5 1 1 2 3 1 1 1

n  24

F x1 40 50 130 70 300 390 79 80 162 246 86 88 90

xx 35,5 25,5

( x  x )2

-12,5 -14,5

156.25 210.25

1260.25 650.25

 (x  x ) 1

STD DEV

s

 ( x  x1 ) 2 n 1

 

 ( x  x1 ) 2 n

s   ______________________________________________________________

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Grade 12 Core Mathematics 2.

Complete the table below:

Height (h) in cm 135  h < 140

Mid points 137,5

Frequency 2

Cumulative Frequency 2

140  h < 145

142,5

145  h < 150

147,5

150  h < 155

155  h < 160

160  h < 165

165  h < 170

170  h < 175

175  h < 180

2.1

Coordinates (140 ; 2)

Calculate the estimated mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.2

Draw a histogram of the data ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

69 Grade 12 Core Mathematics 2.3

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Draw a frequency polygon on the histogram

2.4

State the modal group, median height ,upper and lower quartiles for the data. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 2.5

Sketch the Ogive Curve for the data. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics

Circle Geometry:

Ratio / Proportion and Similar Triangles: Proportion Theorem: Theorem: A line parallel to one side of a triangle divides the other two sides (internally or externally) in the same proportion. A

H K

â&#x2020;&#x2019;

Given: â&#x2C6;&#x2020;ABC , D on AB and E on AC ( or AB and AC produced in either direction) DE | | BC. Required to prove:

đ??´đ??ˇ đ??ˇđ??ľ

đ??´đ??¸ đ??¸đ??ś

Proof: Construction: Draw DK and EH the altitudes for bases AD and AE. Draw DC and BE. đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; â&#x2C6;&#x2020;đ??´đ??ˇđ??¸ đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; â&#x2C6;&#x2020;đ??ľđ??ˇđ??¸

đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; â&#x2C6;&#x2020;đ??´đ??ˇđ??¸ đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; â&#x2C6;&#x2020;đ??śđ??¸đ??ˇ

1 đ??´đ??ˇ.đ??¸đ??ť 2 1 đ??ˇđ??ľ.đ??¸đ??ť 2

1 đ??´đ??¸.đ??ˇđ??ž 2 1 đ??¸đ??ś.đ??ˇđ??ž 2

đ??´đ??ˇ đ??ˇđ??ľ

đ??´đ??¸ đ??¸đ??ś

( đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;&#x2022;đ?&#x2018;Ą đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ??´đ??ˇ)

( đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;&#x2022;đ?&#x2018;Ą đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ??ˇđ??ž)

But đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; â&#x2C6;&#x2020;đ??ľđ??ˇđ??¸ đ??´đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; â&#x2C6;&#x2020;đ??śđ??¸đ??ˇ

â&#x2C6;´ â&#x2C6;´

đ??´đ??¸ đ??¸đ??ś

đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018; đ?&#x2018;&#x2019; đ??ˇđ??¸ ; đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;&#x2022;đ?&#x2018;&#x2019;đ?&#x2018;&#x2013;đ?&#x2018;&#x201D;đ?&#x2018;&#x2022;đ?&#x2018;Ą; đ??ˇđ??¸ | đ??ľđ??ś)

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Grade 12 Core Mathematics Similar Triangle Theorem:

Theorem: If two triangles are equiangular, then the corresponding sides are in proportion, and then the triangles are said to be similar. A ď&#x201A;ˇ D ď&#x201A;ˇ

Given: â&#x2C6;&#x2020;đ??´đ??ľđ??ś and â&#x2C6;&#x2020;đ??ˇđ??¸đ??š with đ??´ = đ??ˇ ; đ??ľ = đ??¸ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ś = đ??š Required to Prove:

đ??ˇđ??¸ đ??´đ??ľ

đ??¸đ??š đ??ľđ??ś

đ??ˇđ??š đ??´đ??ś

Proof: In â&#x2C6;&#x2020;đ??´đ?&#x2018;&#x192;đ?&#x2018;&#x201E; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??ˇđ??¸đ??š đ??´đ?&#x2018;&#x192; = đ??ˇđ??¸ đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;˘đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ??´ = đ??ˇ ( đ??şđ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) đ??´đ?&#x2018;&#x201E; = đ??ˇđ??š ( đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;˘đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;) â&#x2C6;&#x2020;đ??´đ?&#x2018;&#x192;đ?&#x2018;&#x201E; â&#x2030;Ą â&#x2C6;&#x2020;đ??ˇđ??¸đ??š ( đ?&#x2018;&#x2020;đ??´đ?&#x2018;&#x2020;) đ?&#x2018;&#x192;1 = đ??¸ đ?&#x2018;&#x192;1 = đ??ľ

(đ??¸ = đ??ľ ) (đ??śđ?&#x2018;&#x201A;đ?&#x2018;&#x2026;đ?&#x2018;&#x2026;đ??¸đ?&#x2018;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x201A;đ?&#x2018; đ??ˇđ??źđ?&#x2018; đ??ş ď&#x192;? â&#x20AC;˛đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;&#x17E;đ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;)

đ?&#x2018;&#x192;đ?&#x2018;&#x201E;||đ??ľđ??ś đ??´đ?&#x2018;&#x192;

đ??´đ?&#x2018;&#x201E;

â&#x2C6;´ đ??´đ??ľ = đ??´đ??ś

(đ?&#x2018;&#x192;đ?&#x2018;&#x201E;||đ??ľđ??ś đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; â&#x2C6;&#x2020;đ??´đ??ľđ??ś)

đ??ľđ?&#x2018;˘đ?&#x2018;Ą đ??´đ?&#x2018;&#x192; = đ??ˇđ??¸ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??´đ?&#x2018;&#x201E; = đ??ˇđ??š đ??ˇđ??¸

đ??ˇđ??š

â&#x2C6;´ đ??´đ??ľ = đ??´đ??ś

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Grade 12 Core Mathematics Exercise 5.1: Mixed examples on proportion and similarity: 1.

Calculate as requested. A

1.1

1.2

x+ 1

Find x in 1.1

Find BC in 1.2

1.3 8

D 12

1.4 D

x B



y 6

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Grade 12 Core Mathematics 2.

In the figure FE//GA. BA is a tangent at B.

E G

Prove: 2.1 â&#x2C6;&#x2020;đ??´đ??ľđ??ś ď źď źď ź â&#x2C6;&#x2020;ADB . 2.2 đ??´đ??ľ 2 = đ??´đ??ˇ. đ??´đ??ś 2.4 đ??¸2 = đ??ˇ2 . 2.5 â&#x2C6;&#x2020;đ??¸đ??śđ??´|||â&#x2C6;&#x2020;đ??ˇđ??¸đ??´. 2.6 đ??´đ??¸ 2 = đ??´đ??ˇ. đ??´đ??ś. 2.7 đ??´đ??¸ = đ??´đ??ľ

1 2

x 1 2 3

C 2

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 3.

In the diagram , the chords Ad and BC of circle ABCD are produced to meet at F . E is a point on AF such that EC is a tangent to the circle at C and BD//CE.

A 1

Prove with reasons: 3.1 BC = DC

2 3

1 2

3.2

BAF /// DCF B

3.3

BA DE  AF EF

3.4

ECD /// EAC CE 2 

3.5

2 5

2 3 4

AE.BC .EF CF

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______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics 4.

4.1

In the diagram AC is a tangent, FE //AC. Prove with reasons: â&#x2C6;&#x2020;đ??ľđ??¸đ??šď ź ď ź ď ź â&#x2C6;&#x2020;đ??ľđ??šđ??ˇ

4.2

BE 2 ď&#x20AC;˝ BD.BF

D F

1 1

A ______________________________________________________________ B C

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _____________________________________________________________

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Grade 12 Core Mathematics 5.

In the diagram KL // PM and NM = LM

K Q

Prove with reasons: NR 1 5.1  NK 2 5.2

NQL /// PML

5.3

QL.PL = 2NM2

4 4

2 3

1 2

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Grade 12 Core Mathematics 6.

O is the centre ; BP = OB = AO . PT is a tangent and EP  AP. E

Prove with reasons: 6.1 TEPB is a cyclic Quad. 6.2 6.3

ATB  APE . TP = PE.

O B

6.4

ATPEPB.

6.5

2BP2 = BT.BE

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Grade 12 Core Mathematics 7

MN is a tangent at R and PQRS is a cyclic quadrilateral. QT // SR and PS // QR Prove with reasons:

Q 2 1

7.1

QR bisects PQT. P

7.2

MN //PT S M

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Grade 12 Core Mathematics đ??ľđ??ś = đ??śđ??ˇ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??şđ??ľ is a tangent to the larger circle. Hint: đ??żđ?&#x2018;&#x2019;đ?&#x2018;Ą đ??śđ??¸ đ??š = đ??ľ2 = đ??¸ = đ?&#x2018;Ľ

Prove with reasons: 8.1

ď &#x201E;CDF /// ď &#x201E;CED

8.2 8.3 8.4

CDË&#x2020; E ď&#x20AC;˝ CGË&#x2020; B

ď &#x201E;CGB /// ď &#x201E;CDE EC.GB = DE . CD

1 3 1 2 2

3 x

F 1

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Grade 12 Core Mathematics 9.1 AB is a tangent to the smaller circle and BC is a common chord. đ??ťđ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Ą: đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;Ą đ??ˇ1 = đ?&#x2018;Ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ˇ2 = đ?&#x2018;Ś Prove with reasons: B 9.1 AC bisects BCË&#x2020; D 2 9.2

ď &#x201E;ABC /// ď &#x201E;DEC

9.3

BC.DC ď&#x20AC;˝ AC.EC

9.4

ADË&#x2020; C ď&#x20AC;˝ BAË&#x2020; C ď&#x20AC;Ť ACË&#x2020; D

9.5

ď &#x201E;ABE /// ď &#x201E;ACB

9.6

AC 2 ď&#x20AC;˝ AE. AC

A 2 1

1 2

E 1 1 2

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