Grade 12 workbook paper 1 by GoMath Workbooks

Compiled by Chesley Nell

2 GOMATH WORKBOOKS

Grade 12 Core Mathematics

Forward: Welcome to “ GO MATH WORKBOOKS”. This workbook is designed to be 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. Chesley Nell : Mathematics Educator.  Chesley

Nell 2011

3 GOMATH WORKBOOKS

Grade 12 Core Mathematics

GRADE 12 CORE MATHEMATICS Contents: Paper One: Topic:

Page:

Calculus

(4 – 78)

Number Patterns

( 79 – 108)

Financial Maths

(109 – 149)

Functions & Graphs

(150 – 218)

Probability Theory

(219 – 231)

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

5 Grade 12 Core Mathematics

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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. _____________________________________________________ _____________________________________________________ _____________________________________________________

6 Grade 12 Core Mathematics 3.

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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) = x 2 -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) = x 2 -x f(x) = x 2 -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) = x 2 -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

8 Grade 12 Core Mathematics

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Exercise 1.2: 1.1

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

9 Grade 12 Core Mathematics 2.2

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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 GOď&#x201A;ˇMATH 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

2 ď&#x20AC;˝0 x ď&#x201A;Žď&#x201A;Ľ x 3

NB: lim 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: 1. f ( x ) Dx f 2. 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

15 Grade 12 Core Mathematics

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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

16 Grade 12 Core Mathematics

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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

19 Grade 12 Core Mathematics

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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

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

23 Grade 12 Core Mathematics

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Exercise 1.6: 1. 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) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

24 GOMATH WORKBOOKS

Grade 12 Core Mathematics 2.

Determine the equation of the line which touches the parabola y  x 2  2 x  3 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 Grade 12 Core Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

Remainder And Factor Theorems: A: Remainder Theorem: Proof: Let Q(x) be the quotient and R the remainder free of x when f(x) is divided by (x â&#x20AC;&#x201C; k) ď &#x153; f ( x) ď&#x20AC;˝ ( x ď&#x20AC; k )Q( x) ď&#x20AC;Ť R

f (k ) ď&#x20AC;˝ (k ď&#x20AC; k )Q(k ) ď&#x20AC;Ť R f (k ) ď&#x20AC;˝ 0.Q(k ) ď&#x20AC;Ť R f (k ) ď&#x20AC;˝ R. B: Factor Theorem: Statements: đ?&#x2018;? â&#x;š đ??źđ?&#x2018;&#x201C; đ?&#x2018;&#x17D;đ?&#x2018;Ľ â&#x2C6;&#x2019; đ?&#x2018;? đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x17D; đ?&#x2018;&#x201C;đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A; đ?&#x2018;&#x201C; (đ?&#x2018;&#x17D;) = 0 OR đ?&#x2018;? â&#x;š đ??źđ?&#x2018;&#x201C; đ?&#x2018;&#x201C; (đ?&#x2018;&#x17D;) = 0 , đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A; đ?&#x2018;&#x17D;đ?&#x2018;Ľ â&#x2C6;&#x2019; đ?&#x2018;? đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x17D; đ?&#x2018;&#x201C;đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x201C;(đ?&#x2018;Ľ). Example 1: If (đ?&#x2018;Ľ + 2) is a factor of 3đ?&#x2018;Ľ 3 + đ?&#x2018;?đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 2đ?&#x2018;Ľ + 8 , find the value of p. NB. Since đ?&#x2018;Ľ + 2 is a factor of the expression then the remainder theorem states that đ?&#x2018;&#x201C;(â&#x2C6;&#x2019;2) = 0 đ?&#x2018;&#x201C; (â&#x2C6;&#x2019;2) = 3(â&#x2C6;&#x2019;2)3 + đ?&#x2018;?(â&#x2C6;&#x2019;2)2 â&#x2C6;&#x2019; 2(â&#x2C6;&#x2019;2) + 8 = 0 â&#x2C6;&#x2019;24 + 4đ?&#x2018;? + 4 + 8 = 0 4đ?&#x2018;? = 12 đ?&#x2018;?=3 Example 2: If đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = đ?&#x2018;Ľ 3 â&#x2C6;&#x2019; 2đ?&#x2018;Ľ 2 + 5đ?&#x2018;Ľ â&#x2C6;&#x2019; 6 find đ?&#x2018;&#x201C;(2) đ?&#x2018;&#x201C; ( 2) = ( 2) 3 â&#x2C6;&#x2019; 2( 2) 2 + 5( 2) â&#x2C6;&#x2019; 6 đ?&#x2018;&#x201C;(2) = 8 â&#x2C6;&#x2019; 8 + 10 â&#x2C6;&#x2019; 6 đ?&#x2018;&#x201C; ( 2) = 8 Exercise: 1.

Use the remainder theorem to find the remainder when 3x 3 ď&#x20AC; 2 x 2 ď&#x20AC;Ť 5x ď&#x20AC; 10 is divided by x ď&#x20AC; 2 . ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

26 Grade 12 Core Mathematics 2.1

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If 2 x  1 is a factor of px 3  3x 2  3x  p determine the value of

Determine the value of ‘a’ if 5x 3  3x 2  ax  4 is divided by x  2 and leaves a remainder of -6. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.1

If f(x) = x 3 – 3x + 2x – 9 determine f(-1). ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

27 Grade 12 Core Mathematics 3.2

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Prove that x  2 is a factor of f ( x)  2 x 3  3x 2  11x  6 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

28 GOMATH WORKBOOKS

Grade 12 Core Mathematics

Factorising equations to the 3 rd degree [NB it would be to your advantage to purchase a calculator that can factorise equations to the 3 rd 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 2 nd 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 3x 2 – 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)(3x 2 – 4x – 4) = (x-1)(3x+2)(x-2)

ie x – 1 = 0 x = 1

29 GOMATH WORKBOOKS

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 3 rd 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 Test (x-2) f (2)  8  4  44  40 =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)

+40 +40 0

30 Grade 12 Core Mathematics

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Exercise 1. 7: 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.

31 Grade 12 Core Mathematics 4.

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

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 ______________________________________________

32 Grade 12 Core Mathematics 7.

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

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

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

33 Grade 12 Core Mathematics 2.

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

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

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

34 Grade 12 Core Mathematics 5.

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

Factorise f ( x)  2 x 3  3x 2  11x  6 completely. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

6.2

Hence solve 2 x 3  3x 2  11x  6  0 ______________________________________________ _____________________________________________________ _____________________________________________________

35 GOMATH WORKBOOKS

Grade 12 Core Mathematics

Maximum and Minimum Stationary Points The following are two methods that can be used to test for Maximum and Minimum Turning (stationary ) points. 1.

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: 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 

36 GOMATH WORKBOOKS

Grade 12 Core Mathematics 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

-VE

f (0)  11 f ( x)  5

-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 2 nd 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) (2nd derivative) f ( x)  6 x  8 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 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

37 GOMATH WORKBOOKS

Grade 12 Core Mathematics 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)

f (1)  3  12  12  3 positive gradient f (3)  27  36  12  13 positive gradient NB A point of inflection occurs at this

Test

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: 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

Point of inflection

38 Grade 12 Core Mathematics Example: Sketch the graph of

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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.

Sketching curves of Equations of the 3 rd 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.

39 Grade 12 Core Mathematics

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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 2 nd 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  a point of inflection f (4)  6(4)  24  0

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 ( 2 nd 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.

40 GOMATH WORKBOOKS

Grade 12 Core Mathematics y (2 ; 32)

f ( x)  x 3  12 x 2  36 x

Point of Inflection

(0 ; 6)

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 ______________________________________________

41 Grade 12 Core Mathematics 2.

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

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

42 Grade 12 Core Mathematics 4.

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

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

43 Grade 12 Core Mathematics 6.

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

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

44 Grade 12 Core Mathematics

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Exercise 1. 10 : 1. 1.1

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

1.2

Find f (x) ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.3

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

45 Grade 12 Core Mathematics 1.4

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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 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

2.2

46 Grade 12 Core Mathematics 2.3

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2.4

3. 3.1

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

x3  6x 2  9x  4  0

47 Grade 12 Core Mathematics 3.2

Find

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f (x)

3.3

3.4

4. 4.1

f ( x)  x 3  3 x 2 Solve for x if x 3  3x 2  0 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

48 Grade 12 Core Mathematics 4.2

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4.3

4.4

5 5.1

f ( x)  2 x 3  3 x 2 Solve for x if 2 x 3  3x 2  0 ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

49 Grade 12 Core Mathematics 5.2

GOMATH WORKBOOKS

5.3

5.4

50 GOMATH WORKBOOKS

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 a) (i) f (2)  2(2)  6  2 Increasing (iii) f (4)  2(4)  6  2 Decreasing (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

51 GOMATH WORKBOOKS

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 ( x3) 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

52 GOMATH WORKBOOKS

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

53 GOMATH WORKBOOKS

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 = ax 3 +bx 2 + 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

2 Increases for x < 3

2 Decreases for < x <3 3

f’ Increasing for x > 3 x

2 3

54 GOMATH WORKBOOKS

Grade 12 Core Mathematics 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ 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

____________________________________________________ ____________________________________________________ ____________________________________________________

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

55 GOMATH WORKBOOKS

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________

56 GOMATH WORKBOOKS

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.

______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________

57 Grade 12 Core Mathematics

GOMATH WORKBOOKS

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

58 GOMATH WORKBOOKS

Grade 12 Core Mathematics

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

again.

f .

59 GOMATH WORKBOOKS

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 Total surface area (S) = 2r 2  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





60 GOMATH WORKBOOKS

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 Radius = 51,5 mm and h 

50000  2(51,5) 2  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.

61 GOMATH WORKBOOKS

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 5 x 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) = x 2 + 4xh V= x 2 h

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.

62 GOď&#x201A;ˇMATH WORKBOOKS

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

63 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

64 GOMATH WORKBOOKS

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

______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

65 GOMATH WORKBOOKS

Grade 12 Core Mathematics

3. A cylinder has a height of (40 – x) cm, its base has a radius of x cm 3.1 Derive a formula to calculate the volume ( V = r 2 h ) 3.2 Calculate the radius that will give a maximum volume.

40-x m

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

66 GOMATH WORKBOOKS

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 dt

t  .

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? 3.

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

67 GOMATH WORKBOOKS

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: 2.1 v  12t 2  24t  9 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

1 3 s or t  s 2 2

68 Grade 12 Core Mathematics 2.3

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s ď&#x20AC;˝ 4t 3 ď&#x20AC; 12t 2 ď&#x20AC;Ť 9t ď&#x20AC;Ť 28 3

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. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

69 Grade 12 Core Mathematics 1.5

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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 _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

70 Grade 12 Core Mathematics

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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 ? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

71 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

72 Grade 12 Core Mathematics 3.3

lim x x 3

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x3 2  3x ______________________________________________

Determine using standard rules.

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

dy if y  (2 x  5)( x  3) dx

f (x) if

f ( x) 

2  3x 2  4 3 x

73 Grade 12 Core Mathematics 5:

If g ( x)  3x 2

5.1

Determine

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g (x)

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

74 Grade 12 Core Mathematics

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

75 Grade 12 Core Mathematics 6.4

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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.

76 GOMATH WORKBOOKS

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. 10.1 f (x) 10.2

d2y dx 2

11.

INCREASING OR DECREASING FUNCTIONS:

f (c)  stationary If the f (c)  ve then If the f (c)  ve then If the f (c)  0 then it

value

it shows an increasing function at x = c it shows a decreasing function at x = c. has a stationary point at x = c.

77 GOMATH WORKBOOKS

Grade 12 Core Mathematics

12. AVERAGE GRADIENT: 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)

x+h

f ( x  h)  f ( x ) h

78 GOMATH WORKBOOKS

Grade 12 Core Mathematics 13.

GRADIENT FUNCTION OR GRADIENT AT A POINT OR 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.

79 GOMATH WORKBOOKS

Grade 12 Core Mathematics

2. Number Patterns: 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

80 GOMATH WORKBOOKS

Grade 12 Core Mathematics

NOW if I need to find the 20th term simply substitute 20 for n in the formula above: i.e. 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

81 GOMATH WORKBOOKS

Grade 12 Core Mathematics 2.

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

n 2a  (n  1)d  2

common difference Ist term

82 Grade 12 Core Mathematics

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Formal Work: 1. Sum of an Arithmetic Series: đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = đ?&#x2018;&#x17D; + (đ?&#x2018;&#x17D; + đ?&#x2018;&#x2018;) + (đ?&#x2018;&#x17D; + 2đ?&#x2018;&#x2018;) + â&#x2039;Ż + [đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 2)đ?&#x2018;&#x2018;] + â&#x2039;Ż + [đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018;] đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = [đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1) đ?&#x2018;&#x2018;] + [đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 2)đ?&#x2018;&#x2018;] + [đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 3) đ?&#x2018;&#x2018; + [đ?&#x2018;&#x17D; + đ?&#x2018;&#x2018;] + đ?&#x2018;&#x17D;

đ??´đ?&#x2018;&#x2018;đ?&#x2018;&#x2018; 2đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = [2đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018; + [2đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018; ] + [2đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018; ] + â&#x20AC;Ś â&#x20AC;Ś â&#x20AC;Ś + [2đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018; ] = đ?&#x2018;&#x203A;[2đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018;] đ?&#x2018;&#x203A; đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = 2 [2đ?&#x2018;&#x17D; + (đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1)đ?&#x2018;&#x2018;] Example 1. In the series 1; 9; 17 find the sum of the first 20 terms. a = 1 ; d = 8 S20 = ? Sn ď&#x20AC;˝

n ď &#x203A;2a ď&#x20AC;Ť (n ď&#x20AC; 1)d ď ? 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 ď&#x20AC;˝

n ď &#x203A;2a ď&#x20AC;Ť (n ď&#x20AC; 1)d ď ? 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;â&#x20AC;Śâ&#x20AC;Śis equal to 27. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

83 Grade 12 Core Mathematics

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2. In the sequence, 23; 16; 9;… 2.1 Determine the 13th term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ 2.2

Which term in the sequence is -131?

______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ 3. Given the arithmetic sequence: 2; 3½; 5;… 3.1 determine the 53rd term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________

84 Grade 12 Core Mathematics

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3.2 Which term in the sequence is 53? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________ 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ ____________________________________________________

85 Grade 12 Core Mathematics

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

86 Grade 12 Core Mathematics

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

87 Grade 12 Core Mathematics

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11.2 Which term in the sequence is –53? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

12.

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

88 GOMATH WORKBOOKS

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.

89 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.

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 6.

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 ď&#x20AC; 1) Sn ď&#x20AC;˝ for r ď&#x20AC;1

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

a(1 ď&#x20AC; r n ) Sn ď&#x20AC;˝ for 1ď&#x20AC; r

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

Formal Work: Sum of a Geometric Series

đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = đ?&#x2018;&#x17D; + đ?&#x2018;&#x17D;đ?&#x2018;&#x; + đ?&#x2018;&#x17D;đ?&#x2018;&#x; 2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x; đ?&#x2018;&#x203A;â&#x2C6;&#x2019;2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x; đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 đ?&#x2018;&#x; Ă&#x2014; đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = đ?&#x2018;&#x17D;đ?&#x2018;&#x; + đ?&#x2018;&#x17D;đ?&#x2018;&#x; 2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x; đ?&#x2018;&#x203A;â&#x2C6;&#x2019;2 + đ?&#x2018;&#x17D;đ?&#x2018;&#x; đ?&#x2018;&#x203A;â&#x2C6;&#x2019;1 + đ?&#x2018;&#x17D;đ?&#x2018;&#x; đ?&#x2018;&#x203A; đ?&#x2018;&#x;đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; â&#x2C6;&#x2019; đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; = â&#x2C6;&#x2019;đ?&#x2018;&#x17D; + 0 + 0

đ?&#x2018;&#x2020;đ?&#x2018;&#x203A;( đ?&#x2018;&#x; â&#x2C6;&#x2019; 1) = đ?&#x2018;&#x17D;(đ?&#x2018;&#x; đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1) đ?&#x2018;&#x2020;đ?&#x2018;&#x203A; =

+ 0

+ đ?&#x2018;&#x17D;đ?&#x2018;&#x; đ?&#x2018;&#x203A;

đ?&#x2018;&#x17D;(đ?&#x2018;&#x; đ?&#x2018;&#x203A; â&#x2C6;&#x2019;1) (đ?&#x2018;&#x;â&#x2C6;&#x2019;1)

Example 1: In the series 2 + 6 + 18+â&#x20AC;Śâ&#x20AC;Ś find the sum of the first 5 terms. a = 2; r = 3; S5 = ?

a(r n ď&#x20AC; 1) r ď&#x20AC;1 2(35 ď&#x20AC; 1) S5 = 2 S5 = 80. Sn ď&#x20AC;˝

91 GOMATH WORKBOOKS

Grade 12 Core Mathematics Example 2: If 2 + 6 + 18……(to n terms) = 80. i.e. Find n.

Find the number of terms in the series.

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

Exercise 2.2: 1.

Calculate the tenth term of a sequence: 81; 27; 9; … ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

2. If the first term of a sequence is

3 2

and the fourth is –12, determine:

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

92 Grade 12 Core Mathematics

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2.2 The second and third terms if the sequence is geometric. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 3. Find the 10th term in a geometric progression where 1 st term is 5 and the common ratio is 3. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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

93 GOMATH WORKBOOKS

Grade 12 Core Mathematics 5.

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ 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 1 st three terms and then do the T check.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 = 38 - 1 = 6561

Sn 

Exercise 2.3: 1.

Evaluate: 8

1.1

 3k  1 k 1

95 Grade 12 Core Mathematics

GOMATH WORKBOOKS

 5k  2

1.2

k 1

1.3

 4k  2 k 3

1.4

 3(2) k 1

k 1

96 GOMATH WORKBOOKS

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

97 GOď&#x201A;ˇMATH WORKBOOKS

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;Ľ â&#x2C6;&#x2019;1; Do not ď&#x20AC;1 ď&#x20AC;ž

2x ď&#x20AC; 1 ď&#x20AC;ž1 1

forget to invert the inequality signs: Add 1 to each side and simplify.

> 2x > 2

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

98 GOMATH WORKBOOKS

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

99 GOMATH WORKBOOKS

Grade 12 Core Mathematics n

If T2 = 8 and T6 = 24 determine

n if

 Tk

 480

k 1

______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ 5.

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

100 Grade 12 Core Mathematics 6.

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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.

101 Grade 12 Core Mathematics 8.

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

Calculate the first 4 terms of the sequence.

The sum of the first n – terms of an AP is given by 2n2 – n. 9.1

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

102 Grade 12 Core Mathematics 9.2

GOMATH WORKBOOKS

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 20 th term is 130. Determine the tenth term. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

103 Grade 12 Core Mathematics 12.

GOMATH WORKBOOKS

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

104 GOMATH WORKBOOKS

Grade 12 Core Mathematics n

_14.2

 5.3

r 1

 605

r 1

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

105 GOMATH WORKBOOKS

Grade 12 Core Mathematics 16.

The first term of an arithmetic series is 2 The sum of the 3 rd 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

106 GOMATH WORKBOOKS

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

18.

Calculate the value of n if:

k 1

 21

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

107 Grade 12 Core Mathematics 20.

GOMATH WORKBOOKS

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________

108 GOď&#x201A;ˇMATH 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

109 Grade 12 Core Mathematics

GOMATH WORKBOOKS

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 x m = p  logxp = m

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

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.

111 GOď&#x201A;ˇMATH 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. 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??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; 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 (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:

112 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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113 GOď&#x201A;ˇMATH 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)

114 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 S I = Prt A  P(1  ni)

Decay A  P(1  ni)

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.

115 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.



116 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  Jm    100 1   1   or 1  i   1   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.

117 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

118 GOMATH WORKBOOKS

Grade 12 Core Mathematics

Nominal Rate

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



Fv (i)

x[(1  i) n1  1] 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:





x (1  i) n  1 Balance of loan = Pv (1  i) i n

Pv (1 

Jm tm ) m

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

119 GOď&#x201A;ˇMATH WORKBOOKS

Grade 12 Core Mathematics

Effective Rate: Nominal 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&#x192;ˇ ď&#x20AC; 1ď&#x192;ş mď&#x192;¸ ď&#x192;Şď&#x192;Ťď&#x192;¨ ď&#x192;şď&#x192;ť

ď &#x203A;

Jm = Nominal rate Jeff = effective rate

ď ?

Jm ď&#x20AC;˝ 100m m 1 ď&#x20AC;Ť Jeff ď&#x20AC; 1

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

120 Grade 12 Core Mathematics

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Deferred Payments: đ??šđ?&#x2018;Ł =

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

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; â&#x201E;&#x17D;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 2600 ď&#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 )12 Fv ď&#x20AC;˝ 0.16 12 12 Fv ď&#x20AC;˝ R698150,10 Notes:

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

121 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 x[(1  i) n1  1] Fv  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

122 Grade 12 Core Mathematics

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Exercise 3.1: 1.

Alida decides to start saving for a car. Starting immediately on her 16 th 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 24 th 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? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________

123 Grade 12 Core Mathematics 3.

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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? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

124 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________ ____________________________________________________

125 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? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

126 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 ) ] 12 Pv  0,14 12 Pv  R31487,44 1250[1  (1 

127 Grade 12 Core Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

128 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? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

129 Grade 12 Core Mathematics 5.

GOMATH WORKBOOKS

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 ‘bakkie’ 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

130 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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

131 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 18 th to 21st payments. Calculate the balance of the loan at the end of the 17 th 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

x[1  (1  i)  n ] NB: Pv  i a)

x

n5

Pv (i )

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

132 GOď&#x201A;ˇMATH WORKBOOKS

Grade 12 Core Mathematics

Balance of loan: Pv (1 ď&#x20AC;Ť i) n ď&#x20AC;

ď &#x203A;

ď ?

x (1 ď&#x20AC;Ť i) n ď&#x20AC; 1 i

0.07 17 ď&#x192;š ď&#x192;Š 2574.16ď&#x192;Ş(1 ď&#x20AC;Ť ) ď&#x20AC; 1ď&#x192;ş 0,07 17 12 ď&#x192;Ť ď&#x192;ť Balance ď&#x20AC;˝ 130000(1 ď&#x20AC;Ť ) ď&#x20AC; 0.07 12 12 Balance of loan = R 97 647,50

đ?&#x2018;?)

đ?&#x2018;&#x192;đ?&#x2018;&#x17D;đ?&#x2018;Śđ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x2122;đ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;Ą đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; 21 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018; = 97647.50 (1 +

xď&#x20AC;˝

0.07 4 12

) = đ?&#x2018;&#x2026;99945.96

Pv (i ) 1 ď&#x20AC; (1 ď&#x20AC;Ť i ) ď&#x20AC; n

ď&#x192;Ś 0.07 ď&#x192;ś 99945.96ď&#x192;§ ď&#x192;ˇ 12 ď&#x192;¸ ď&#x192;¨ xď&#x20AC;˝ ď&#x20AC;39 ď&#x192;Ś 0.07 ď&#x192;ś 1 ď&#x20AC; ď&#x192;§1 ď&#x20AC;Ť ď&#x192;ˇ 12 ď&#x192;¸ ď&#x192;¨ x ď&#x20AC;˝ R 2872,74 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. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

1.2

After she has paid 45 installments she decided to settle the balance on the loan.

133 Grade 12 Core Mathematics

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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. ______________________________________________ _____________________________________________________ _____________________________________________________

2.3

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

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

3.2

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

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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 1 st time period.] i.e DELAYED PAYMENTS.





x (1  i) n1  1 Fv   (1  i) n2 i

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

Pv (1  i) n2 



x 1  (1  i)  n1 i OR

x

Pv (1  i ) (i ) n2

1  (1  i )  n1

EXAMPLES: 1.



Pv 



x 1  (1  i)  n1 i(1  i) n2



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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  i)  n1 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 Pv (1  i) 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 8 th year.





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

Exercise 3.4:

n1  88 n2  6

137 Grade 12 Core Mathematics 1.

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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

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?

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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? ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

FinalPayment on a loan:

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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

T37



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







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

FIXED PAYMENT ANNUITIES.

y T38

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

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



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



y



5000 1  (1.015)  n  y (1.015) 155 0.015

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



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

PV 





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



 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)

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.

Exercise 3.5:

2500

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________

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

143 Grade 12 Core Mathematics

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

Depreciation and Sinking Funds

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

1.4

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.

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ 2.

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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________

146 Grade 12 Core Mathematics 2.2

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

147 Grade 12 Core Mathematics 3.2

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________

148 Grade 12 Core Mathematics 4.

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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. ______________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ ______________________________________________

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

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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Grade 12 Core Mathematics

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

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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Grade 12 Core Mathematics

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

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

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

-12

-10

-8

-6

-4

-2

-4

-6

-8

155 GOMATH WORKBOOKS

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

156 GOMATH WORKBOOKS

Grade 12 Core Mathematics

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

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Grade 12 Core Mathematics

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

158 GOMATH WORKBOOKS

Grade 12 Core Mathematics

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

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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 .

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

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Grade 12 Core Mathematics

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

162 GOMATH WORKBOOKS

Grade 12 Core Mathematics

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

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Grade 12 Core Mathematics

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

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

164 GOMATH WORKBOOKS

Grade 12 Core Mathematics 10.

y  3x 2  2 x  1

-14

-12

-10

-8

-6

-4

-2

2 -2

-4

-6

-8

-10

-12

-14

165 GOMATH WORKBOOKS

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 2

p1  3  5 p1  2

New equation

y  2( x  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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______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ Vertical Shifting ( upwards and downwards) To shift vertically using the completed square form of the equation simply affect the “q” value of the turning point. 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 upwards. q  8 2

q1  8  5 q1  3

y  2( x  3)  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. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

168 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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Grade 12 Core Mathematics

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=ax 2 +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 – intercepts (1;0) and (-3;0) passing through (-1;-4 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

x – intercepts (2;0) and (-4;0) passing through (3;-14) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

172 GOMATH WORKBOOKS

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.

-10

-5

-2

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Grade 12 Core Mathematics 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 . ________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

174 Grade 12 Core Mathematics

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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.

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

______________________________________________________________ ______________________________________________________________

1.8 Sketch the graph. 4

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

-12

1.9

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

176 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 .

-12

-10

-8

-6

-4

-2

-4

-6

177 Grade 12 Core Mathematics

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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. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

178 Grade 12 Core Mathematics 4.1

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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. ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

179 GOMATH WORKBOOKS

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. ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

181 Grade 12 Core Mathematics 5.5

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

182 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

-2

-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

-2

-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 x 1 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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Grade 12 Core Mathematics

Shifting the exponential graph: 1. 1.

Vertical Shifts: y  a x or y  2 x Original graph

y  ax

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 (2 x) 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

185 GOMATH WORKBOOKS

Grade 12 Core Mathematics Exercise 4.7:

Sketch the graph of y  3 x on a Cartesian plane.

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

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.

186 GOMATH WORKBOOKS

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

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

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.

187 GOMATH WORKBOOKS

Grade 12 Core Mathematics 2.

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.

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

188 GOMATH WORKBOOKS

Grade 12 Core Mathematics Exercise 4.8:

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

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.

189 GOMATH WORKBOOKS

Grade 12 Core Mathematics 2

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.

190 GOMATH WORKBOOKS

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.

191 GOMATH WORKBOOKS

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

General equation: y  Sketching the graphs:

Method 1. Table Method y

1. X Y

4 x

-4 -1

-2 -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

192 GOMATH WORKBOOKS

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)

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 as follows: x

y

4 x3

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

194 GOMATH WORKBOOKS

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

Straight line 1 – 1

y  ax  bx  c

Parabola

y   r 2  x2 y  ax

Circle

Function

1 – M Function

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

195 GOMATH WORKBOOKS

Grade 12 Core Mathematics 2.

where

fx = x2-2x-3 4

-5

gy = y2-2y-3 -2

-4

196 GOMATH WORKBOOKS

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 ______________________________________________________________ _______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

197 Grade 12 Core Mathematics

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

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

198 Grade 12 Core Mathematics

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

199 GOMATH WORKBOOKS

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

200 GOMATH WORKBOOKS

Grade 12 Core Mathematics 2.2

gx = -x2+2

-5

-2

2.3 hx = 2x 2

-5

-2

201 GOMATH WORKBOOKS

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

202 GOMATH WORKBOOKS

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

203 GOMATH WORKBOOKS

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

204 GOMATH WORKBOOKS

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 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

205 GOMATH WORKBOOKS

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

206 Grade 12 Core Mathematics

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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 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

207 GOMATH WORKBOOKS

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 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

208 GOMATH WORKBOOKS

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 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

209 GOMATH WORKBOOKS

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

210 Grade 12 Core Mathematics

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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 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Interpretation of Trigonometric Functions: Question 1: Sketched below are the functions of đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = cos(đ?&#x2018;Ľ + 30Â°) đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ ) = 2đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Ľ For đ?&#x2018;Ľđ?&#x153;&#x2013;[â&#x2C6;&#x2019;180Â°; 180Â°]

1.1

Write down the period of f. ______________________________________________________________ ______________________________________________________________

1.2

Give the new range of g if g undergoes a positive vertical shift of 1 unit. ______________________________________________________________ ______________________________________________________________

1.3

Write down thw new equation of f if it is shifted 30ď&#x201A;° horizontally to the right. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics Question 2: The graphs of đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = 3đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;Ľ for đ?&#x2018;Ľ â&#x2C6;&#x2C6; [â&#x2C6;&#x2019;180Â°; 90Â°]

2.1

đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) = sin(đ?&#x2018;Ľ â&#x2C6;&#x2019; 60Â°) is sketched below

Write down the range of f. ______________________________________________________________

2.2

`If đ??´(â&#x2C6;&#x2019;97,3 7Â° ; â&#x2C6;&#x2019;0,38) write down the coordinates of B. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.3

Write down the period of g(3x). ______________________________________________________________ ______________________________________________________________

2.4

Write down the value of x for which g(x) â&#x20AC;&#x201C; f(x) is a maximum. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Question 3: The graphs of the functions đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = tan(đ?&#x2018;Ľ â&#x2C6;&#x2019; 45Â°) đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) = đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;2đ?&#x2018;Ľ for đ?&#x2018;Ľ â&#x2C6;&#x2C6; [â&#x2C6;&#x2019;135Â°; 90Â°] is sketched below.

Write down the period of g. ______________________________________________________________ ______________________________________________________________

For which value(s) of x willf have an asymptote if đ?&#x2018;Ľ â&#x2C6;&#x2C6; [â&#x2C6;&#x2019;135Â°; 90Â°]? ______________________________________________________________ ______________________________________________________________

3.3

Write down the equation of k if k is the reflection of g in the x-axis. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Question 4: Sketched below are the graphs of the functions đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = tan(đ?&#x2018;Ľ â&#x2C6;&#x2019; 45Â°) đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ ) = 3đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Ľ For đ?&#x2018;Ľ â&#x2C6;&#x2C6; [â&#x2C6;&#x2019;180Â°; 180Â°]

4.1

Write down the equations of the asymptotes of đ?&#x2018;Ś = đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) đ?&#x2018;&#x201C;đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;Ľ â&#x2C6;&#x2C6; [â&#x2C6;&#x2019;90Â°; 180Â°] ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.2

Describe the transformation of the graph of f to h if â&#x201E;&#x17D;(đ?&#x2018;Ľ ) = tan(45Â° â&#x2C6;&#x2019; đ?&#x2018;Ľ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

4.3

The period of g is reduced to 180Â° and the amplitude and tghe y â&#x20AC;&#x201C; intercept remains the same. Write down the equation of the resulting function. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Question 5. Given : đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = tan(đ?&#x2018;Ľ â&#x2C6;&#x2019; 30Â°) 5.1 Sketch the graph of đ?&#x2018;Ś = đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) đ?&#x2018;&#x201C;đ?&#x2018;&#x153;đ?&#x2018;&#x; â&#x2C6;&#x2019; 90Â° â&#x2030;¤ đ?&#x2018;Ľ â&#x2030;¤ 90Â° on the grid below.

5.2

Write down the equation of an asymptote of f. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.3

Describe in words the transformation of f to g if đ?&#x2018;&#x201D;(đ?&#x2018;Ľ ) = tan(30Â° â&#x2C6;&#x2019; đ?&#x2018;Ľ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 12 Core Mathematics Question 6. Consider: đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ) = 2đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Ľ 6.1

6.2

Write down the range of â&#x201E;&#x17D; (đ?&#x2018;Ľ ) = 2đ?&#x2018;&#x201C;(đ?&#x2018;Ľ ). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.3

đ?&#x2018;Ľ

Write down the period of â&#x201E;&#x17D;(đ?&#x2018;Ľ ) = đ?&#x2018;&#x201C; ( ). 2

Give a value of đ?&#x153;&#x192; if đ?&#x2018;&#x201C; (đ?&#x2018;Ľ + đ?&#x153;&#x192;) = 2đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;Ľ. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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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 1 st 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.

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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 1 st ) 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 5.1: 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

______________________________________________________________ ______________________________________________________________ ______________________________________________________________

222 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)!

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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;&#x2DC;wordsâ&#x20AC;&#x2122; 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!

remaining (n â&#x20AC;&#x201C; r) are all different is d) e)

đ?&#x2018;&#x203A;! đ?&#x2018;&#x;!

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.

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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 1 st 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. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________

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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.

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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

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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

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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

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Grade 12 Core Mathematics

Exercise 5.3: 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: ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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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? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ____________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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