GOMATHMEMOGR12

Page 1

Compiled by Chez Nell

Grade 12 Core Mathematics

2

CONTENT:

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PAGE.

1.

Calculus

3

2.

Number Patterns

31

3.

Financial Maths

43

4.

Functions & Graphs

55

5.

Linear Programming

85

6.

Probability Theory

98

7.

Analytical Geometry

101

8.

Transformation Geometry

104

9.

Trigonometry

106

10. Data Handling

115

11. Circle Geometry

125

ďƒ“ Norma Nell 2011

Grade 12 Core Mathematics

3

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Memo: Grade12 : Calculus: Workbook: Exercise 1.1: Page 5 1. Find the average gradients for f ( x )  x 2 between the following points: 1.1 x = 3 and 5 y = 9 and 25 y AveM  x 25  9 AveM  2 AveM  8 1.2

x= 2 and 7 y = 4 and 49 y AveM  x 49  4 AveM  5 AveM  9

1.3

points ( -2;4) and (3;9) y AveM  x 94 AveM  5 AveM  1

2.

Find the average gradient on f ( x )  x 2  4 2.1 between points (1;5) and (4;20) y AveM  x 20  5 AveM  3 AveM  5 2.2

x= 3 and 8 y = 13 and 68 y AveM  x 68  13 AveM  5 AveM  11

Grade 12 Core Mathematics

4

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2.3

x= -1 and 5 y = 5 and 29 y AveM  x 29  5 . AveM  6 AveM  4

3. Find the average gradients between the following points for f ( x )  x 3 : 3.1 x = -3 and 1 y = -27 and 1 y AveM  x 1  (27) AveM  4 AveM  7 3.2 x = -5 and -2 y = -125 and -8 AveM 

y x

 8  ( 125) 3 117 AveM  3 AveM  39 AveM 

3.3 x = 2 and h = 3 points (2; 125) and (1;1) y AveM  x 125  8 AveM  3 AveM  39 4.

Find the average gradients between the following points for f ( x )  x 3  2 : 4.1 x = 2 and x+h = 4. y =6 and 62 y AveM  x 62  6 AveM  2 AveM  28

Grade 12 Core Mathematics

4,2

4.3

5

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x = 3 and x+h = 5 y = 25 and 123 y AveM  x 123  25 AveM  2 AveM  49 points ( -2;-10) and (3;25)

AveM 

y x

25  10 5 AveM  7 AveM 

Exercise 1.2: Page 8 1.1

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

= = = = =

f ( x  h)  f ( x ) h [( x  h) 2  3( x  h)  2]  [ x 2  3x  2] h 2 2 x  2 xh  h  3x  3h  2  x 2  3x  2 h 2 2 xh  h  3h h 2x + h + 3

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

x = 2 and 6

AveM  2 x  h  3 AveM  2(2)  4  3 AveM  11

1.2.2

x = 4 and –2

AveM  2 x  h  3 AveM  2(2)  6  3 AveM  5

Grade 12 Core Mathematics

6

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

f ( x  h)  f ( x ) h [2( x  h) 2  ( x  h)  1]  [2 x 2  x  1] = h 2 2 2 x  4 xh  2h  x  h  1  2 x 2  x  1 = h 2 4 xh  2h  h = h = 4x +2 h - 1 2.2 Use the equation above to find the average gradients between the following Average Gradient

=

points: 2.2.1

x = -5 and –2

AveM= 4x + 2h -1 = 4(-5) +2(3) -1 = -20 +6 -1 = -15 2.2.2

x = 3 and 7

AveM= 4x + 2h -1 = 4(3) +2(4) -1 = 12 +8 -1 = 19

3. Derive an equation that will help you to find the average gradients between any two given points on the following curves: 3 3.1 f ( x)  x

Average Gradient

= = = = =

f ( x  h)  f ( x ) h 3 ( x  h)  x 3 h 3 x  3x 2 h  3xh 2  h 3  x 3 h 2 2 3x h  3xh  h 3 h 3x2 +3x h – h2

Grade 12 Core Mathematics

7 f ( x)  x 3  x

3.2

Average Gradient

f ( x  h)  f ( x ) h [( x  h) 3  x  h]  [ x 3  x] h 3 2 x  3x h  3xh 2  h 3  x  h  x 3  x h 2 2 3 3x h  3xh  h  h h 3x2 +3x h +h2-1

= = = = =

3.3 Exercise 1.3: Page 12 1.

Limit Concept:

Find the following limits: 1.1 lim( x  5)  8 x 3

x 2  3x  2 x  2 x2 ( x  1)( x  2)  lim x  2 x2  lim x  1 lim

1.2

x  2

 1 3x 2  x  1 x  2 x 2  x  6 3  2

lim

1.3

1.4

 x2  9   lim  x 3 x  3    ( x  3)( x  3)   lim   x 3 x3    lim x  3 x 3

9

1.5

1.6

lim (3x  6)  0

x 2

lim

x 

5x  2 5 x 1

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Grade 12 Core Mathematics

8

Exercise 1.4 : Page 16: Use First Principles to differentiate the following:

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

1.

h 0

 2x

f ( x)  x 2  2 f ( x  h)  f ( x ) h 0 h 2 [( x  h)  2]  [ x 2  2]  lim h 0 h 2 x  2 xh  h 2  2  x 2  2  lim h 0 h  lim 2 x  h f ( x)  lim

2.

h 0

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

3.

h 0

 6x

4.

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Grade 12 Core Mathematics

9

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f ( x)  x 2  3 x  4 f ( x  h)  f ( x ) f ( x)  lim h 0 h 2 [( x  h)  ( x  h)  lim h 0 h 2 [( x  2 xh  h 2 )  3( x  h)  4]  [ x 2  3x  4]  lim h 0 h 2 2 x  2 xh  h  3x  3h  4  x 2  3x  4]  lim h 0 h  lim 2 x  h  3 h 0

 2x  3 Exercise 1.5: Page 19 Find the derivatives of the following expressions using standard rules.

1.

f ( x)  x 3  3 x 2  5 x  7 f ( x)  3x 2  6 x  5

f ( x)  ( x  3)(2 x  5) 2.

f ( x)  2 x 2  11x  15 f ( x)  4 x  11 ( x  4)( x 2  4 x  16) x4 2 f ( x)  x  4 x  16 f ( x)  2 x  4 f ( x) 

3.

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

4.

8 x 3  27 2x  3 (2 x  3)(4 x 2  6 x  9) f ( x)  2x  3 2 f ( x)  4 x  6 x  9 f ( x)  8 x  6 3 f ( x)  3 x f ( x )  3 x 3 f ( x) 

5.

6.

f ( x)  9 x 4

Grade 12 Core Mathematics

10

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f ( x)  33 x 1

7.

f ( x)  3 x 3 f ( x)  x

2 3

2x3  4 x2 f ( x)  2 x  4 x  2 f ( x) 

8.

f ( x)  2  8 x 3

1 15 x 5 x 5 f ( x)  15 1 f ( x )   6 3x f ( x) 

9.

10

f ( x)  5 x 4  6 x 3  2 x 2  7 x  20 f ( x)  20 x 3  18 x 2  4 x  7

Page 23: Exercise 1.6

In each of the following determine the equation of the tangent at the point indicated

1.1

f ( x)  2 x 2  x  3 f ( x)  4 x  1 f (1)  4(1)  1 m  3 y  mx  c 4  3(1)  c 1 c y  3 x  1

1.2

Grade 12 Core Mathematics

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y  x3  x2 f ( x)  3 x 2  2 x f (1)  3(1) 2  2(1) m5 y  5x  c 2  5(1)  c 3  c y  5x  3

1.3 y  x 2  2 x  3 point ( -2 ; y)

f ( x)  x 2  2 x  3 f ( x)  2 x  2 f (2)  2(2)  2 m  6 y  mx  c y  6 x  c sub(2;5) 5  6(2)  c 5  12  c c  7 y  6 x  7

f (2)   2)   2(2)  3  443 5 2

2.

f ( x)  x 2  2 x  3 f ( x)  2 x  2 m  2 y  mx  c y  2 x  c 3  2(2)  2 c  1

4x  2 y  4  0 y  2 x  2 m  2

f ( x)  2 x  2  2  2x  2 2 x  4 x  2 y  (2) 2  2(2)  3 y3 (2;3)

y  2 x  1

3..

f ( x)  6 x  8 m2 y  2x  c  11  2(1)  c c  9 y  2x  9

2  6x  8

1 y   x3 2 m2

6 x  6 x  1 y  3(1) 2  8(1)  6 y  11

Grade 12 Core Mathematics

12

Page 27:Exercise 1.7 1. f ( x)  3x 3  7 x 2  4 and (x –2) is a factor 3 -7 + 0 + 4 6 -2 -4 3 -1 -2 0 f ( x)  ( x  2)(3x 2  x  2) f ( x)  ( x  2)(3x  2)( x  1) 2

2.

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

1

-1 -22 + 40 -5 -5 30 -40 1 -6 +8 0 2 f ( x)  ( x  5)( x  6 x  8) f ( x)  ( x  5)( x  2)( x  4) 3.

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

-1

4.

4 + 0 -19 -15 -4 + 4 +15 4 -4 -15 0 f ( x)  ( x  1)(4 x 2  4 x  15) f ( x)  ( x  1)(2 x  3)(2 x  5)

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

3

1 -6 +3 1 -3

+11 -9 +2

-6 +6 0

f ( x)  ( x  3)( x 2  3x  2) f ( x)  ( x  3)( x  1)( x  2) 5.

f ( x )  x 3  2 x 2  5x  6

1 -2 1

and (x + 2) is a factor

-2 -5 +6 -2 8 -6 -4 +3 0

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

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Grade 12 Core Mathematics

6.

13

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

1 +3 -1 -3 -3 -3 0 +3 1 0 -1 0

f ( x)  ( x  3)( x 2  1) f ( x)  ( x  3)( x  1)( x  1) 7.

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

1

-4

+3 -6 -8 -4 +4 +8 1 -1 -2 0 f ( x)  x  4)( x 2  x  2) f ( x)  ( x  4)( x  2)( x  1)

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

x-5

1 -6 +3 5 +5 -5 1 -1 -2

+10 -10 0

f ( x)  ( x  5)( x 2  x  2) f ( x)  ( x  5)( x  2)( x  1) Page 29: Exercise 1.8

1.

f ( x)  x 3  x 2  x  1 Test (x+1) f (1)  1  1  1  1  0 1 +1 -1 -1 x+1 -1 -1 0 +1 1 0 -1 0 f ( x)  ( x  1)( x 2  1) f ( x)  ( x  1) 2 ( x  1)

2. f ( x)  x 3  2 x 2  9 x  18 Test(x +2) f (2)  8  8  18  18  0 1 +2 -9 -18 x+2 -2 -2 0 +18 1 0 -9 0

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

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Grade 12 Core Mathematics

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3. f ( x)  2 x 3  x 2  13x  6 test (x-2) f ( x)  16  4  26  6  0 2 +1 -13 +6 x-2 2 4 +10 -6 2 +5 -3 0

f ( x)  ( x  2)(2 x 2  5 x  3) f ( x)  ( x  2)(2 x  1)( x  3)

4. f ( x)  2 x 3  5x 2  23x  10 test (x+2) f (2)  16  20  46  10  0 2 -5 -23 -10 x+2 -2 -4 18 +10 2 -9 -5 0

f ( x)  ( x  2)(2 x 2  9 x  5) f ( x)  ( x  2)(2 x  1)( x  5) 5. f ( x)  4 x 3  8 x 2  x  2 test(x-2) f (2)  32  32  2  2  0 4 -8 -1 +2 x-2 2 8 0 -2 4 0 -1 0

f ( x)  ( x  2)(4 x 2  1) f ( x)  ( x  2)(2 x  1)(2 x  1) Exercise 1.9 ; Page 37: Curve Sketching 1.

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

f x   3x 2  6 x

f x   6 x  6

Roots:

x  1 x2

y-intecept @ y  4

Grade 12 Core Mathematics

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Stationary points @ f x   0

3x 2  6 x  0 3x( x  2)  0 x = 0 or x = 2 y = 4 or y = 0 TP’s (0:4) & (2:0) Test for max/min @

f x   6 x  6

f 0  60  6  6 max tp

f 2  62  6  6

min tp y - axis (0 ; 4

f ( x)  ( x  1)( x  2) 2

x - axis

(2 ; 0)

2. f ( x)  (2  x)( x  1) 2 f ( x)   x 3  3 x  2 f x   3x 2  3

Roots: x  1 / 2

f x   6 x

Max/min turning points @ f x   0

 3x 2  3  0  3( x  1)( x  1)  0 x = 1 or x = -1 y= 9 or y = 0 Test for max/min TP’s @ f x   6 x

f 1  61  6 max TP f  1  6(1)  6 min TP y ( 1; 9)

f ( x)  (2  x)( x  1) 2

( -1; 0 )

(2 ; 0 )

x

Grade 12 Core Mathematics

f ( x)  x 3  3x 2

3.

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f ( x)  3x 2  6 x f ( x)  6 x  6

Roots:

3x(x  2)  0 TP’s @ x  0 x  -2 y0 y4 f (0)  6 min tp f (2)  6 max tp.

x 2 ( x  3)  0 x  0 / 3

-2 ; 4

f ( x)  x 3  3 x 2

0

-3

f ( x)  6 x 2  x 3

4.

TP’s @

f ( x)  12 x  3x 2 f ( x)  12  6 x

roots:

x 2 (6  x )  0 x  0/6

12 x  3x 2  0 4 ; 32

3x(4  x)  0 x0 x4 0r y0 y  32

f ( x)  6 x 2  x 3

f (0)  12 min tp f (4)  12 max tp

0

6

Grade 12 Core Mathematics

f ( x)   x 3  6 x 2  9 x

5.

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17

f ( x)  3x 2  12 x  9 f ( x)  6 x  12

roots:

 x( x 2  6 x  9)  x( x  3) 2

 3x 2  12 x  9  0

TP’s @ x 2  4 x  3  0 ( x  1)( x  3)  0 x 1 x3 or y  4 y0 f (1)  6 min tp f (3)  6 max tp

f ( x)   x 3  6 x 2  9 x

3;0

0

1 ;- 4

f ( x)  x 3  3x  2

6.

f ( x)  3x 2  3 f ( x)  6 x

TP’s @ 3( x  1)( x  1)  0 x 1 x  1 or y  4 y0 f (1)  6 min tp f (1)  6 max tp

roots:

x  1 / 2 syntheticd ivision

-1 ; 0 2

f ( x)  x 3  3 x  2

1 ; -4

Grade 12 Core Mathematics

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18

7. f ( x )  ( x  1) 3  8  0 f ( x )  ( x  1)( x 2  2 x  1)  8 f ( x)   x 3  3 x 2  3 x  7 f ( x )  3 x 2  6 x  3 f ( x )  6 x  6

TP(‘s) at f ( x )  0 x2  2x  1  0 ( x  1) 2  0 x  1 y8 TP ( 1;8)

f ( x )  6 x  6 Test for Max/Min TP @ f ( 1)  0 Point of Inflection as f (x)  0 Using factors of difference of cubes x 3  y 3  ( x  y)( x 2  xy  y 2 ) x 3  y 3  ( x  y)( x 2  xy  y 2 )

f ( x)   ( x  1) 3  8  0 f ( x)  ( x  1) 3  8  0 f ( x)  [( x  1)  2][( x  1) 2  2( x  1)  4] x- intercepts @ f ( x)  [( x  1)  2][( x 2  2 x  1  2 x  2  4]

f ( x)  ( x  1)( x 2  4 x  7) x 2  4x  7

no solution

x  1 y (-1;8 )

(0;7)

Point of inflection

f(x)=-(x+1)3+8

(1;0)

x

Summary: the graph has a point of inflection at (-1;8) and has only one root(x-intercept). The fact that the graph has only one stationary point is normally a clue that a point of inflection is prevalent. Also the fact that f ( x )  0 .

Grade 12 Core Mathematics

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Page 40 Exercise 1.10: 1. 1.1

f ( x)  x 3  x 2  8x  12 Solve for x if

x 3  x 2  8x  12  0

x-2

-1

-8

+12

+2

+2

-12

+1

-6

0

1

2 1

( x  2)( x 2  x  6)  0 ( x  2)( x  2)( x  3)  0

x = 2 or -3 1.2

Find f (x) f ( x)  3 x 2  2 x  8

1.3

Determine the coordinates of the turning points of f. 3x 2  2 x  8  0 (3x  4)( x  2)  0

x = 2 or x   y = 0 or y  1.4

4 3

500 27

Draw a neat sketch of f and label it correctly.

y - axis (-1,3;18,5) 12

(-3;0)

(2 ; 0)

x - axis

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f ( x)  3 x 2  2 x 3

2 2.1

Solve for x if 3x 2  2 x3  0 đ?‘Ľ 2 3 − 2đ?‘Ľ = 0 3

đ?‘Ľ = 0 đ?‘œđ?‘&#x; đ?‘Ľ = 2 2.2

Find

f (x)

đ?‘“ ËŠ đ?‘Ľ = 6đ?‘Ľ − 6đ?‘Ľ 2 2.3

Determine the coordinates of the turning points of f. 6đ?‘Ľ − 6đ?‘Ľ 2 = 6đ?‘Ľ(1 − đ?‘Ľ) đ?‘Ľ = 0 đ?‘œđ?‘&#x; 1 đ?‘Ľ = 0 đ?‘œđ?‘&#x; 1 Turning points : 0 ; 0

2.4

��� (1 ; 1)

Draw a neat sketch of f and label it correctly.

y

(1 ; 1)

f x  = 3ďƒ—x 2-2ďƒ—x 3

(0 ;0)

f ( x)  x 3  6 x 2  9 x  4

3. 3.1

Solve for x if

x3  6x 2  9x  4  0

đ?‘Ľâˆ’1

1 −6 +9 −4 1 1

+1

-5

+4

-5

+4

0

đ?‘Ľ − 1 đ?‘Ľ 2 − 5đ?‘Ľ + 4 = 0 đ?‘Ľâˆ’1 đ?‘Ľâˆ’1 đ?‘Ľâˆ’4 = 0 đ?‘Ľ = 1 đ?‘œđ?‘&#x; 4 3.2

Find

f (x)

đ?‘“ ′ đ?‘Ľ = 3đ?‘Ľ 2 − 12đ?‘Ľ + 9

(1

1 2

; 0)

x

Grade 12 Core Mathematics

3.3

21

Determine the coordinates of the turning points of f. 3đ?‘Ľ 2 − 12đ?‘Ľ + 9 = 0

đ?‘Ľ 2 − 4đ?‘Ľ + 3 = 0 đ?‘Ľâˆ’1 đ?‘Ľâˆ’3 =0 đ?‘Ľ = 1 đ?‘œđ?‘&#x; đ?‘Ľ = 3 đ?‘Ś = 0 or đ?‘Ś = −4

3.4

Draw a neat sketch of f and label it correctly. y

f x  = x 3-6ďƒ—x 2+9ďƒ—x -4

(0 ;0)

(4 ; 0)

x

(3 ; -4)

4. 4.1

f ( x)  x 3  3 x 2 Solve for x if

x 3  3x 2  0

đ?‘Ľ2 đ?‘Ľ + 3 = 0 đ?‘Ľ = 0 đ?‘œđ?‘&#x; đ?‘Ľ = −3 4.2

Find

f (x)

� ′ � = 3� 2 + 6� 4.3

Determine the coordinates of the turning points of f. 3đ?‘Ľ 2 + 6đ?‘Ľ = 0

3đ?‘Ľ đ?‘Ľ + 2 = 0 đ?‘Ľ = 0 đ?‘œđ?‘&#x; đ?‘Ľ = −2 đ?‘Ś=0

đ?‘œđ?‘&#x; đ?‘Ś = 4

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4.4

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22

Draw a neat sketch of f and label it correctly. f x  = x 3+3ďƒ—x 2

(-2 ; 4) y

(0 ;0)

(-3 ; 0)

. 5. 1

x

f ( x)  2 x 3  3 x 2 Solve for x if

2 x 3  3x 2  0

đ?‘Ľ 2 2đ?‘Ľ − 3 = 0 đ?‘Ľ = 0 đ?‘œđ?‘&#x; đ?‘Ľ =

5.2

Find

3 2

f (x)

đ?‘“ ′ đ?‘Ľ = 6đ?‘Ľ 2 − 6đ?‘Ľ

5.3

Determine the coordinates of the turning points of f. 6đ?‘Ľ 2 − 6đ?‘Ľ = 0

6đ?‘Ľ đ?‘Ľ − 1 = 0 đ?‘Ľ = 0 đ?‘œđ?‘&#x; đ?‘Ľ = 1 đ?‘Ś = 0 đ?‘œđ?‘&#x; đ?‘Ś = −1 5.4

Draw a neat sketch of f and label it correctly. y

f x  = 2ďƒ—x 3-3ďƒ—x 2

(0 ;0)

x

1 (1

(1 ; -1)

2

; 0)

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Page 27: Exercise 1.11 1. The curve of y  ax 3  24 x  b has a local minimum point at (2 ; -17) Calculate: The values of a and b f ( x )  3ax 2  24

f ( 2)  3a 2  24 12a  24 2

a2 y  ax 3  24 x  b  17  (2)(2) 3  24(2)  b  17  16  48  b b  15 y  2 x 3  24 x  15

The co-ordinates of the maximum turning point on the curve. 6 x 2  24  0

6( x  2)( x  2)  0 x  2orx  2 y  17ory  47 TP’s (2;17) and (-2;47) 2.

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

f ( 0 )  2 2.2

m2 Where is f ( x ) increasing?

2 -2

f ( x ) increasing where  2  x  1

1

Grade 12 Core Mathematics

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24

f ( x )  ax 3  bx 2  cx .

3.

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

y=f’((x )

y

12

0

1

2

x

1.1 Prove that a = 2; b = -9 and c = 12. đ?‘Ś = đ?‘Ž đ?‘Ľ − đ?‘Ľ1 đ?‘Ľ − đ?‘Ľ2 đ?‘Ś=đ?‘Ž đ?‘Ľâˆ’1 đ?‘Ľâˆ’2 12 = đ?‘Ž 0 − 1 0 − 2 2đ?‘Ž = 12 đ?‘Ž=6 đ?‘“ ′ (đ?‘Ľ) = 6 đ?‘Ľ − 1 (đ?‘Ľ − 2) đ?‘“ ′ (đ?‘Ľ) = 6đ?‘Ľ 2 − 18đ?‘Ľ + 12 đ?‘“ đ?‘Ľ = 2đ?‘Ľ 3 − 9đ?‘Ľ 2 + 12đ?‘Ľ đ?‘Ž = 2 ; đ?‘? = −9 đ?‘Žđ?‘›đ?‘‘ đ?‘? = 12

3.

Sketched is the graph of f ( x )  x 3  4 x 2  11 x  30 , 3.1 Determine the coordinates of A and B.

f ( x )  3 x 2  8 x  11 3 x 2  8 x  11  0 ( 3 x  11)( x  1)  0 x

11 y  -14,81 3

or x  1 y  36 ďƒŚ 11 ďƒś Aďƒ§ ;14,81 ďƒˇ ďƒ¨ 3 ďƒ¸ B  1;36 

Grade 12 Core Mathematics

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3.2 conclude the turning points of g if g( x )  f ( x  2). (Phase shift)

 17  (1;36) and  ;14,81   3  3.3 Determine the average rate of change of the function f from A to B.

36  ( 14,81) 11 1 3 50,8 Average Rate of change=   4 ,6  10,89 3.4 Determine the equation of the tangent to the graph of f at x  1.

f (1)  3(1) 2  8(1)  11 m  16 y  16(1)  c 16  16  c c  32 y  16 x  32 3.5 Determine the x-coordinate of the point at which the tangent in 3.4 cuts the graph of f

again.

 16 x  32  x 3  4 x 2  11x  30 0  x 3  4 x 2  5x  2 0  ( x  1)( x  1)( x  2) x 1 or x2 The tangent cuts at x = 2 again. 3.6 Determine the values of k for which x 3  4 x 2  11 x  30  k will have only one real root.K k > 36 or k < -14,81 3.7 Determine the point(s) of inflection of f ( x)  6 x  8 3.8 6 x  8  0 4 286 x  ;y  3 27

 4 286  Point of inflection is at (  ;   3 27 

f.

Grade 12 Core Mathematics

26

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Page 58:Exercise 1.12 1. 1.1

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

l  30  2 x

b  30  2 x ht  x

Show that the volume V cm3 of the box in terms of x is given by the equation V = 4x3 – 120x2 + 900x volume(V )  l.b.h V  x(30  2 x)(30  2 x)

1.2

V  4 x 3  120 x 2  900 x 1.3 Determine the value of x for which V is a maximum. Max volume @ V ( x)  0 V ( x)  12 x 2  240 x  900 12 x 2  240 x  900  0 x 2  20 x  75  0 ( x  5)( x  15)  0 x  5 or x  15 Test for max volume @ V ( x)  24 x  240 V (5)  120 V (15)  120 For max volume x = 5

1.4 Calculate the maximum volume of the box. v  4(5) 3  120(5) 2  900(5)  2000 2.2. =  x 2  6x  x 2  4x  3 =  2 x 2  10 x  3 PQ(x) =  4 x  10  4 x  10 = 0 Max PQ @ 4x = 10 5 x= 2 2 5 5 y  2   10   3  2  2 25 y    25  3 2 y  9,5 Max PQ = 9,5 units PQ

3.1

Derive a formula to calculate the volume ( V = r 2 h ) V   ( x) 2 (40  x) V  40x 2  x 3

Grade 12 Core Mathematics

3.2

27

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Calculate the radius that will give a maximum volume.

V  40ď °x 2  ď °x 3 V ( x)  80ď °x  3ď °x 2 Max volume at V ( x)  0 80ď °x  3ď °x 2  0

ď °x(80  3x)  0 x = 0 or x 

80  26,7cm 3

Page 64: Exercise 1.13 1. The motion of a particle is given by: đ?‘ = 9 − 6đ?‘Ą + đ?‘Ą 2 . 1.1 Find s at đ?‘Ą = 0 ; 1 ; 2 ; 3. đ?‘ = 9 đ?‘Žđ?‘Ą đ?‘Ą = 0 đ?‘ = 4 đ?‘Žđ?‘Ą đ?‘Ą = 1 đ?‘ = 1 đ?‘Žđ?‘Ą đ?‘Ą = 2 đ?‘ = 0 đ?‘Žđ?‘Ą đ?‘Ą = 3 1.2

What is the particle doing during the interval (0 ; 3) for t ? The particle is moving downwards

1.3

Find s at đ?‘Ą = 4 ; 5 ; 6. đ?‘ = 1 đ?‘Žđ?‘Ą đ?‘Ą = 4 đ?‘ = 4 đ?‘Žđ?‘Ą đ?‘Ą = 5 đ?‘ = 9 đ?‘Žđ?‘Ą đ?‘Ą = 6

1.4 Describe the motion of the particle relative to 0 after đ?‘Ą = 3. The particle is moving upwards again 1.5

Find an expression for v in terms of t. đ?‘Ł = 2đ?‘Ą − 6

1.6 Find v at đ?‘Ą = 1 ; đ?‘Ą = 2. (v should be negative). What does this imply about the motion of the of the particle? (compare with 1.2) đ?‘Ł = −4 đ?‘Žđ?‘Ą đ?‘Ą = 1 đ?‘Ł = −2 đ?‘Žđ?‘Ą đ?‘Ą = 2 đ?‘–đ?‘Ą đ?‘ đ?‘˘đ?‘”đ?‘”đ?‘’đ?‘ đ?‘Ąđ?‘ đ?‘Ąđ?‘•đ?‘Žđ?‘Ą đ?‘Ąđ?‘•đ?‘’ đ?‘?đ?‘Žđ?‘&#x;đ?‘Ąđ?‘–đ?‘?đ?‘™đ?‘’ đ?‘–đ?‘ đ?‘šđ?‘–đ?‘Łđ?‘–đ?‘›đ?‘” đ?‘‘đ?‘œđ?‘¤đ?‘›đ?‘¤đ?‘Žđ?‘&#x;đ?‘‘đ?‘ 1,7 Find v at đ?‘Ą = 4 ; đ?‘Ą = 5. What do the positive values obtained imply about the motion of the particle ? đ?‘Ł = 2 đ?‘Žđ?‘Ą đ?‘Ą = 4 đ?‘Ł = 4 đ?‘Žđ?‘Ą đ?‘Ą = 5 It suggests that the particle is moving upwards again 2.

A cricket ball is thrown vertically up into the air. After x seconds, its height is y metres where đ?‘Ś = 50đ?‘Ľ − 5đ?‘Ľ 2 . Determine: 2.1 the velocity of the ball after 3 seconds đ?‘Ł = 50 − 10đ?‘Ľ đ?‘Ł = 20

Grade 12 Core Mathematics

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2.2

the maximum height reached by the ball. Particle is at rest when đ?‘Ł = 0 At rest when đ?‘Ľ = 5 đ?‘Ś = 50 5 − 5 5 2 = 125 Maximum height is 125 metres 2.3

2.4

the acceleration of the ball. đ?‘Ž = đ?‘Ł ′′ đ?‘Ľ = −10đ?‘š/đ?‘ −2

the total distance travelled by the ball when it returns to the ground. Total distance travelled is 250m

Page 66: Exercise 1.14

Question 1.

(2;9) and (5;51) ď „y 51  9 AveM    14 ď „x 5  2

Question 2. From First Principles determine f (x) of : f ( x  h)  f ( x ) f ( x)  lim h 0 h 4( x  h) 2  4 x 2  f ( x)  lim h 0 h 2 4 x  8 xh  4h 2  4 x 2 f ( x)  lim h 0 h f ( x)  lim 8 x  4h h 0

f ( x)  8 x Question 3.

3.1

3.2

Determine:

x 2  4 x  21 lim x3 x 3 ( x  7)( x  3)  lim x3 x 3  10

lim 2 x

2

 5x  3

x 1

limđ?‘Ľâ†’−1 2(−1)2 − 5 −1 + 3 = 10

f ( x)  4 x 2

Grade 12 Core Mathematics

lim x x 3

3.3

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x3 2  3x

( x  3) x 3 x ( x  3) 1  lim ( ) x x 3 1  3  lim

Question 4. 4.1

29

Determine using standard rules.

f (x) if

f ( x)  3x 3  2 x 2  3x 1  6 f ( x)  9 x 2  4 x  3x 2

y  (2 x  5)( x  3) 4.2

y  2 x 2  x  15 dy  4x  1 dx

2  3x 2  4 3 x f ( x )  2 x 3  6 x  4 f ( x) 

4.3

f ( x)  6 x  4  6

Question 5:

If g ( x)  3x 2

5.1

Determine g ( x)  6 x

5.2

Calculate the value of g (2)  6(2)  12 Explain what g (2) represents . It represents the gradient of the tangent at the point x = -2 Find the co-ordinates of the point on the curve of g where the gradient is equal to 6. 6  6 x 5.3

5.4

x  1 (2;3)

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

6.2

Find the intercepts on the axes. x3  6x 2  9x  0 x( x  3) 2  0 x = 0 or x = 3 Find the co-ordinates of the stationery points. TP’s @ f ( x)  0

3x 2  12 x  9  0 x 2  4x  3  0 ( x  1)( x  3)  0 x = 1 or x = 3 y = 4 or y = 0

6.3

Make a neat sketch of the graph. 6

(1 ; 4) 4

f(x)=x 3 -6x2 +9x

2

-10

-5

0

(3 ;0)

-2

-4

6.4

Find the equation of the tangent to the curve when x = 2 f ( x)  3x 2  12 x  9 f (2)  3(2) 2  12(2)  9 m  3

y  y1  m( x  x1 ) Pt(-3 ; 12)

y  12  3( x  2) y  3x  18

5

10

Grade 12 Core Mathematics

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Question 7: f ( x)  x 3  4 x 2  4 x

f ( x)  3 x 2  8 x  4 If f ( x)  0 (3 x  2)( x  2)  0 2 x or x  2 3

2

2

7.1 Increasing đ?‘Ľ < 3 7.2 Decreasing

2

3

2 3

or x  2

<đ?‘Ľ<2

Page 80: Exercise 2.1 1. Determine which term in the arithmetic sequence 3; 5; 7;‌‌is equal to 27. đ?‘‡đ?‘› = đ?‘Ž + đ?‘› − 1 đ?‘‘ 27 = 3 + đ?‘› − 1 2 27 = 3 + 2đ?‘› − 2 2đ?‘› = 26 đ?‘› = 13 27 is the 13th term. 2. In the sequence, 23; 16; 9;‌ 2.1 Determine the 13th term. đ?‘‡13 = 23 + 12(−7) đ?‘‡13 = −61 2.2 Which term in the sequence is -131? đ?‘‡đ?‘› = đ?‘Ž + đ?‘› − 1 đ?‘‘ −131 = 23 + đ?‘› − 1 (−7) −131 = 23 − 7đ?‘› + 7 7đ?‘› = 161 đ?‘› = 23 -131 is the 23rd term. 3. Given the arithmetic sequence: 2; 3½; 5;‌ 3.1 determine the 53rd term. đ?‘‡53 = 2 + 52 1,5 = 80

3.2

Which term in the sequence is 53?

53 = 2 + đ?‘› − 1 1,5 1,5đ?‘› = 52,5 đ?‘› = 35 53 is the 35th term

Grade 12 Core Mathematics

32

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4. Determine the 1st 3 terms in an arithmetic sequence with the 4th term equals 25 and the 11th term is 81. đ?‘‡11 = đ?‘Ž + 10đ?‘‘ = 81 đ?‘‡4 = đ?‘Ž + 3đ?‘‘ = 25 7đ?‘‘ = 56 đ?‘‘=8 ∴đ?‘Ž=1 AP: 1 ; 9 ; 17‌.. 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. đ?‘‡5 = đ?‘Ž + 4đ?‘‘ = 2 đ?‘†10 = 5 2đ?‘Ž + 9 đ?‘‘ = 30 đ?‘Ž = 2 − 4đ?‘‘ 30 = 10đ?‘Ž + 45đ?‘‘ 30 = 10 2 − 4đ?‘‘ + 45đ?‘‘ 30 = 20 − 40đ?‘‘ + 45đ?‘‘ 5đ?‘‘ = 10 đ?‘‘=2 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. đ?‘› 140 = 2 [10 + đ?‘› − 1 2] 280 = 10đ?‘› + 2đ?‘›2 − 2đ?‘› 2đ?‘›2 + 8đ?‘› − 280 = 0 đ?‘›2 + 4đ?‘› = 140 = 0 đ?‘› + 14 đ?‘› − 10 = 0 đ?‘› = 10 đ?‘œđ?‘&#x; − 14 đ?‘› = 10 Number of terms is 10. 7. Evaluate the sum of the series: 1 – 4 – 9 - ‌‌‌- 239. −239 = 1 + đ?‘› − 1 (−5) 5đ?‘› = 245 đ?‘› = 49 đ?‘†49 =

49 2

2 + 48(−5) = −5831

8. In an arithmetic series, the sum to n terms (Sn ) is equal to n2- 2n. Determine: 8.1 The sum to 8 terms. đ?‘†đ?‘› = đ?‘›2 − 2đ?‘› đ?‘†8 = 82 − 2 8 = 48 8.2

The eighth term. đ?‘†1 = −1 đ?‘‡1 = −1 đ?‘†2 = 0 đ?‘‡2 = 1 đ?‘†3 = 3 đ?‘‡3 = 3 đ?‘‘=2 đ?‘‡8 = −1 + 7 2 = 13

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. đ?‘˜ + 4 − đ?‘˜ − 1 = 4đ?‘˜ + 1 − đ?‘˜ − 4 3đ?‘˜ = 6 đ?‘˜=2 AP: 3 ; 6 ; 9 đ?‘†80 = 40 6 + 79 3

= 9720

10. The second term of an arithmetic progression is 4 and the sixteenth term is 25. Find the first term and the common difference 4=đ?‘Ž+đ?‘‘

25 = đ?‘Ž + 15đ?‘‘

đ?‘Ž =4−đ?‘‘ 25 = 4 − đ?‘‘ + 15đ?‘‘ 14đ?‘‘ = 21 3

đ?‘‘=2 đ?‘Ž = 4 − 1,5 = 2,5 11. In an arithmetic progression 23; 19; 15; ‌. 11.1 Determine the twelfth term. đ?‘‡12 = 23 + 11 −4 = −21 Which term in the sequence is –53?

11.2

−53 = 23 + đ?‘› − 1 (−4) 4đ?‘› = 80 đ?‘› = 20 12. If x+4; 3x – 1; 4x – 3 are the first three terms in an arithmetic progression determine: 12.1 the value of x 3đ?‘Ľ − 1 − đ?‘Ľ − 2 = 4đ?‘Ľ − 1 − 3đ?‘Ľ + 1 đ?‘Ľ=3 12.2

the first three terms of the sequence. �1 = 7 ; �2 = 8 ; �3 = 9

Geometric Sequences: Page 89:Exercise 2.2 1.

Calculate the tenth term of a sequence: 81; 27; 9; ‌ 27 3 đ?‘&#x; = 81 = 4 đ?‘Ž = 81 đ?‘‡10 = 81

1 9 3

1

= 243

Grade 12 Core Mathematics

34

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3 and the fourth is –12, determine: 2 the second and third terms if the sequence is arithmetic. 3 đ?‘‡4 = 2 + 3đ?‘‘ = −12 3đ?‘‘ = −13,5 đ?‘‘ = −4,5 đ?‘‡2 = −3 đ?‘Žđ?‘›đ?‘‘ đ?‘‡3 = −7,5

2. If the first term of a sequence is 2.1

2.2

The second and third terms if the sequence is geometric. 3 đ?‘‡4 = 2 đ?‘&#x; 3 = −12 đ?‘&#x; 3 = −8 đ?‘&#x; = −2 đ?‘‡2 = −3 đ?‘Žđ?‘›đ?‘‘ đ?‘‡3 = 6

3. Find the 10th term in a geometric progression where 1st term is 5 and the common ratio is 3. đ?‘‡10 = 5(3)9 = 98415 If the 7th term is 192 and the 2nd term is 6 find the geometric sequence. đ?‘‡7 = đ?‘Žđ?‘&#x; 6 = 192 đ?‘‡2 = đ?‘Žđ?‘&#x; = 6 đ?‘&#x; 5 = 32 đ?‘&#x;=2 đ?‘Ž=3 GP: 3 ; 6 ; 12;‌‌ 4.

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. 2đ?‘Ľ+2 4đ?‘Ľ+1 = 3đ?‘Ľâˆ’2 2đ?‘Ľ+2 (2đ?‘Ľ + 2)2 = 3đ?‘Ľ − 2 (4đ?‘Ľ + 1) 4đ?‘Ľ 2 + 8đ?‘Ľ + 4 = 12đ?‘Ľ 2 − 5đ?‘Ľ − 2 8đ?‘Ľ 2 − 13đ?‘Ľ − 6 = 0 8đ?‘Ľ + 3 đ?‘Ľ − 2 = 0 đ?‘Ľ=2 5.2

Determine the common ratio of the sequence đ??şđ?‘ƒ = 4 ; 6 ; 9 3 đ?‘&#x;=2

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Page 92: Exercise 2.3 1.

Evaluate: 8

1.1

ďƒĽ 3k  1 k 1

�1 = 2 ; �2 = 5 ; �3 = 8 �=3 8 �� = 2 4 + 7 3 = 336 10

ďƒĽ 5k  2

1.2

k 1

�10 = 5 14 + 9 5

= 295

8

1.3

ďƒĽď€˝ 4k  2 k 3

�3 = 10 �=4 �6 = 3 20 + 5 4

= 60

10

1.4

ďƒĽ 3(2) k 1

k 1

đ?‘‡1 = 3 ; đ?‘‡2 = 6 ; đ?‘&#x; = 2 đ?‘†10 =

3(210 −1) 1

= 3069

10

1.5

ďƒĽ 5(3) n1 n 1

đ?‘‡1 = 5 ; đ?‘‡2 = −15 ; đ?‘&#x; = −3 đ?‘†10 =

5[ −3 10 −1] −4

= −73810

Page 95 :Exercise 2.4 1.

In a GP with first 3 terms: 5k + 1 ; 2k + 2; k + 1‌‌.. Find the the value of k. 2đ?‘˜+2 5đ?‘˜+1

đ?‘˜+1

= 2đ?‘˜+2

(2đ?‘˜ + 2)2 = đ?‘˜ + 1 (5đ?‘˜ + 1) 4đ?‘˜ 2 + 8đ?‘˜ + 4 = 5đ?‘˜ 2 + 6đ?‘˜ + 1 đ?‘˜ 2 − 2đ?‘˜ − 3 = 0 đ?‘˜âˆ’3 đ?‘˜+1 =0 đ?’Œ = đ?&#x;‘ đ?‘œđ?‘&#x; đ?‘˜ = −1

Grade 12 Core Mathematics

2.

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36

Use a suitable formula to find which term in an Arithmetic Series –61 – 58 – 55 ---- is the first term to exceed 10. đ?‘Ž + đ?‘› − 1 đ?‘‘ > 10 −61 + đ?‘› − 1 3 > 10 −61 + 3đ?‘› − 3 > 10 3đ?‘› > 74 đ?‘› > 24,6 đ?‘› = 25 n

3.

Find the largest number for

ďƒĽ (2r  3) ďƒĄ 48

r 1

đ?‘Ž = −1 ; đ?‘‡2 = 1 ; đ?‘‘ = 2 đ?‘›

đ?‘†đ?‘› = 2 [2đ?‘Ž + (đ?‘› − 1đ?‘‘] 48 =

đ?‘› −2 + đ?‘› − 1 2 2

96 < −2đ?‘› + 2đ?‘›2 − 2đ?‘› 2đ?‘›2 − 4đ?‘› − 96 > 0 đ?‘›2 − 2đ?‘› − 48 > 0 đ?‘›âˆ’8 đ?‘›+6 >0 đ?‘›>8 đ?‘›=9 n

4.

If T2 = 8 and T6 = 24 determine n if

ďƒĽ Tk

 480

k 1

đ?‘›

�6 = � + 5� = 24

480 = 2 [8 + đ?‘› − 1 4]

�2 = � + � = 8

960 = 8đ?‘› + 4đ?‘›2 − 4đ?‘› 4đ?‘›2 + 4đ?‘› − 960 = 0

4đ?‘‘ = 16

đ?‘›2 + đ?‘› − 240 = 0

đ?‘‘=4

đ?‘› + 16 đ?‘› − 15 = 0 đ?‘› = 15 5.

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

đ?‘› 2

30 + đ?‘› − 1 (−2)

−72 = 30đ?‘› − 2đ?‘›2 + 2đ?‘› 2đ?‘›2 − 32đ?‘› − 72 = 0 đ?‘›2 − 16đ?‘› − 36 = 0 đ?‘› − 18 đ?‘› + 2 = 0 đ?‘› = 18

Grade 12 Core Mathematics

6.

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37

Given the sequence 23 ; 27 ; 31 ------Find: (a) The number in the sequence which will be greater than 5000. 46 + đ?‘› − 1 4 > 5000 46 + 4đ?‘› − 4 > 5000 4đ?‘› > 4958 đ?‘› > 1239,5 đ?‘› = 1240 (b) How many terms must be added in the sequence so that the sum is greater than 5000. đ?‘› 2

46 + đ?‘› − 1 4 > 5000 46đ?‘› + 4đ?‘›2 − 4đ?‘› > 10000 2đ?‘›2 + 21đ?‘› − 5000 > 0 đ?‘› > 45,2 đ?‘›=4

7.

If T3 = 8 and T8 =

1 in a GP Find: 4

7.1 The common ratio đ?‘‡8 = đ?‘Žđ?‘&#x; 7 =

1 4

đ?‘‡đ?‘› = đ?‘Žđ?‘&#x; 2 = 8 đ?‘&#x;5 =

1 32

1 5 đ?‘&#x; = 2 1 đ?‘&#x;= 2 5

7.2 The sum of the first 8 terms.

8.

18 32 2 − 1 đ?‘†8 = = 63,75 1 −2 The sum of the first n- terms of an arithmetic sequence is Sn = n2 + 4n. 8.1 Calculate the first 4 terms of the sequence.

�1 = 12 + 4 1 = 5

∴ đ?‘‡1 = 5

�2 = 12

∴ đ?‘‡2 = 7

�3 = 21

∴ đ?‘‡3 = 9

�4 = 32

∴ đ?‘‡4 = 11

Grade 12 Core Mathematics

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38

8.2 Write down the value of the 100th term. �100 = 5 + 99 2 = 203 9.

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. đ?‘†3 = 15 9.2 Determine a formula for the n –th term of this sequence. đ?‘‡đ?‘› = đ?‘Ž + đ?‘› − 1 đ?‘‘ đ?‘‡đ?‘› = 1 + đ?‘› − 1 5 = 5đ?‘› − 4

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. đ?‘Ľ − 3 − 2đ?‘Ľ + 4 = 8 − 2đ?‘Ľ − đ?‘Ľ + 3 2đ?‘Ľ = 10 đ?‘Ľ=5 đ?‘†20 = 10 12 + 19 −4

11.

= −640

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

7 2đ?‘Ž + 6đ?‘‘ = 126 2

252 = 14đ?‘Ž + 42đ?‘‘ đ?‘‡20 = đ?‘Ž + 19đ?‘‘ = 130 đ?‘Ž = 130 − 19đ?‘‘ 252 = 14 130 − 19đ?‘‘ + 42đ?‘‘ đ?‘‘=7 đ?‘Ž = −3 đ?‘‡10 = 60 12.

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.

đ?‘†12 = 6 2đ?‘Ž + 11đ?‘‘ = 186 12đ?‘Ž + 66đ?‘‘ = 186 đ?‘‡6 = đ?‘Ž + 5đ?‘‘ = 14 đ?‘Ž = 14 − 5đ?‘‘ 168 − 60đ?‘‘ + 66đ?‘‘ = 186 6đ?‘‘ = 18 đ?‘‘=3

∴ đ?‘Ž = −1 AP : -1 ; 2 ; 5‌‌..

Grade 12 Core Mathematics

13.

39

The sum of three consecutive terms of an AP is 18. Their product is 192. Calculate the numbers. đ?‘Ž + đ?‘Ž + đ?‘‘ + đ?‘Ž + 2đ?‘‘ = 18 đ?‘‘ =6−đ?‘Ž đ?‘Ž đ?‘Ž + đ?‘‘ đ?‘Ž + 2đ?‘‘ = 192 đ?‘Ž đ?‘Ž + 6 − đ?‘Ž đ?‘Ž + 12 − 2đ?‘Ž = 192 đ?‘Ž 6 12 − đ?‘Ž = 192 72đ?‘Ž − 6đ?‘Ž2 − 192 = 0 đ?‘Ž2 − 12đ?‘Ž + 32 = 0 đ?‘Žâˆ’4 đ?‘Žâˆ’8 =0 đ?‘Ž = 4 đ?‘œđ?‘&#x; đ?‘Ž = 8

14.

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Calculate the value the value of n if : n

14.1

ďƒĽ (4k  6)  240

k 1

đ?‘Ž = −2 ; đ?‘‘ = 4 đ?‘› −4 + đ?‘› − 1 4 = 240 2 −4đ?‘› + 4đ?‘›2 − 4đ?‘› = 480 đ?‘›2 − 2đ?‘› − 120 = 0 đ?‘› − 12 đ?‘› + 10 = 0 đ?‘› = 12 n

` 14.2

ďƒĽ 5(3 r 1 )ďƒą 605

r 1

đ?‘Ž=5 ; đ?‘&#x;=3 5 3đ?‘› − 1 = 605 2 3đ?‘› − 1 = 242 3đ?‘› = 35 đ?‘›=5

Grade 12 Core Mathematics

15.

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40

The 3rd term of a geometric series is

2 2 and the 8th term is  . Find the first 9 2187

term and the common ratio. 2 2187 2 đ?‘‡3 = đ?‘Žđ?‘&#x; 2 = 9 1 đ?‘&#x;5 = − 243

đ?‘‡8 = đ?‘Žđ?‘&#x; 7 = −

1 đ?‘&#x; = − 3 1 đ?‘&#x;=− 3

5

5

16.

The first term of an arithmetic series is 2 The sum of the 3rd and 11th term is 40. 16.1

Find n if the n-th term of the series is 212. đ?‘Ž=2

đ?‘Ž + 2đ?‘‘ + đ?‘Ž + 10đ?‘‘ = 40 2đ?‘Ž + 12đ?‘‘ = 40 4 + 12đ?‘‘ = 40 đ?‘‘=3 212 = 2 + đ?‘› − 1 3 3đ?‘› = 213 đ?‘› = 71

16.2

Determine the sum of the first 71 terms of the series. �71 = 7597

ďƒŚ1ďƒś Solve for n , the number of terms, if ďƒĽ 4ďƒ§ ďƒˇ k 1 ďƒ¨ 2 ďƒ¸ n

17.

k 1

 7 63 64

đ?‘Ž = 4;

đ?‘&#x;=

1 2

1đ?‘› 4 2 −1 511 = 1 64 −2 1đ?‘› 511 −1=− 2 512 1đ?‘› 1 = 2 512 1đ?‘› 19 = 2 2 đ?‘›=9

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41 ďƒŚ 1ďƒś ďƒĽ 27ďƒ§ďƒ¨ 3 ďƒˇďƒ¸ k 1 n

18.

Calculate the value of n if:

k 1

ďƒĄ 21

đ?‘Ž = 27 đ?‘&#x; = −

1 3

1đ?‘› 27 − 3 − 1 > 21 4 −3 1đ?‘› 28 −1=− 3 27

−

−

1đ?‘› 1 =− 3 3

3

đ?‘›=3 19.

The sum to infinity of a geometric series is 81 and the sum of the first 3 terms of this series is 57. Find the first term and the common ratio. đ?‘Ž

đ?‘†âˆž = 1−đ?‘&#x; = 81

�3 =

đ?‘Ž = 81(1 − đ?‘&#x;)

�3 =

đ?‘Ž(đ?‘&#x; đ?‘› −1) đ?‘&#x;−1

= 57

81(1−đ?‘&#x;)(đ?‘&#x; đ?‘› −1) đ?‘&#x; −1

= 57

57

đ?‘&#x; đ?‘› − 1 = − 81 đ?‘&#x;đ?‘› =

8 27 3

2 đ?‘&#x; = 3 2 đ?‘&#x;= 3 đ?‘›

đ?‘Ž = 27

20.

The sum of the first 3 terms of a geometric sequence of which the terms are 8 . If the first term is 1 , find : positive is 1 49

20.1

the common ratio. 57 1(đ?‘&#x; 3 − 1) = 49 đ?‘&#x;−1 57 đ?‘&#x;2 + đ?‘&#x; + 1 = 49

�3 =

49đ?‘&#x; 2 + 49đ?‘&#x; − 8 = 0 1

đ?‘&#x;=7 20.2

The sum to infinity. đ?‘†âˆ? =

1

7

1 7

1−

=6

Grade 12 Core Mathematics

21.

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42

Consider the infinite geometric series: ( x  2) 2  ( x  2) 3  ( x  2) 4  ........

21.1

Write down the common ratio in terms of x. đ?‘&#x; =đ?‘Ľâˆ’2

21.2

Determine the value(s) of x for which the series will converge. đ?‘&#x; =đ?‘Ľâˆ’2 −1 < đ?‘Ľ − 2 < 1 1<đ?‘Ľ<3

21.3

If the sum to infinity of this series is

đ?‘†âˆž =

x2 , calculate the value(s) of x. 3

đ?‘Ľâˆ’2 đ?‘Ľâˆ’2 = 3 1− đ?‘Ľâˆ’2

3đ?‘Ľ − 6 = −đ?‘Ľ 2 + 5đ?‘Ľ − 6 đ?‘Ľ 2 − 2đ?‘Ľ = 0 đ?‘Ľ = 0 đ?‘œđ?‘&#x; 2

Grade 12 Core Mathematics

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43

Page 118 :Exercise 3.1:

Future Value Formula:

1.

5000

5000

T0

T1

T2

T32 18% (quarterly)

x (1  i ) n1  1 i 33   0.18  5000 1    1 4    Fv  0.18 4 Fv  R363781,13 Fv 

2.

x

x

T

T

T

0

1

2

T12 25% (monthly) x

0

Fv = R1 000 000

Fv (i)

(1  i) n  1 0.25 1000000( ) 12 x 0.25 120 (1  ) 1 12 x  R1915,96 3.

x T0

x T1

T2

T60 13% (monthly) Fv = R250 000

x

Fv (i)

[(1  i) n1  1] 0.13 250000( ) 12 x 0.13 61 [(1  )  1] 12 x  R 2913,64

Grade 12 Core Mathematics

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44

4.

500 T0

T1

500 T2

T60 16% (monthly) Fv = R10 000

n   0.16  500 1    1 12    10000  0.16 12

n

10000 0.16  0.16    1   1 500 12 12   n

19  0.16   1   15  12  19 0.16 log  n log(1  ) 15 12

19 log 15 n 0.16 log(1  ) 12

12n  17,85months n  1,5 years

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

Grade 12 Core Mathematics

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45

5.

2000 T0

2000 T1

T2

T60 16% (half-yearly) Fv = R100 000

n 1   0.16  2000 1   1  2    100000  0.16 2 0.16 n1 4  (1  ) 1 2

0.16 n1 ) 2 0.16 (n  1) log(1  )  log 5 2 log 5 n 1 0.16 log(1  ) 2 n  1  20,09 5  (1 

n  19,9 19,9 2 n  9,96 yrs n

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

6.

5000

400

T0

T1

400

400

T2

T21 14% (half-yearly)

0.14 20   400[(1  )  1]  0.14 21  2 5000(1  )   0 . 14 2     2 Fv  R37101,01

Grade 12 Core Mathematics

7.

46

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Mr and Mrs Mosepele are newly – 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 repayments 

 0.085  300000   12 

 360    0.085  1   1    12     repayments  R 2306,74

7.2 repayments 

 0.085  300000   12 

 240    0.085  1   1    12     repayments  R 2603,47

7.3 7.3.1 7.3.2

total  2306,74  360  R830426,40 total  2603,47  240  R624832,80

Page 123: Exercise 3.2:

Present Value Formula

1. 0.18  40 ) ] 4 Pv  0.18 4 Pv = R36803,17 2000[1  (1 

2.

 0.186  100000  2   x  25  0.186  1  1   2   x = R10429,13 3. 0.18 36 ) ] 12 Pv  0.18 12 Pv = R55321,37 2000[1  (1 

Grade 12 Core Mathematics

47

4.

 0.14  1000000  12   x  240  0.14  1  1   12   x = R 12 435,21 5.

x

 0.08  500000   4 

0.08   1  1   4   x = R18 277,87

 40

6. 0.11 ) 12 x 0.11  48 i  (1  ) 12 x = R 2481,27 96000(

7. 7.1

 0.09  2500000  12   x 361  0.09  1   1 12   x = R 1354,67 7.2

0.09 361 )  1] 12 Fv  0.09 12 Fv  R 2500008,34 1354,67[(1 

 0.07  2500008,34  12   x  240  0.07  1  1   12   x = R19382,47

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Grade 12 Core Mathematics

48

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Page 128: Exercise 3.3:

0,12 ) 12 x 0,12  60 1  (1  ) 12 x  R 4270,93 192000(

1.

0.12 45   4270.93(1  )  1 0,12 45 12   192000(1  )   R59216.947 0.12 12 12 0,095 ) 12 x 0,095  240 1  (1  ) 12 x  R6338.49 680000(

2.1

2.2

 6338.49  240  R1521237,60

2.3 0.095 84   6338.49(1  )  1 0,095 84 12    R566659,43 680000(1  )  0.095 12 12

2.4 Total Amount paid = R532433,16 3. 3.1

Calculate his repayments if the duration of the loan is 5 years. 0,18 350000( ) 12 x 0,18  60 1  (1  ) 12 x  R8887,70

He decides to settle the loan after 3 years. Calculate the balance of the loan. 0.18 36   8887.70(1  )  1 0,18 36 12    R178024,21 Balance of Loan = 350000(1  )  0.18 12 12

Grade 12 Core Mathematics

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49

Page 133: Exercise 3.4:

1.

32

3 Gap

T0

T3

Pv (1  i) n2 

x 1  (1  1)  n1 i

T4

T36

Pv 

OR

Pv 

x 1  1  i  n1

i(1  i) n2

6000 1  1.093321

0.093(1.093) Pv  R 46538.67

3

2.

14

47 Gap

T0

T47 x

T48

T61

120454 (1.0125) 47 (0.0125)

1  (1.0125) 14 x  R16911.34

3.

12

19 T0

T19

Gap

T20

7000 (1.06)12  1 Fv   (1.06)19 0.06 Fv  R357291.80

T32

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50

4.

99

9

12

T9 T10

T0

T108

T120

0.16 99   2600(1  )  1 12    (1  0.16 ) 9 (1  0.16 )12 Fv  0.16 12 12 12 Fv  R698150,10

Page 138: Exercise 3.5: 1 1.1

the number of payments.

Pv  

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

 Pv (i )  log 1  x   n log(1  i )  4000(0.01025)  log 1   270   n log(1.01025)  n  16,15049835 n  16,15049835 There will be 16 payments of R270 and a final payment less than R270.

1.2

the final payment.

4000 T0

27 0

T1

27 0

27 0

T2

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

T16 T17

 270 1  (1.01025) 16 4000   0.01025  y (1.01025) 17 y  R 40,81077309

y

 

Grade 12 Core Mathematics

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51

Final Payment of R40,81

2.

ďƒŠ Pv (i ) ďƒš log ďƒŞ1  x ďƒşďƒť ďƒŤ n  log(1  i ) 0.16 ďƒš ďƒŠ ďƒŞ 9000( 12 ) ďƒş log ďƒŞ1  ďƒş 270 ďƒŞ ďƒş ďƒŞ ďƒŤ ďƒťďƒş n  0.16 log(1  ) 12  n  11.90590354 n  11.90590354

11 payments of R1200 and a final payment less than R1200 đ?‘Ś = đ?‘“đ?‘–đ?‘›đ?‘Žđ?‘™ đ?‘?đ?‘Žđ?‘Śđ?‘šđ?‘’đ?‘›đ?‘Ą

ď ›

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

ď ?

ď ›

ď ?

ďƒŠ1200 1  (1.08) 11 ďƒš 9000  ďƒŞ ďƒş 0.08 ďƒŞ ďƒşďƒť ďƒŤ y 12 (1.08) y  R1090,98 Final payment of R1090,98

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. 

ďƒŠ Pv (i ) ďƒš log ďƒŞ1  x ďƒşďƒť ďƒŤ n log(1  i ) ďƒŠ 8000(0.039) ďƒš log ďƒŞ1  ďƒş 650 ďƒŤ ďƒť n log(1.039)  n  17.09222374 n  17.09222374 17 payments of R650 plus a final one less than R650

Grade 12 Core Mathematics

3.2

52

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

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 650 1  (1.039) 17 8000   0.039  y (1.039) 18 y  R60,99 Final Payment of R60,99

 

Page 140: Exercise 3.6 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. 1.1

Find the scrap value of the old press. ScrapV  340000(1  0.1)8  R146358.85

1.2

Find the cost of a new press in 8 years time. NewV  340000(1  0.15)8  R1040067.77

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. R893708.92

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. 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. 0.125 893708.92( ) 12 Payment   R5462.48 0.125 96 (1  ) 1 12

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 2.2

Find the scrap value of the old machine in 6 years time. ScrapV  450000(1  0.18) 6  R136803.00 Find the cost of a new machine in 6 years time. NewV  450000(1  0.12) 6  R888220.21

Grade 12 Core Mathematics

2.3

53

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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. 0.095 751417.21( ) 12  R7783.20 Payment  0.095 72 (1  ) 1 12

3

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. ScrapV  300000(1  0,15) 5  R133111,59 3.2 Find the cost of a new vehicle in 5 years time. NewV  300000(1  0.12) 5  R528702,51 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. R395 590,92 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. 0.125 395590.92( ) 12 Payment   R 4779,24 0.125 60 (1  ) 1 12

4.

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. ScrapV  450000(1  0,10) 6  R239148,45 4.2 Find the cost of a new vehicle in 6 years time. NewV  450000(1  0.12) 6  R797202,45 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. R558054,00

Grade 12 Core Mathematics

54

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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. 0.145 558054( ) 12 Payment   R 4905,90 0.145 72 (1  ) 1 12

5.

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 5.2 5.3

5.4

Calculate the scrap value of the harvester in 6 years time. ScrapV  2500000(1  0,10) 6  R1328602,50 Find the cost of a new harvester in 6 years time. NewV  2500000(1  0.12) 5  R4934556,71 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. R3605954,21 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.

0.155 ) 12 Payment   R30653,57 0.155 72 (1  ) 1 12 3605954.21(

Grade 12 Core Mathematics

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55

Page 146: Exercise 1: Sketch the graphs of the functions below: 1. y  x 2  3x  4 đ?‘Ľâˆ’4 đ?‘Ľ+1 =0 đ?‘Ľ = 4 đ?‘œđ?‘&#x; − 1 đ?‘Śđ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘?đ?‘Ą = −4 3

1

TP(2 ; −6 4) 4

f x  = x 2-3ďƒ—x-4

2

-10

-5

5

10

-2

-4

-6

y  x 2  4x  5

2.

đ?‘Ľâˆ’5 đ?‘Ľ+1 =0 đ?‘Ľ = 5 đ?‘œđ?‘&#x; − 1 đ?‘Ś − đ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘?đ?‘Ą = −5 Turning point = (2 ; −9) gx  = x 2-4ďƒ—x-5

-10

-5

5

-2

-4

-6

-8

10

Grade 12 Core Mathematics

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56

y  x2  x  6 𝑥+3 𝑥−2 =0 𝑥 = 2 𝑜𝑟 − 3

3.

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −6

-10

-5

5

10

-2

y = x2 +3x-6 -4

-6

-8

y  x 2  3x  10 𝑥+5 𝑥−2 =0 𝑥 = −5 𝑜𝑟 2

4.

𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −10 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = −

-10

-5

5

-2

y = x2 +3x-10 -4

-6

-8

-10

-12

-14

3 1 ; −12 2 4

10

15

20

Grade 12 Core Mathematics

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57

5. y  x 2  2x  8 𝑥−4 𝑥+2 =0 𝑥 = 4 𝑜𝑟 − 2 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −8 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = (1 ; −9) 2

-10

-5

5

10

15

20

10

15

20

-2

y = x2 -2x-8 -4

-6

-8

-10

-12

6. y  x 2  4 x  12 𝑥+6 𝑥 =2 =0 𝑥 = −6 𝑜𝑟 2 𝑦 = 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −12 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = (−2 ; −16) -10

-5

5

-2

y = x2 +4x-12 -4

-6

-8

-10

-12

-14

-16

Grade 12 Core Mathematics

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58

7. y  2x 2  7 x  6 2𝑥 − 3 𝑥 − 2 = 0 3 𝑥 = 2 𝑜𝑟 2 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 6 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 =

7 4

1

; −8 12

10

8

6

4

2

f x  = 2x 2-7x +6

-10

-5

5

10

15

20

-2

8. y  2 x 2  5x  3 2𝑥 − 1 𝑥 + 3 = 0 1 𝑥 = 2 𝑜𝑟 − 3 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −3 5

1

𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = − 4 ; −6 8 2

-10

-5

5

-2

-4

-6

-8

y = 2x2 +5x--3

10

Grade 12 Core Mathematics

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59

9. y  x 2  6x  7 đ?‘Ľ+7 đ?‘Ľâˆ’1 =0 đ?‘Ľ = −7 đ?‘œđ?‘&#x; 1 đ?‘Ś − đ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘?đ?‘’đ?‘?đ?‘Ą = −7 đ?‘‡đ?‘˘đ?‘&#x;đ?‘›đ?‘–đ?‘›đ?‘” đ?‘?đ?‘œđ?‘–đ?‘›đ?‘Ą = (−3 ; −16) -10

-5

5

-2

10

15

20

y = x2 +6x--7

-4

-6

-8

-10

-12

-14

-16

Page 155: Exercise 4.2 1. y  x 2  6 x  9 . a) đ?‘Ś = (đ?‘Ľ + 3)2 b) (-3 ; 0) c) đ?‘Ľ = −3 đ?‘Ś − đ?‘–đ?‘›đ?‘Ą = 9 d) 12

10

y = x2 +6x+9 8

6

4

2

-10

-5

5

-2

10

15

20

Grade 12 Core Mathematics

2. y  (2 x  1)( x  1) 𝑦 = −2𝑥 2 − 𝑥 + 1 1 1 a) 𝑦 = −2[𝑥 2 + 2 𝑥 − 2]

1 2

1

𝑦 = −2 𝑥 2 + 2 𝑥 + 𝑦 = −2

1 2

𝑥+4

1 2

𝑦 = −2 𝑥 + 4 b)

1

−4 ; 1

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60

4

1 2

−

4

1

−2

9

− 16 9

+8

9 8

c) 𝑥 = 2 𝑜𝑟 = 1

𝑦 − 𝑖𝑛𝑡 = 1 6

4

2

-10

-5

5

10

15

20

-2

y =-2 x2 -x+1 -4

-6

-8

-10

3. y  x 2  2 x  3 a) 𝑦 = 𝑥 − 1 2 − 4 b) (1 ; −4)

𝑦𝑖𝑛𝑡 = −3

𝑐) 𝑥 − 1 = ±2 𝑥 = 3 𝑜𝑟 − 1 d)

8

6

4

2

y = x2 -2x-3

-10

-5

5

-2

-4

-6

-8

10

15

20

Grade 12 Core Mathematics

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61

4. y  2 x 2  4 x  6 a) đ?‘Ś = 2[đ?‘Ľ 2 − 2đ?‘Ľ + 3] đ?‘Ś = 2[ đ?‘Ľ − 1 2 + 2] đ?‘Ś = 2(đ?‘Ľ − 1)2 + 4 b) (1 ; 4)

���� = 6

c) (đ?‘Ľ − 1)2 = 4 No x-intercepts

d) 10

8

6

f x  = 2ďƒ—x 2-4ďƒ—x +6

4

2

-10

-5

5. y   x 2  2 x  3 a) đ?‘Ś = −1[đ?‘Ľ 2 − 2đ?‘Ľ − 3] đ?‘Ś = −1[ đ?‘Ľ − 1

2

− 4]

đ?‘Ś = −(đ?‘Ľ − 1)2 + 4 b)

(1 ; 4)

c) đ?‘Ľ − 1 = Âą2 đ?‘Ľ = 3 đ?‘œđ?‘&#x; − 1

���� = 3

5

10

Grade 12 Core Mathematics

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62

d) 6

4

2

y = -x2 +2x+3

-10

-5

5

-2

-4

-6

-8

6.

y  3x 2  2 x  1

a)

đ?‘Ś = −3 đ?‘Ľ 2 + 3 đ?‘Ľ − 3

2

1 2

đ?‘Ś = −3

đ?‘Ľ+3

1 2

đ?‘Ś = −3 đ?‘Ľ + 3

b)

1

−3 ;

4 3

1 2

đ?‘?) đ?‘Ľ + 3

4

=9

1

2

đ?‘Ľ + 3 = Âą3 đ?‘Ľ=

1

−1Âą2 3

đ?‘Ľ = −1 đ?‘œđ?‘&#x;

1 3

4

−9 4

+3

10

15

20

Grade 12 Core Mathematics

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63

d) 6

4

2

-10

-5

5

-2

y = -3x2 -2x+1 -4

-6

-8

Page 161: Exercise 4.3: 1. đ?‘? = −3 đ?‘?1 = −3 − 4 = −7 đ?‘Ś = (đ?‘Ľ + 7)2 1

2. đ?‘? = − 4 1

1

đ?‘? = − 4 − 4 = −4 4 đ?‘Ś = −2 đ?‘Ľ +

17 2 4

9

+8

3. đ?‘? = 1 đ?‘?1 = 1 − 4 = −3 đ?‘Ś = (đ?‘Ľ + 3)2 − 4 4. đ?‘? = 1 đ?‘?1 = 1 − 4 = 3 đ?‘Ś = 2(đ?‘Ľ + 3)2 − 8 5. đ?‘? = 1 đ?‘?1 = 1 − 4 = −3 đ?‘Ś = −1(đ?‘Ľ + 3)2 + 4

10

15

20

Grade 12 Core Mathematics

64

Page 163: Exercise 4.4: 1. đ?‘ž = 0 đ?‘ž1 = 0 + 3 = 3 đ?‘Ś = (đ?‘Ľ + 3)2 + 3 9

2. đ?‘ž = 8

1

1

đ?‘ž1 = 1 8 + 3 = 4 8 1 2

đ?‘Ś = −2 đ?‘Ľ + 4

1

+ 48

3. đ?‘ž = −4 đ?‘ž1 = −4 + 3 = −1 đ?‘Ś = (đ?‘Ľ − 1)2 − 1 4. đ?‘ž = 4 đ?‘ž1 = 4 + 3 = 7 đ?‘Ś = 2(đ?‘Ľ − 1)2 + 7 5. đ?‘ž = 4 đ?‘ž =4+3=7 đ?‘Ś = −(đ?‘Ľ − 1)2 + 7 Page 166: Exercise 4.5: Find the equations of the following given: 1.

Turning Point (2;10) passing through (0 ;2) đ?‘Ś = đ?‘Ž(đ?‘Ľ − 2)2 + 10 2 = đ?‘Ž(0 − 2)2 + 10 2 = 4đ?‘Ž + 10 4đ?‘Ž = −8 đ?‘Ž = −2 đ?‘Ś = −2(đ?‘Ľ − 2)2 + 10

2.

Turning point ( -1;5) passing through (1;13) đ?‘Ś = đ?‘Ž(đ?‘Ľ + 1)2 + 5 13 = đ?‘Ž(1 + 1)2 + 5 13 = 4đ?‘Ž + 5 4đ?‘Ž = 8 đ?‘Ž=2 đ?‘Ś = 2(đ?‘Ľ + 1)2 + 5

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Grade 12 Core Mathematics

3.

65

Turning point ( -4;-1) passing through (-3;2) đ?‘Ś = đ?‘Ž(đ?‘Ľ + 4)2 − 1 2 = đ?‘Ž(−3 + 4)2 − 1 2=đ?‘Žâˆ’1 đ?‘Ž=3 đ?‘Ś = 3(đ?‘Ľ + 4)2 − 1

4.

x – intercepts (1;0) and (-3;0) passing through (-1;-4 đ?‘Ś = đ?‘Ž đ?‘Ľ − 1 (đ?‘Ľ + 3) −4 = đ?‘Ž −1 − 1 (−1 + 3) −4 = −4đ?‘Ž đ?‘Ž=1 đ?‘Ś = đ?‘Ľ − 1 (đ?‘Ľ + 3) đ?‘Ś = đ?‘Ľ 2 + 2đ?‘Ľ − 3

5.

x – intercepts (2;0) and (-4;0) passing through (3;-14) đ?‘Ś = đ?‘Ž đ?‘Ľ − 2 (đ?‘Ľ + 4) −14 = đ?‘Ž 3 − 2 (3 + 4) −14 = 7đ?‘Ž đ?‘Ž = −2 đ?‘Ś = −2 đ?‘Ľ − 2 (đ?‘Ľ + 4) đ?‘Ś = −2đ?‘Ľ 2 − 4đ?‘Ľ + 16

6.

x – intercepts (1;0) and 5;0) and y –intercept (0;-5) đ?‘Ś = đ?‘Ž đ?‘Ľ − 1 (đ?‘Ľ − 5) −5 = đ?‘Ž 0 − 1 (0 − 5) −5 = 5đ?‘Ž đ?‘Ž = −1 đ?‘Ś = −1 đ?‘Ľ − 1 (đ?‘Ľ − 5) đ?‘Ś = −đ?‘Ľ 2 + 6đ?‘Ľ − 5

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66

Page 168: Exercise 4.6 1.1

Sketch the graphs of y   x 2  x  12 and y  3x  12 on the same system of axes.. đ?‘Ľ 2 − đ?‘Ľ − 12 = 0

đ?‘Žđ?‘Ą đ?‘Ľ = 0 đ?‘Ąđ?‘•đ?‘’đ?‘› đ?‘Ś = 12

đ?‘Ľâˆ’4 đ?‘Ľ+3 =0

đ?‘Žđ?‘Ą đ?‘Ś = 0 đ?‘Ąđ?‘•đ?‘’đ?‘› đ?‘Ľ = 4

đ?‘Ľ = 4 đ?‘œđ?‘&#x; − 3 đ?‘Śđ?‘–đ?‘›đ?‘Ą = 12 đ?‘Ąđ?‘˘đ?‘&#x;đ?‘›đ?‘–đ?‘›đ?‘” đ?‘?đ?‘œđ?‘–đ?‘›đ?‘Ą =

1 2

1

; 12 4

12

10

8

6

y = -x2 +x+12 4

2

-10

-5

5

-2

1.2

10

15

y = -3x+12

Write down the co-ordinates of the points of intersection. 4 ; 0 ��� (0 ; 12)

1.3 Calculate the distance between the two graphs at x = -2 đ?‘“ −2 = −(−2)2 + −2 + 12 = 6 đ?‘“ −2 = −3 −2 + 12 = 18 đ?‘‘đ?‘–đ?‘ đ?‘Ąđ?‘Žđ?‘›đ?‘?đ?‘’ = 18 − 6 = 12

1.5

Write y  x 2  2 x  8 in the form of y  a( x  p) 2  q . đ?‘Ś = (đ?‘Ľ + 1)2 − 9

1.6

Write down the co-ordinates of the turning point. (−1 ; −9)

20

Grade 12 Core Mathematics

1.7

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67

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

(đ?‘Ľ + 1)2 = 9 đ?‘Ľ + 1 = Âą3 đ?‘Ľ = −1 Âą 3 đ?‘Ľ = −4 đ?‘œđ?‘&#x; 2 Write down the co-ordinates of the y – intercept.

1.8 (0 ; -8) 1.9

Sketch the graph.

4

2

y = x2 +2x-8

-10

-5

5

10

15

20

-2

-4

-6

-8

-10

1.10

Find the new equation if y  x 2  2 x  8 is moved 5 units to the left. đ?‘Ś = (đ?‘Ľ + 1)2 − 9 đ?‘?1 = −1 − 5 = −6 đ?‘Ś = (đ?‘Ľ + 6)2 − 9 đ?‘Ś = đ?‘Ľ 2 + 12đ?‘Ľ + 27

3.

Write y   x 2  4 x  5 in the form y  a( x  p) 2  q đ?‘Ś = −(đ?‘Ľ 2 − 4đ?‘Ľ − 5) đ?‘Ś = −(đ?‘Ľ − 2)2 + 9

Grade 12 Core Mathematics

3.1.

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68

y  x 2  4x  5 .

Sketch the graph of 8

6

4

f 2x  = x-5

y = -x2 +4x+5 2

-10

-5

5

10

15

20

-2

-4

-6

3.2.

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

3.3.

Write down the co-ordinates of the points of intersection of the two graphs. 5 ; 0 đ?‘Žđ?‘›đ?‘‘ (−2 ; −7)

3.4.

Find the equation if y   x 2  4 x  5 is moved 4 units downwards.

4.

đ?‘Ś = −(đ?‘Ľ − 2)2 − 13 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. đ?‘Ś = −2(đ?‘Ľ − 2)2 + 18 đ?‘Ś = −2(đ?‘Ľ + 1)2 + 18 đ?‘Ś = −2đ?‘Ľ 2 − 4đ?‘Ľ+16

4.1

4.2

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. đ?‘Ś = −2(đ?‘Ľ − 2)2 + 18 đ?‘Ś = −2(đ?‘Ľ − 2)2 + 24 đ?‘Ś = −2đ?‘Ľ 2 + 8đ?‘Ľ + 16 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. đ?‘Ś = −2(đ?‘Ľ − 2)2 + 18 đ?‘Ś = −2(đ?‘Ľ − 7)2 + 14 đ?‘Ś = −2đ?‘Ľ 2 + 28đ?‘Ľ − 84

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69

5. The graphs above are of f: y   x 2  8x  20 and g: y  mx  c . a.

Find the co-ordinates of A ; B ; H and D. đ?‘Ľ 2 − 8đ?‘Ľ − 20 = 0 đ?‘Ľ − 10 đ?‘Ľ + 2 = 0 đ?‘Ľ = 10 đ?‘œđ?‘&#x; − 2 đ??´ −2 ; 0 ; đ??ľ 10 ; 0 ; đ??ť 0 ; 20 ; đ??ˇ(4 ; 36)

b.

Write down the lengths of ; AO ; OH ; OB ; AB and SD đ??´đ?‘‚ = 2 ; đ?‘‚đ??ť = 20 ; đ?‘‚đ??ľ = 10 ; đ??´đ??ľ = 12 ; đ?‘†đ??ˇ = 36

c.

Calculate the length of AC. đ??´đ??ś 2 = 82 + 322 (đ?‘?đ?‘Śđ?‘Ąđ?‘•đ?‘Žđ?‘”đ?‘œđ?‘&#x;đ?‘Žđ?‘ ) đ??´đ??ś = 32,98

d.

Write down the equation of g. đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘? đ?‘Ś = 4đ?‘Ľ + đ?‘? 32 = 4(6) + đ?‘? đ?‘?=8 đ?‘Ś = 4đ?‘Ľ + 8

e.

EF = 16 units in length. Calculate the length of OG. 16 = −đ?‘Ľ 2 + 8đ?‘Ľ + 20 − 4đ?‘Ľ − 8 đ?‘Ľ 2 − 4đ?‘Ľ + 4 = 0 đ?‘Ľâˆ’2 đ?‘Ľâˆ’2 =0 đ?‘Ľ=2 đ?‘‚đ??ş = 2

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Page 181: Exercise 4.7 Sketch the graph of y  3 x on a Cartesian plane.

1.

8

6

4

2

-10

f x  = 3x

-5

5

10

Shift y  3 x by 2 units upwards and sketch this graph.

1.1

8

6

f x  = 3x+2 4

2

-10

1.2

-5

5

10

Shift y  3 x by 4 units downwards and sketch this graph.

4

2

-10

-5

f x  = 3x-4

5

-2

-4

10

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1.3

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Write down the equations of the asymptotes after the shifts in question 1.1 and 1.2 1.11 đ?‘Ś = 2 1.12 đ?‘Ś = −4 2. Sketch the graph of y  4  x on a Cartesian plane.

8

f x  = 4-x

6

4

2

-10

-5

5

10

Shift y  4  x by 3 units upwards and sketch this graph.

2.1

8

f x  = 4-x+3

6

4

2

-10

2.2

-5

5

10

Shift y  4  x by 4 units downwards and sketch this graph. 4

2

-10

-5

5

f x  = 4-x-4 -2

-4

10

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2.3

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Write down the equations of the asymptotes after the shifts in 2.1 & 2.2 2.1 đ?‘Ś=3 2.2 đ?‘Ś = −4

Page 184: Exercise 4.8 1. Sketch the graph of y  2 x on a Cartesian plane. 8

6

4

f x  = 2x

2

-10

-5

1.1

5

10

Shift y  2 x by 4 units to the left and sketch this graph. 8

6

4

f x  = 2x+4 2

-10

-5

5

10

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

8

6

4

f x  = 2x-4

2

-10

-5

5

10

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2. Write down the equations of the asymptotes after the shifts in question 1. đ?‘Ś=0 đ?‘Ś=0

1.1 1.2

3. Sketch the graph of y  4  x on a Cartesian plane.

8

6

f x  = 4-x

4

2

-10

-5

5

10

3.1 Shift y  4  x by 3 units to the left and sketch this graph.

8

6

f x  = 4-x+3 

4

2

-10

-5

5

10

3.2 Shift y  4  x by 4 units to the right and sketch this graph.

8

6

f x  = 4-x-4 4

2

-10

-5

5

10

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4. Write down the equations of the asymptotes after the shifts in 3. 3.1 3.2

đ?‘Ś=0 đ?‘Ś=0

5. Sketch the graph of y  2 x

8

6

f x  = 2x 4

2

-10

5.1

-5

5

10

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

8

6

f x  = 2x-4+3 4

2

-10

-5

5

10

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Page 192: 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 A. đ?’™ = đ?&#x;?đ?’š + đ?&#x;‘ đ?&#x;?đ?’š = đ?’™ − đ?&#x;‘ đ?&#x;?

đ?&#x;‘

đ?’š = đ?&#x;?đ?’™ − đ?&#x;? B. f x  = 2ďƒ—x+3 4

2

gx  =

-10

-5

 1 2

ďƒ—x-

3 2

5

10

-2

-4

1.2 A. đ?’™ = −đ?&#x;’đ?’š + đ?&#x;? −đ?&#x;’đ?’š = đ?’™ − đ?&#x;? đ?&#x;?

đ?&#x;?

đ?’š = −đ?&#x;’đ?’™ + đ?&#x;?

f x  = -4ďƒ—x+2 4

2

-10

-5

5

gx  = -2

-4

10

  -1 4

ďƒ—x+

1 2

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1.3 A. đ?’™ = đ?’šđ?&#x;? đ?’š=Âą đ?’™ B.

6

4

f x  = x 2

hy  = y 2

2

-10

-5

5

10

-2

-4

1.4 A. đ?’™ = đ?’šđ?&#x;? + đ?&#x;?đ?’š − đ?&#x;‘ đ?’šđ?&#x;? + đ?&#x;?đ?’š + đ?&#x;? = đ?’™ + đ?&#x;’ (đ?’š + đ?&#x;?)đ?&#x;? = đ?’™ + đ?&#x;’ đ?’š+đ?&#x;? =Âą đ?’™+đ?&#x;’ đ?’š = −đ?&#x;? Âą đ?’™ + đ?&#x;’

B. 6

4

f x  = x 2+2ďƒ—x -3

2

-10

-5

hy  = y 2+2ďƒ—y -3

5

-2

-4

10

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1.5 A. đ?’™ = đ?&#x;?đ?’š đ?’š = đ?’?đ?’?đ?’ˆđ?&#x;? đ?’™ B. 6

y = 2x 4

2

-10

y = log2 x

-5

5

10

-2

-4

1.6 A. đ?’™ = đ?&#x;?−đ?’š đ?’š = −đ?’?đ?’?đ?’ˆđ?&#x;? đ?’™ B. 6

4

y = 2-x

2

-10

-5

5

-2

-4

10

y = -log2 x

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78

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

-5

5

-2

The inverse of 2.1 is a function as it is a 1 to 1 function. i.e. the x-value is not repeated. 2.2

2

gx = -x2+2

-5

5

-2

The inverse of 2.2 is a non – function as it is a 1 to many function. i.e. the x –value is repeated. 2.3 hx = 2x 2

-5

5

-2

The inverse of 2.3 is a function as it is a 1 to 1 function. i.e. the x-value is not repeated.

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Page 198: 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 ďƒŽ [0 ď Ż ;360 ď Ż ] .

From the sketch find the following: 1.1 the period of y  sin x 360° 1.2 the range of y  cos x đ?‘Ś ∈ [−1 ; 1] 1.3 1

the amplitude of y  sin x

1.4 the value for x for sin x  cos x đ?‘Ľ = 45° đ?‘Žđ?‘›đ?‘‘ 225° 2. Sketch the graphs of y  2 sin x and y  cos x on the same set of axes for the interval x ďƒŽ [0 ď Ż ;360 ď Ż ] .

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From the sketch find the following: 2.1 the period of y  2 sin x 360° 2.2 the range of y  cos x đ?‘Ś ∈ [−1 ; 1] 2.3 2

the amplitude of y  2 sin x

2.4 the value for x for 2 sin x  cos x 27° đ?‘Žđ?‘›đ?‘‘ 207° 3. Sketch the graphs of y  sin 2 x and y  cos x on the same set of axes for the interval x ďƒŽ [0 ď Ż ;360 ď Ż ] .

From the sketch find the following: 3.1 3.2

3.3 1 3.4

the period of y  sin 2 x 180° the range of y  cos x đ?‘Ś ∈ [−1 ; 1] the amplitude of y  sin 2 x the value for x for sin 2 x  cos x đ?‘Ľ = 30° đ?‘Žđ?‘›đ?‘‘ 90° đ?‘Žđ?‘›đ?‘‘ 270°

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4. Sketch the graphs of y   sin x and y  cos 2 x on the same set of axes for the interval x ďƒŽ [0 ď Ż ;360 ď Ż ] .

From the sketch find the following: 4.1

the period of y   sin x

360° 4.2 the range of y  cos 2 x đ?‘Ś ∈ [−1 ; 1] 4.3 the amplitude of y   sin x

4.4

1 the value for x for  sin x  cos 2 x đ?‘Ľ = 90° đ?‘Žđ?‘›đ?‘‘ 210° đ?‘Žđ?‘›đ?‘‘ 330°

5. Sketch the graphs of y  sin x  1 and y  cos x on the same set of axes for the interval x ďƒŽ [0 ď Ż ;360 ď Ż ] .

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From the sketch find the following: 5.1 the period of y  sin x  1 360° 5.2 the range of y  cos x đ?‘Ś ∈ [−1 ; 1] 5.3 5.4

the amplitude of y  sin x  1 1 the value(s) for x for sin x  1  cos x 90° đ?‘Žđ?‘›đ?‘‘ 180°

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

From the sketch find the following: 6.1

the period of y  2 sin x 360°

6.2

the range of y  cos x  1

đ?‘Ś ∈ [0 ; 2] 6.3

the amplitude of y  2 sin x 2

6.4

the value(s) for x for 2 sin x  cos x  1

53° and 180°

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83

7. Sketch the graphs of y  sin( x  30 ď Ż ) and y  cos x on the same set of axes for the interval x ďƒŽ [0 ď Ż ;360 ď Ż ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30 ď Ż 3

2

1

gx  = cos x  -1

-2

From the sketch find the following: 7.1 the period of y  sin( x  30 ď Ż ) 360° 7.2 the range of y  cos x đ?‘Ś ∈ [−1 ; 1] 7.3

the amplitude of y  sin( x  30 ď Ż )

7.4

1 the value for x for sin( x  30ď Ż )  cos x

30°

and 210°

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

Grade 12 Core Mathematics

84

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3

2

1

gx  = cos 2ďƒ—x  -1

-2

From the sketch find the following: 8.1

the period of y   sin( x  30 ď Ż ) 360°

8.2 the range of y  cos 2 x đ?‘Ś ∈ [−1 ; 1] 8.3

the amplitude of y   sin( x  30 ď Ż ) 1

8.4

the value for x for  sin( x  30 ď Ż )  cos 2 x

đ?&#x;”đ?&#x;Ž° ; đ?&#x;?đ?&#x;Žđ?&#x;Ž° ; đ?&#x;?đ?&#x;?đ?&#x;Ž° ; 330°

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Page 212: Exercise 5.1: 1. Sketch the following inequalities on the grids supplied below. Use arrows to show which side of the line the solution should lie. 1.1

x y  6

8

6

4

2

0

5

10

15

1.2 3 y  2 x  24

8

6

4

2

0

5

10

15

Grade 12 Core Mathematics

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1.3 x  2 y  0

8

6

4

2

0

5

10

15

5

10

15

1.4 4  x  8

8

6

4

2

0

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87

1ď‚Ł y ď‚Ł 5

1.5

8

6

4

2

0

5

10

15

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

8

Feasible Region

6

4

2

0

5

10

15

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y _

90 _

B(0;40)

Feasible Region

Maximization Point

_ ;

45 _ 40 _

_C(10;35)

_D(30;15)

_40 _45

A(0;0) _

80 _

x _

2.1 Write down the inequalities represented in the graph above. Answer: x  y  45 ; x  2 y  80 ; 4 x  9 y  360 ; 2 y  x  0 x; y  0 2.2 Shade in the feasible region. 2.3 Given that the maximisation equation is M  6 x  4 y , Sketch the optimisation line on the graph above and ascertain which point (A;B;C or D) will maximise the objective function.

M  6x  4 y m 6  x 4 4 3 m 2

Answer: y 

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89

3.1 Represent the following inequalities on a set of axes:

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

16 14 12 10 8 6 4 2

0

2

4

6

8

10

12

14

16

Grade 12 Core Mathematics

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90

y

Line 1

Line 2

30 Line 5 Line 6 T y 20 p e 15

Line 3

B 10 Line 4

5

0

5

10

15

20

25

30

40

50

60

x

Type A

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

Write down these constraints as inequalities.

15  x  40 5  y  20 x  y  30 x  2 y  60 4.2

If the shopkeeper makes R2 profit on every article of A and R5 on every article of

B, write down an equation in terms of x and y which will represent the profit (P) that the shopkeeper makes.

P  2x  5 y 4.3

If the shopkeeper desires to maximise his profit, how many articles of each type

should he sell. 4.4 What is his maximum profit?

(20;20) P  2(20)  5(20)  R140

4.5 What is his minimum profit subject to the same constraints and how many articles of each are sold. (25;5) P  2(25)  5(5)  R75

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4.6 If the shopkeeper increases his prices and now makes R3 on each article of A and R6 on each of B. P  3x  6 y 4.6.1 How many of each type must he sell? (30;15) He will maximise his profit if he sells between 20 and 40 of type A and 5 and 20 of type B. I.e. Any combination between these points will maximise the profit. 4.6.2 What is his new profit? At point (30;15)

P  3(30)  6(15)  R180

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

1 1 hour of engine work and hour of body work is required to produce a 12 seater 2 2

bus. 

1 1 hour of engine work and hour of bodywork is required to produce a 16 seater bus 3 5

The ratio of 16 seater busses to 12 seater busses produced per week must be 3 : 2 A minimum of 200 12 seater busses must be produced per week. Let the number of 12 seaters be x Let the number of 16 seaters be y. If two of the constraints are x  200 and 3x  2 y  2160 5.1 Write down the remaining constraints in terms of x and y, from the information above. 1 1 x  y  480 2 5

 

y

3 x 2

 5x  2 y  4800

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5.2 Use the graph below to represent the constraints graphically.

2400

Feasible Region 16 S e a t e r s 1080

0

200

720

960

12 Se aters

5.3 Clearly indicate the feasible region by shading it. 5.4 If the profit on a 12 seater is R4000 and 16 seater is R12000 write down an equation that will represent the profit on minibuses.

P  4000 x  12000 y 5.5 If the aim is to maximise the profit : 5.5.1 How many of each type of bus is sold? and, 600 of the 12 seaters 900 of the 16 seaters. 5.5.2 What is the maximum profit?

P  4000(600)  12000(900) p  2400000  10800000 P  R13200000 [Hint let each square represent 100 units]

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Page 221: Exercise 5.2 1.

A toy factory produces wooden motor-cars and trains. The profit on a car is R12,00 and on a train is R8,00. The owner wants to maximize his profit but there are certain constraints that he must consider. Under contract he has to deliver at least 50 trains to the wholesaler. Market research has shown that he will not sell more than 120 motor-cars per week. He must not produce more than 2 trains for every motor-car to ensure that he does not waste wood.

If the following constraints in the form of inequalities are given y  50 ; x  120 ; x  0 and y  0 where x, y  Z .

y  2x Draw the graphs and determine the feasible region.

B

200 T R A 150 I N S 100

50

0

50 C

A

100 R S

150

200

How many motor-cars and how many trains must he produce in order to maximize his profit. P  12 x  8 y 3 m 2 Must produce 120 cars and 240 trains (Point B on graph) Determine the maximum profit. P  12 x  8 y

P  12(120)  8(240) P  R3360

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

If the following constraints in the form of inequalities are given 100 x  200 y  1200 ; 500 x  200 y  2000 ; x  0 and y  0 where x, y  Z . Write down the remaining constraint as an inequality. x y 5 Draw the graphs and determine the feasible region.

12

10

S 8

P O N G E

6

B 4

2

5

F

R

U

10

I

15

20

T

-2

Using the graphs, make a list of all possible combinations she could bake. No of Fruit 1 1 2 2 2 3 0 Cakes No of Sponge 4 5 3 4 5 2 5 Cakes

0 6

The profit on a fruit cake is R8 and on a sponge cake is R4. How many of each must she bake to ensure a maximum profit? P  8x  4 y

m  2 Max profit at point B(2 ;5) She must bake 2 fruit cakes and 5 sponge cakes.

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95

In the sketch , the shaded area , including the boundary lines, represents the feasible region of a set of inequalities. The equation of the objective function is given by P  mx  y . 3.1 Write down all the inequalities.

y  2 x  10 ; 2 y  x  8 ; x  6 and y  8 3.2

Determine the minimum value of P, if m = 0. At C: namely x = 6 and y = 1: thus P = 1

3.3

Determine the possible value(s) of m if P is minimized at D. 2m

1 2

4. A Certain industrial problem can be reduced to the following set of inequalities: x  10 ; y  8 ; 2 x  y  12 ; x  2 y  12 where x, y   4.1 Sketch the set of inequalities graphically in order to determine the feasible region 14

12

10

8

A

6

4

2

5

10

15

20

4.2 The objective function K  5x  2 y must be minimized. K  5x  2 y 5 m 2 4.2.1 Draw a line representing the objective function if K = 10. 4.2.2 Explain how you would minimize K. At point A(2 ; 8) 4.2.3 Write down the minimum value of K. Min value of K = 5(2) +2(8) = 26

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5 A factory has a contract to deliver at least 90 units of a certain piece of furniture per week. There may not be more than 18 employees. An artisan , who earns R600 per week, can produce 7 units per week. While an apprentice, who earns R300 per week, can only produce 4 units per week. The labour laws specify that at least one apprentice should be employed for every 5 artisans. The labour union , however , insists that the ratio of apprentices to artisans should not exceed 1: 2. 5.1 Let the number of artisans be x and the number of apprentices be y. 4 of the constraints are given by the following inequalities: x  0 ; y  0 ; 5 y  x and 7 y  4 x  90 . Write down two more constraints in terms of x and y. 2 y  x and y  x  18 5.2 Represent all the constraints graphically and indicate the feasible region. 24 22,5 22 20 18 16 14 12 10 8 6 B 4

(15 ; 3 )

2

0

2

4

6

8

10

12 12,8

16

18

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

L  600 x  300 y m  2

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97 5.3.1

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

Minimize the bill at point B(10 ; 5) 5.3.2

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

At point B(10 ;5) 10 artisans and 5 apprentices 6. 12

10

Feasible Region 8

Profit Line

Maximum profit point 6

4

(900 ; 300) 2

(600 ; 200)

5

10

15

20

-2

6.1

Extrapolate the set of inequalities that represent the above constraints.

x  y  800 ; x  y  1200 ; x  400 ; x  3 y 6.2

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

P  x

4 y 5

5 5P y   x Max profit at x = 900 and y = 300 4 4 5 m 4 6.3

Hence determine the maximum daily profit. Profit = 900 +

4 (300) =R1140 5

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Page 231: Exercise 6.1 1. Use your calculator to calculate the following: 1.1 8 = 40320 1.2

18 = 6,402373706× 1015

1.3

24 = 6,204484017× 1023

2. Write 15 in expanded mode instead of scientific notation as the calculators answer. 15! = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 3. Use the n Pr key on the calculator to calculate the product: 3.1 60 × 59 × 58 60 P3 = 205320 3.2

18 × 17 × 16 × 15 × 14 18 P5  1028160

4. Check if the answers in 3 are the same as follows: 60! 18! 4.1 4.2 (60−3)! (18−5)! P3  205320 18 P5  1028160 5. State whether the following statements are true or false, without using a calculator: 5.1 10 × 9! = 9! 60

False 5.2

5.3

20! 19!

= 20 True

15! 4!×3!

= 1! False

5.4

10! 6!×4!

= 210 True

5.5

3! × 20 = 5! True

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Page 235: 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. 6 5 4 3 = 6 P 3  120

1.2

repetition is allowed. 6 6 6 6 = 64 = 1296

2. How many ways can a captain and then a vice-captain be chosen from a rugby team of 15 members? 15 (15−2)

= 210

đ?‘‚đ?‘…

15

P 2  210

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 × 1 = 6 OR 3 = 6 3.2

3.3

NAIL. 4 Ă— 3 Ă— 2 Ă— 1 = 24 TIMBER 6! = 720

OR

4 = 24

[6 Ă— 5 Ă— 4 Ă— 3 Ă— 2 Ă— 1]

Page 240: Exercise 6.3 1. The Matric Dance Committee has decided on the menu below for the 2008 Matric Dance. A person attending the dance must choose only ONE item from each category, that is starters, main course and dessert.

STARTERS Crumbed Mushroom Garlic Bread Fish 1.1.1

DESSERT Ice-Cream Mulva Pudding

How many different meal combinations can be chosen? 3 Ă— 4 Ă— 2 = 24

1.1.2

1.2

MAIN COURSE Fried Chicken Beef Bolognaise Chicken Curry Vegetable Curry

A particular person wishes to have chicken as his main course. How many different meal combinations does he have? 3Ă—1Ă—2=6

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:

1.2.1

Any learner may be seated in any chair 6 Ă— 5 Ă— 4 Ă— 3 Ă— 2 Ă— 1 = 6! = 720

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Two particular learners wish to be seated next to each other 4 Ă— 3 Ă— 2 Ă— 1 = 4! = 24

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?

2.2

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

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PAPER TWO Page 7: Exercise 1.1 A:

Analytical Geometry.

Equations of circles of the form: (x  a) 2  (y  b) 2  r 2 1. Determine the equation of the circle with: 1.1 centre (2 ; 3) and radius 5 units. (đ?‘Ľ − đ?‘Ž)2 + (đ?‘Ś − đ?‘?)2 = đ?‘&#x; 2 (đ?‘Ľ − 2)2 + (đ?‘Ś − 2)2 = 25 1.2

Centre (4; -5) and radius 10 units (đ?‘Ľ − đ?‘Ž)2 + (đ?‘Ś − đ?‘?)2 = đ?‘&#x; 2 (đ?‘Ľ − 4)2 + (đ?‘Ś + 5)2 = 100

1.3

Centre (-2 ;-3) passing through (-2; 4) đ?‘&#x; 2 = (0)2 + (−3 − 4)2 đ?‘&#x; 2 = 49 (đ?‘Ľ + 2)2 + (đ?‘Ś + 3)2 = 49

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

đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘&#x;đ?‘’ 4 ; −7 & đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = 8.

2.2

đ?‘Ľ 2 + đ?‘Ś 2 − 6đ?‘Ś + (3)2 = 27 + 9 đ?‘Ľ 2 + (đ?‘Ś − 3)2 = 36 đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘&#x;đ?‘’ 0 ; 3 & đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = 6

2.3 đ?‘Ľ 2 + 4đ?‘Ľ + (2)2 + đ?‘Ś 2 = 5 + 4 . (đ?‘Ľ + 2)2 +đ?‘Ś 2 = 9 đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘&#x;đ?‘’ 2 ; 0 & đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = 3 2.4

đ?‘Ľ 2 − 4đ?‘Ľ + (2)2 + đ?‘Ś 2 + 2đ?‘Ś + (1)2 = 20 + 4 + 1 (đ?‘Ľ − 2)2 + (đ?‘Ś + 1)2 = 25 đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘&#x;đ?‘’ 2 ; −1 & đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = 5

2.5

đ?‘Ľ 2 + 6đ?‘Ľ + (3)2 + đ?‘Ś 2 − 4đ?‘Ś + (2)2 = 12 + 9 + 4

(đ?‘Ľ + 3)2 + (đ?‘Ś − 2)2 = 25 đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘&#x;đ?‘’ −3 ; 2 & đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = 5

2.6

đ?‘Ľ 2 + 4đ?‘Ľ + (2)2 + đ?‘Ś 2 + 6đ?‘Ś + (3)2 = 3 + 4 + 9

(đ?‘Ľ + 2)2 + (đ?‘Ś + 3)2 = 16 đ?‘?đ?‘’đ?‘›đ?‘Ąđ?‘&#x;đ?‘’ −2 ; −3 & đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = 4

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Grade 12 Core Mathematics

3.

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102

Find the equations of the tangents in the following: 3.1

to circle đ?‘Ľ 2 + đ?‘Ś 2 = 5 at the point (3 ; 4). ∆đ?‘Ś

4

đ?‘”đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘’đ?‘›đ?‘Ą đ?‘œđ?‘“ đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = ∆đ?‘Ľ = 3 3

đ?‘”đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘’đ?‘›đ?‘Ą đ?‘œđ?‘“ đ?‘Ąđ?‘Žđ?‘›đ?‘”đ?‘’đ?‘›đ?‘Ą = − 4 đ??¸đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘Ąđ?‘Žđ?‘›đ?‘”đ?‘’đ?‘›đ?‘Ą: đ?‘Ś − đ?‘Ś1 = đ?‘š đ?‘Ľ − đ?‘Ľ1 3

đ?‘Ś − 4 = −4 đ?‘Ľ −3 4

đ?‘Ś = 3đ?‘Ľ + 3.2

25 4

to circle đ?‘Ľ 2 + đ?‘Ś 2 = 36 at the point (−2 ; 3). ∆đ?‘Ś

3

đ?‘”đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘’đ?‘›đ?‘Ą đ?‘œđ?‘“ đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘˘đ?‘ = ∆đ?‘Ľ = − 2 2

đ?‘”đ?‘&#x;đ?‘Žđ?‘‘đ?‘–đ?‘’đ?‘›đ?‘Ą đ?‘œđ?‘“ đ?‘Ąđ?‘Žđ?‘›đ?‘”đ?‘’đ?‘›đ?‘Ą = 3 đ??¸đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘œđ?‘“ đ?‘Ąđ?‘Žđ?‘›đ?‘”đ?‘’đ?‘›đ?‘Ą: đ?‘Ś − đ?‘Ś1 = đ?‘š đ?‘Ľ − đ?‘Ľ1 2

đ?‘Śâˆ’3=3 đ?‘Ľ+2 2

đ?‘Ś = 3đ?‘Ľ +

13 3

3.3

to (đ?‘Ľ + 1)2 + đ?‘Ś 2 = 20 which is parallel to 2đ?‘Ś − đ?‘Ľ = 0

3.4

to (đ?‘Ľ − 2)2 + (đ?‘Ś + 3)2 = 16.

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

the length of AD. = 100  25  11.18

1.2

the mid-point of DC = (6;4 12 )

1.3

The gradient of BC

1.4

the inclination of BC =

=

ď „y 5 1   ď „x 10 2

tanď ą  0.5

ď ą  26,57

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103 y  12 x  c

1.5

the equation of BC

=

4 

1 2

 4  c

c  2 y  12 x  2

1.6

Prove that ABCD is a parallelogram.

AD  125 y 1 mAD   x 2 2.

BC  125 y 1 mBC   x 2

p3 p3  3 5  5 p  15  3 p  9 p  12

AB  ( x1  x 2 ) 2  ( y1  y 2 ) 2  (4  1) 2  (12  3) 2

3.1

 90  9,49

MidPtAC 

( x1  x 2 ) ( y1  y 2 ) ; 2 2

Mdpt (4;7)

3.3

AD // BC ABCD is a parm.

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

3.2

AD = BC

mAB = mBE

9 6  3 1 p 9  9 p  18 9p  9 p  1 3.4

1 3 tan A  0.333333

mBC  

A  161,6 

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3.5

cos BCˆ D 

10  20  50 2 10 20

cos BCˆ D  0,707 BCˆ D  135  3.6 9 3 3 1 mBC   3 mAB.mBC  1 mAB 

AB  BC Page 17: Exercise 2.1: Calculate the co-ordinates of Pˊ of the point ( 2 ; 4) after rotation about the origin, through an angle of: 1.

60° 𝑃′ (𝑥𝑐𝑜𝑠60° − 𝑦𝑠𝑖𝑛60°; 𝑦𝑐𝑜𝑠60° + 𝑥𝑠𝑖𝑛60°) 𝑃′ (1 − 2 3; 2 + 3)

2.

150° 𝑃′ (𝑥𝑐𝑜𝑠150° − 𝑦𝑠𝑖𝑛150°; 𝑦𝑐𝑜𝑠150° + 𝑥𝑠𝑖𝑛150°) 𝑃′ (−2𝑐𝑜𝑠30° − 4𝑠𝑖𝑛30°; −4𝑐𝑜𝑠30° + 2𝑠𝑖𝑛30°) 𝑃′ (−2 − 3; 1 − 2 3)

3.

210° 𝑃′ (𝑥𝑐𝑜𝑠210° − 𝑦𝑠𝑖𝑛210°; 𝑦𝑐𝑜𝑠210° + 𝑥𝑠𝑖𝑛210°) 𝑃′ (−2𝑐𝑜𝑠30° + 4𝑠𝑖𝑛30°; −4𝑐𝑜𝑠30° − 2𝑠𝑖𝑛30°) 𝑃′ (2 − 3; −1 − 2 3)

4.

−30° 𝑃′ (2𝑐𝑜𝑠(−30°) − 4𝑠𝑖𝑛(−30°); 4𝑐𝑜𝑠(−30°) + 2𝑠𝑖𝑛(−30°) 𝑃′ (2𝑐𝑜𝑠 30° + 4𝑠𝑖𝑛 30° ; 4𝑐𝑜𝑠 30° − 2𝑠𝑖𝑛(30°) 𝑃′ (2 + 3; ; −1 + 2 3)

5.

−225° 𝑃′ (𝑥𝑐𝑜𝑠(−225°) − 𝑦𝑠𝑖𝑛(−225°); 𝑦𝑐𝑜𝑠(−225°) + 𝑥𝑠𝑖𝑛(−225°) ′ 𝑃 (−2𝑐𝑜𝑠45° − 4𝑠𝑖𝑛45°; −4𝑐𝑜𝑠45° − 2𝑠𝑖𝑛45°) 𝑃′ (−3 2; − 2)

Grade 12 Core Mathematics

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105

𝑥ˊ = 𝑥𝑐𝑜𝑠𝜃 − 𝑦𝑠𝑖𝑛𝜃

6.

= 2𝑐𝑜𝑠75 − 4𝑠𝑖𝑛75 = 2 cos 30 + 45° − 4(sin30° + 45°) = 2𝑐𝑜𝑠30°𝑐𝑜𝑠45° − 2𝑠𝑖𝑛30°𝑠𝑖𝑛45° − 4𝑠𝑖𝑛30°𝑐𝑜𝑠45° − 4𝑠𝑖𝑛45°𝑐𝑜𝑠30° =

3

2

2

2

2 6

= −2 =−

2 2

4

3 2 2

− 6−3 2

= 7.

2

−

2

2

−6

4 6

1

−2

2

𝑥ˊ = 𝑥𝑐𝑜𝑠𝜃 − 𝑦𝑠𝑖𝑛𝜃 3𝑐𝑜𝑠𝜃 − 𝑠𝑖𝑛𝜃 =

3− 3 2

-----------A

𝑦ˊ = 𝑦𝑐𝑜𝑠𝜃 + 𝑥𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 + 3𝑠𝑖𝑛𝜃 =

1+3 3 2

𝑨 × 𝟑 𝒂𝒏𝒅 𝑩 × 𝟏 9𝑐𝑜𝑠𝜃 − 3𝑠𝑖𝑛𝜃 = 𝑐𝑜𝑠𝜃 + 3𝑠𝑖𝑛𝜃 =

9−3 3 2

1+3 3 2

𝑨 + 𝑩 10𝑐𝑜𝑠𝜃 = 5 𝑐𝑜𝑠𝜃 = 0,5 𝜽 = 𝟔𝟎°

8. 𝑥ˊ = 𝑥𝑐𝑜𝑠𝜃 − 𝑦𝑠𝑖𝑛𝜃 = 3𝑐𝑜𝑠120 − 2𝑠𝑖𝑛120 = 3 −𝑐𝑜𝑠60° − 2𝑠𝑖𝑛60° 1

= 3 −2 −2 3

= −2 − 3 =

−3−2 3 2

3 2

----- B

−4

1 2

2 2

−4

3 2

2 2

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đ?‘ŚËŠ = đ?‘Śđ?‘?đ?‘œđ?‘ đ?œƒ + đ?‘Ľđ?‘ đ?‘–đ?‘›đ?œƒ = 2đ?‘?đ?‘œđ?‘ 120ď‚° + 3đ?‘ đ?‘–đ?‘›120ď‚° = 2 −đ?‘?đ?‘œđ?‘ 60° + 3đ?‘ đ?‘–đ?‘›60° 1

= 2 −2 +3 = −1 + =

3 2

3 3 2

−2+3 3 2

Trigonometry: Page 23: Exercise 3.1 1.

sin 2 x cos x  sin x cos 2 x  sin(2 x  x)  sin 3x 2. sin( x  90)  sin x cos 90  sin 90 cos x

 Sinx (0)  1(cos x)  cos x 3. sin 50 cos10  sin 10 cos 50

 sin(50  10)  sin 60 

3 2

Exercise: Use the compound angle formulae to simplify each expression to one term only: cos 3x. cos 2x  sin 3x. sin 2x  cos 5x . 1.1 1.2

sin 3x. cos 2x  cos 3x.sin 2x  sin 5x .

1.3

cos 5x. cos 2x  sin 5x sin 2x  cos 3x .

1.4

sin 3x. sin 2x  cos 3x. cos 2x  cos x .

1.5

sin 2x.sin x  cos 2x. cos x   cos 3x .

1.6

sin 2x. cos x  cos 2x. sin x  sin x .

1.7

sin 50ď Ż. cos 10ď Ż  cos 50ď Ż. sin 10ď Ż  sin 40 .

1.8

sin 81ď Ż. cos 23ď Ż  sin 23ď Ż . cos 81ď Ż  sin 104 .

1.9

cos18ď Ż.sin 31ď Ż  sin18ď Ż cos 31ď Ż  sin 49

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107

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

cos(2 x  10 )  cos 2 x.cos10  sin 2 x.sin10 .

2.3

sin(a  2b)  sin a cos 2b  sin 2b cos a .

2.4

cos(a  2b)  cos a cos 2b  sin a sin 2b .

2.5

sin(2a  20 )  sin 2a cos 20  sin 20 cos 2a .

2.6

cos(a  30 )  cos a cos 30  sin a sin 30 .

Evaluate the following without a calculator: 3.1

sin 40. cos 20  cos 40.sin 20  sin 60 

3 2

3.2

cos 40.cos 20  sin 40.sin 20  cos 60 

1 2

cos100 cos 280  sin 100.sin 280 3.3

  sin 10 sin 10  cos10 cos10   cos(10  10)  1 sin 80.sin 40  sin 10.sin 50  cos10 cos 50  sin 10 sin 50

3.4

 cos 60 

1 2

Prove the following using compound angle formula: 4.1

sin(180   )   sin  LHS  (sin 180. cos   sin  . cos180)  0  sin    sin   RHS 4.2

cos(360   )  cos  LHS  cos 360 cos   sin 360 sin   cos   0  cos   RHS

Grade 12 Core Mathematics

108

5.1 3 sin( x  60 )  sin( x  30 )  cos x LHS  3 sin x cos 60  sin 60 cos x   (sin x cos 30  sin 30 cos x   1 3 3 1  3  sin x    cos x    cos x   cos x 2 2 2   2 3 3 3 1 sin x  cos x  cos x  cos x 2 2 2 2  cos x 

 RHS

5.2

cos(a  60 )  cos(a  60)  cos a

LHS  cos a cos 60  sin a sin 60  cos a cos 60  sin a sin 60) 1 3 1 3  cos x  sin x  cos a  sin a 2 2 2 2  cos a  RHS

5.3

sin(a  30 )  sin(a  30 )  cos a

LHS  sin a cos 30  sin 30 cos a  sin a cos 30  sin 30 cos a 3 1 3 1 sin a  cos a  sin a  cos a 2 2 2 2  cos a 

 RHS 5.4

cos( P  Q)  cos( P  Q)  2 cos P cos Q LHS  cos P cos Q  sin P sin Q  cos P cos Q  sin P sin Q  2 cos P cos Q  RHS

sin 5 A  sin 3A  2 sin 4 A.cos A 5.5 LHS  sin(4 A  A)  sin(4 A  A)  sin 4 A cos A  sin A cos 4 A  sin 4 A cos A  sin A cos 4 A  2 sin 4 A cos A  RHS

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6. sin 75  cos 105  sin 15  cos 15  0

LHS  sin(45  30)  cos(60  45)  sin(45  30)  cos(45  30)  sin 45 cos 30  sin 30 cos 45  cos 60 cos 45  sin 60 sin 45  sin 45 cos 30  sin 30 cos 45  cos 45 cos 30  sin  2 sin 30 cos 45  cos 60 cos 45  sin 60 sin 45  cos 45 cos 30  sin 45 sin 30 1 1 1 1 3 1 1 3 1 1  2(  )        2 2 2 2 2 2 2 2 2 2 1 1 3 3 1     2 2 2 2 2 2 2 2 2 0 

 RHS Page 33: Exercise 3.2:

1.

Proving Identities.

2 cos x sin x  sin 2 x 1 sin 2 x 2 cos x sin 2 x LHS   sin x 1  2 sin x cos x  sin 2 x

Simplify as a fraction

 RHS

2.

1 sin 2 x sin x  cos x   cos 2 x cos 2 x cos x  sin x 1  2 sin x cos x LHS  cos 2 x 2 cos x  2 sin x cos x  sin 2 x  cos 2 x  sin 2 x (cos x  sin x)(cos x  sin x)  (cos x  sin x)(cos x  sin x) sin x  cos x  cos x  sin x  RHS

Sin double L 1 = sin2x + cos2x

Choose correct cos double L 2 suit numerator

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3.

110

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sin x  sin 2 x  tan x 1  cos x  cos 2 x sin x  2 sin x cos x LHS  1  cos x  2 cos 2 x  1 sin x(1  2 cos x)  cos x(1  2 cos x) sin x  cos x  tan x  RHS

Compound L expansion

4.

sin 3 x  sin x  2 sin x 1  cos 2 x sin( 2 x  x)  sin x LHS  1  2 cos 2 x  1 sin 2 x cos x  sin x cos 2 x  sin x  2 cos 2 x 2 sin x cos 2 x  sin x cos 2 x  sin x  2 cos 2 x sin x(2 cos 2 x  cos 2 x  1)  2 cos 2 x sin x(2 cos 2 x  2 cos 2 x  1  1)  2 cos 2 x sin x(4 cos 2 x)  2 cos 2 x  2 sin x  RHS

5.

sin 2 x  tan x 1  cos 2 x 2 sin x cos x LHS  1  2 cos 2 x  1 2 sin x cos x  2 cos 2 x sin x  cos x  tan x  RHS

Double L’s and factorise

Grade 12 Core Mathematics

6.

sin 4 x  sin 2 x. cos 2 x  1  cos x 1  cos x sin 2 x(sin 2 x  cos 2 x) LHS  1  cos x 2 1  cos x  1  cos x (1  cos x)(1  cos x)  1  cos x  1  cos x  RHS

7.

sin 2 x  tan x  tan x cos 2 x 2 sin x cos x sin x  1 cos x 2 2 sin x cos x  sin x cos x LHS  cos 2 x sin x(2 cos 2 x  1) cos x  cos 2 x sin x(2 cos 2 x  1) 1   cos x 2 cos 2 x  1 sin x  cos x  tan  RHS

8.

sin 2 x  cos x cos x  sin x  cos 2 x sin x  1 2 sin x cos x  cos x LHS  sin x  (1  2 sin 2 x) cos x(2 sin x  1)  sin x  1  2 sin 2 x cos x(2 sin x  1)  (2 sin x  1)(sin x  1) cos x  sin x  1  RHS

111

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112

9. 1  cos 2 x tan 2 x  cos 2 x tan x

sin 2 x cos 2 x sin x cos x 2 sin x cos x cos x   cos 2 x sin x 2 2 cos x  cos 2 x RHS 

1  2 cos 2 x  1 cos 2 x 2 2 cos x  cos 2 x LHS 

LHS  RHS Page 39: Exercise 3.3: A. Find the general solution for the following: 1.

sin x  0,235 đ??žđ?‘’đ?‘Ś đ??ż = 13,6°

-ve 3/4

đ?’™ = đ?&#x;?đ?&#x;—đ?&#x;‘, đ?&#x;”° + đ?’Œ. đ?&#x;‘đ?&#x;”đ?&#x;Ž° đ?’™ = đ?&#x;‘đ?&#x;’đ?&#x;”, đ?&#x;’° + đ?’Œ. đ?&#x;‘đ?&#x;”đ?&#x;Ž° 2.

3 cos x  1,2066 đ?‘?đ?‘œđ?‘ đ?‘Ľ = 0,4022 đ??žđ?‘’đ?‘Ś đ??ż = 66,3° +ve 1/4 đ?’™ = đ?&#x;”đ?&#x;”, đ?&#x;‘° + đ?’Œ. đ?&#x;‘đ?&#x;”đ?&#x;Ž° đ?’™ = đ?&#x;?đ?&#x;—đ?&#x;‘, đ?&#x;•° + đ?’Œ. đ?&#x;‘đ?&#x;”đ?&#x;Ž°

3.

tan 2 x  4,302 đ??žđ?‘’đ?‘Ś (2đ?‘Ľ) = 76,9° -ve 2/4 2đ?‘Ľ = 103.10° + đ?‘˜. 180° đ?’™ = đ?&#x;“đ?&#x;?, đ?&#x;”° + đ?’Œ. đ?&#x;—đ?&#x;Ž° 2đ?‘Ľ = 283.10° + đ?‘˜. 180° đ?’™ = đ?&#x;?đ?&#x;’đ?&#x;?, đ?&#x;”° + đ?’Œ. đ?&#x;—đ?&#x;Ž°

4.

2 tan 3x  2,3648 3 đ?‘Ąđ?‘Žđ?‘›3đ?‘Ľ = 3,5472 đ??žđ?‘’đ?‘Ś 3đ?‘Ľ = 74,3° 3đ?‘Ľ = 74,3° + đ?‘˜. 180° đ?’™ = đ?&#x;?đ?&#x;’, đ?&#x;–° + đ?’Œ. đ?&#x;”đ?&#x;Ž°

3đ?‘Ľ = 285,7° + đ?‘˜. 180° đ?’™ = đ?&#x;—đ?&#x;“, đ?&#x;?° + đ?’Œ. đ?&#x;”đ?&#x;Ž°

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113

B: Find the specific solutions of the above equations if x  [360 ;360  ] 1. 𝑰𝒇 𝒌 = 𝟎 𝒕𝒉𝒆𝒏 𝒙 = 𝟏𝟗𝟑, 𝟔° 𝒐𝒓 𝟑𝟒𝟔, 𝟒° 𝑰𝒇 𝒌 = −𝟏 𝒕𝒉𝒆𝒏 𝒙 = −𝟏𝟔𝟔, 𝟒° 𝒐𝒓 − 𝟏𝟑, 𝟔°

2. 𝑰𝒇 𝒌 = 𝟎 𝒕𝒉𝒆𝒏 𝒙 = 𝟔𝟔, 𝟑° 𝒐𝒓 𝟐𝟗𝟑, 𝟕° 𝑰𝒇 𝒌 = −𝟏 𝒕𝒉𝒆𝒏 𝒙 = −𝟐𝟗𝟑, 𝟕° 𝒐𝒓 − 𝟔𝟔, 𝟑°

3. 𝑰𝒇 𝒌 = 𝟎 𝒕𝒉𝒆𝒏 𝒙 = 𝟓𝟏, 𝟔° 𝒐𝒓 𝟏𝟒𝟏, 𝟔° 𝑰𝒇 𝒌 = 𝟏 𝒕𝒉𝒆𝒏 𝒙 = 𝟐𝟑𝟏, 𝟔° 𝑰𝒇 𝒌 = 𝟐 𝒕𝒉𝒆𝒏 𝒙 = 𝟑𝟐𝟏, 𝟔° 𝑰𝒇 𝒌 = −𝟏 𝒕𝒉𝒆𝒏 𝒙 = −𝟑𝟖, 𝟒° 𝑰𝒇 𝒌 = −𝟐 𝒕𝒉𝒆𝒏 𝒙 = −𝟏𝟐𝟖, 𝟒° 𝑰𝒇 𝒌 = −𝟑 𝒕𝒉𝒆𝒏 𝒙 = −𝟐𝟏𝟖, 𝟒°

Page 41 : Exercise 3.4 Equations: 1.

or or or

2.

4sin2x –3 = 0 3 sin2x = 4 3 sinx =  2 key angle = 60 x = 60 + 360k x = 120  + 360k x = 240 + 360k x = 300  + 360k

ve {1;2;3;4}

60 + 180k/ 120 + 180k

cos2x – 7cosxtanx = 4  sin x  1-2sin2x – 7cosx   =4  cos x  2sin2x + 7sinx + 3 = 0 (sinx + 1)(sinx + 3) = 0 sinx = -0,5 or sinx = - 3 Key L = 30 reject(invalid) x = 210 + 360k or 330 + 360k x = -30 or –150.

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. 3.

4sinxcosx = 1 2(2sinxcosx) = 1 sin2x = 0,5 key L = 30 2x = 30 + 360k x = 15 + 180k OR 2x = 150 + 360k x = 75 + 180k x = 15 /75/ 195/ 255.

Solution

4.

4cos2x – 6cosx + 5 = 0 4(2cos2x – 1) – 6cosx + 5 = 0 8cos2x – 6cosx + 1 = 0 (4cosx – 1)(2cosx – 1) = 0 cosx = 0,25 or cosx = 0,5 key L = 75,5 key L = 60 x = 75,5 +360k or 60 + 360k x = 284,5 + 360k or 300 + 360k

2 sin 2 x  sin x  1  0 5. 𝟐𝒔𝒊𝒏𝒙 + 𝟏 𝒔𝒊𝒏𝒙 − 𝟏 = 𝟎 𝟏

𝒔𝒊𝒏𝒙 = − 𝟐

where x  0 ;360

or 𝒔𝒊𝒏𝒙 = 𝟎

𝑲𝒆𝒚𝑳 = 𝟑𝟎°

𝑲𝒆𝒚𝑳 = 𝟏𝟖𝟎°

𝒙 = 𝒌. 𝟏𝟖𝟎° or 𝒙 = 𝟐𝟏𝟎° + 𝒌. 𝟑𝟔𝟎° or 𝒙 = 𝟑𝟑𝟎 + 𝒌. 𝟑𝟔𝟎 Solution: 𝒙 = 𝟏𝟖𝟎 𝒐𝒓 𝟐𝟏𝟎 𝒐𝒓 𝟑𝟑𝟎 2 6. Solve for: 2 sin x  5 cos x  4 𝟐 − 𝟐𝒄𝒐𝒔𝟐 𝒙 + 𝟓𝒄𝒐𝒔𝒙 − 𝟒 = 𝟎

𝟐𝒄𝒐𝒔𝟐 𝒙 − 𝟓𝒄𝒐𝒔𝒙 + 𝟐 = 𝟎 𝟐𝒄𝒐𝒔𝒙 − 𝟏 𝒄𝒐𝒔𝒙 − 𝟐 = 𝟎 𝟏

𝒄𝒐𝒔𝒙 = 𝟐

or 𝒄𝒐𝒔𝒙 = 𝟐(reject)

𝒌𝒆𝒚𝑳 = 𝟔𝟎 𝒙 = 𝟔𝟎° 𝒐𝒓 𝟑𝟎𝟎

where x  0 ;360

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4. DATA HANDLING: Page 49: Exercise 4.1:

1.

The marks , out of 150, for 30 learners were as follows:

97

100

109

122

118

124

127

105

112

128

107

114

115

121

135

98

111

117

120

130

123

141

107

113

116

119

121

131

129

139

Organise the marks using a stem & leaf diagram.

9

78

10

0 5 7 7 9

11

1 23 4 56789

12

011234789

13

0159

14

1

Draw a Box & Whisker diagram to illustrate the dispersion of the marks. Determine the mean for the above data. 118,3 Determine the standard deviation, ‘s’ (correct to 1 decimal place ) 11,4 What percentage of calls lie within one standard deviation of the mean. 70% What can the teacher conclude about these marks. The majority of the class have marks clustered around the mean

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2.1 Do a stem and leaf diagram for the data 1

5

2

034589

3

000145567

4

00234555567 8

5

02342

6

0

7

5

2.2 Find the median, mode and mean for the data

x  39,7 Q2 40 Mode = 45 2.3 Find the lower and upper quartile Q1 = 30 and Q3 = 47 2.4 Calculate: 2.4.1

the interquartile range.

IQR = 17

2.4.2

the semi-interquartile range. SIQR = 8,5

2.4.3

the range for the class.

Range = 60

2.5

Write down the maximum and minimum scores. Min = 15 Max = 75

2.6

Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H)

100

P 75 E R C

50

E N T 25

0

15

75 40

30

0

1

2

3

S

47

4

C

O

5

R

E

6

S

7

8

9

10

Grade 12 Core Mathematics

2.7

117

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Standard Deviation = 12,4

2.7.1 What % of scores lie within 1 standard deviation from the mean. 25  71,4% 35

2.7.2 What % of scores lie within 2 standard deviations of the mean. 34  97,1% 35

2.9.

State whether the data is negatively or positively skewed and give a reason for

your decision. Data is negatively skewed as the median is larger than the mean

3.1 Do a stem and leaf diagram for the data

3.2

5

035

6

24455678

7

2 2 2 3 4 4 (74,5) 5 5 5 5 8 9

8

0 00 002489

9

00

Find the median, mode and mean for the data

Mean = 73,4 Median = 74,5 Mode = 80 3.3

Find the lower and upper quartile Q1 = 66

and Q3 = 80

3.4

Calculate:

3.4.1

the interquartile range.

3.4.2

the semi-interquartile range. SIQR = 7

3.4.3

the range for the class. Range = 40

3.5

IQR = 14

Write down the maximum and minimum scores. Min = 50 and Max = 90

Grade 12 Core Mathematics

3.6

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118

Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H)

100

P 75 E R C

50

E N T 25

0

90

66

1

0

2

3

S

3.7 3.8 3.9

4

C

Standard Deviation.

5

R

O

E

6

74.5

7

S

ď ł  10,1

What % of scores lie within 1 standard deviation from the mean. 73,5 % What % of scores lie within 2 standard deviations of the mean. 94,1 %

4.1 Do a back to back stem and leaf diagram for the data GIRLS

BOYS 5

7

44

6

24

6 5 5 73.5 2 0

7

2 2 25559

9854320

8

00013589

5200

9

5

80

8

9

10

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119

4.2 Find the median, mode and mean for both sets of data GIRLS: Median = 82 Mode = 64 ; 75 & 90 Mean = 80,3 BOYS: Median = 79 Mode = 75 & 80 Mean = 72,8 4.3 Find the lower and upper quartile of each set of data Girls :Q1 = 73,5 & Q3 = 89 Boys: Q1 = 72 & Q3 = 83 4.4 Calculate: 4.4.1

the interquartile ranges for:

4.4.2

4.4.1.1

girls

15,5

4.4.1.2

boys 11

4.4.1.3

class 13

the semi-interquartile ranges for: 4.4.2.1 girls 7,8 4.4.2.2 boys 5,5 4.4.2.3

4.4.3

class 6.5

the ranges for: 4.4.3.1 girls

4.5

31

4.4.3.2

boys 38

4.4.3.3

class

38

Write down the maximum and minimum scores of each set of data

Girls: Min = 64 ; Max = 95. Boy: Min = 57 ; Max = 95

4.6

Do separate box and whisker diagrams for the girls and the boys

Girls : Box & Whisker Diagram. 6.5

6

5.5

5

4.5

4

3.5

3

2.5

2

1.5

1

95

64 0.5

73.5

1

2

3

4

5

6

7

82

8

89 9

10

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Boys : Box & Whisker Diagram: 6.5

6

5.5

5

4.5

4

3.5

3

2.5

2

1.5

1

95

57 0.5

72

1

4.7

2

3

4

5

6

7

83

79

8

Standard Deviation : 4.7.1 the girls. 9,3 4.7.2 the boys 9,9 4.7.3

4.8

4.9

the class as a whole 9.7

What % of scores lie within 1 standard deviation from the mean for: 4.8.1 girls 59% 4.8.2 boys 65% 4.8.3 class 68% What % of scores lie within 2 standard deviations of the mean for: 4.9.1 girls 82% 4.9.2 boys 94% 4.9.3 class 97%

9

10

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5. Class 0 to 9 10 to 19 20 to 29 30 to 39 40 to 49 50 to 59 60 to 69 70 to 79 80 to 89 90 to 99 100 to 109 Totals 5.1 5.2 5.3 5.4 5.5

Frequency(f) 15 10 17 40 35 22 20 20 15 5 1 200

Mid-points(X) 4.5 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5

fX 67.5 145 416.5 1380 1557.5 1199 1290 1490 1267.5 472.5 104.5 9390

Complete the last column of the table i.e (fX) Find the modal class 30 to 39 Find the median class 40 to 49 Find the interval where Q1 and Q3 lie. Q1 lies in 30 to 39 and Q3 lies in 60 to 69 Calculate the estimated mean. ďƒĽ fX = 47 NB estimated mean = n

5.6

45 F R E Q U E N C Y

Cum Freq 15 25 42 82 117 139 159 179 194 199 200

35

25

15

5

0

10

20

30

40

C LASS

50

60

70

80

90

100 110

Grade 12 Core Mathematics

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Page 65: Exercise 4.2 1

x1

f

40 50 65 70 75 78 79 80 81 82 86 88 90 x  75,5

1 1 2 1 4 5 1 1 2 3 1 1 1

n  24

F x1 40 50 130 70 300 390 79 80 162 246 86 88 90

xx

( x  x )2

35,5 25,5 10.5 5.5 0.5 -2.5 -3.5 -4.5 -5.5 -6.5 -10.5 -12,5 -14,5

1260.25 650.25 110.25 30.25 0.25 6.25 12.25 20.25 30.25 42.25 110.25 156.25 210.25

 (x  x ) 1

2

=

2639.25 109.97 10.4

Variance STD DEV 2. Height (h) in cm 135  h < 140

Mid points 137,5

Frequency

Fx 275

Cumulative Frequency 2

Coordinates (140 ; 2)

2

140  h < 145

142,5

5

712.5

7

(145 ;7)

145  h < 150

147,5

10

1475

17

(150 ;17)

150  h < 155

152.5

17

2592.5

34

(155 ;34)

155  h < 160

157.5

19

2992.5

53

(160 ; 53)

160  h < 165

162.5

15

2437.5

68

(165 ; 68)

165  h < 170

167.5

4

670

72

(170 ; 72)

170  h < 175

172.5

2

345

74

(175 ;74)

175  h < 180

177.5

1

177.5

75

(180 ; 75)

2.1

75 Calculate the estimated mean. Estimated mean = 155.7

11677.5

Grade 12 Core Mathematics

123

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2.2Draw a histogram of the data

Question 8.2 & 8.3 20 18 16 Frequency

14 12 10 8 6 4 2 0 130 £ h < 135 £ h < 140 £ h < 145 £ h < 150 £ h < 155 £ h < 160 £ h < 165 £ h < 170 £ h < 175 £ h < 180 £ h < 135 140 145 150 155 160 165 170 175 180 185

Height in cm

2.4

State the modal group, median height ,upper and lower quartiles for the data. Modal Group = 155 – 160 ; Median Height =157.5 ; Upper Quartile =170 Lower quartile = 145 2.5

Sketch the Ogive Curve for the data.

80 70 60 50 40 30 20 10 0 130 £ 135 £ 140 £ 145 £ 150 £ 155 £ 160 £ 165 £ 170 £ 175 £ 180 £ h< h< h< h< h< h< h< h< h< h< h< 135 140 145 150 155 160 165 170 175 180 185

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CIRCLE GEOMETRY: Exercise 5.1 1.1 đ?‘Ľ+1 4

1. 2

3

24

=đ?‘Ľ

12

đ?‘Ľ 2 + đ?‘Ľ = 12 đ?‘Ľ 2 + đ?‘Ľ − 12 = 0 đ?‘Ľ+4 đ?‘Ľâˆ’3 =0 đ?‘Ľ = −4 đ?‘œđ?‘&#x; 3

2.1

2.2 2.3

2

2.4

1

RTP

(alt L’s GF//EA)

2

ď „ECA /// ď „DEA Proof: Aˆ1  Aˆ1 ( common) Eˆ 2  Dˆ 2 ( proved in c)

ECˆ A  DEˆ A ( L sum ď „) ď „AEC /// ď „ADE (AAA) AE EC AC   AD DE AE 2.5 AE 2  AD. AC (FROM 2.3:

AE AC  AD AE

2.6 ď œAE2 = AB2 ( from 2.2 & 2.5) AE = AB

22

= 33

2đ??ľđ??¸ = 72 đ??ľđ??¸ = 36 đ??ľđ??ś = 69

ACˆ B  CBˆ D ( L sum ď „) ď „ABCď‚˝ď‚˝ď‚˝ď „ADB ( AAA) AB BC AC   AD BD AB AB2 = AC.AD Eˆ 2  Fˆ1 ˆ Fˆ1  Dˆ 2 Eˆ  Dˆ

24 đ??ľđ??¸

2đ??šđ??ś = 22 đ??šđ??ś = 11

RTP: ď „ABCď‚˝ď‚˝ď‚˝ď „ADB (common) Aˆ 2  Aˆ 2 Bˆ  Dˆ (Alt segt thm) 1

22

= đ??šđ??ś

Grade 12 Core Mathematics

3.1

125

BC = DC Cˆ  Dˆ ( Alt Segt Thm) 5

2

Cˆ 5  Bˆ 2 ( Alt L,s BD//CE) BC  DC ( sides opp = L’s) 3.2

BAF /// DCF Fˆ  Fˆ ( common) ABˆ C  Dˆ ( Ext L cyclic quad ABCD) 3

BAˆ F  DCˆ F ( L sum triangle) FBA /// FDC ( AAA) FB BA FA   FD DC FC 3.3

BA DE  AF EF CE // BD

DE BC  EF CF BA DC  (BAF /// DCF ) AF CF BC  ( DC  BC ;3.1) CF DE DE BC  (  ; proved ) EF EF CF 3.4

ECD /// EAC Eˆ1  Eˆ1 ( common) Cˆ  Aˆ ( Alt segt thm) 3

2

ˆ  D ˆ ( L sum triangle) AD 3 ECD /// EAC ( AAA) EC CD ED   EA AD ED

3.5 CE 2 

AE.BC .EF CF

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126

4.1 RTP: BEF /// BFD Proof. Bˆ  Bˆ ( common) ABˆ F  Dˆ ( Alt Segt Thm) ABˆ F  EFˆB ( Alt L’s Ef //CA)  EFˆB  Dˆ BEˆ F  BFˆD ( L sum ) BFE /// BDF ( AAA) 4.2 BF FE BE   BD DF BF ( from 4.1) BF 2  BD.BE

6. 6.1 RTP: TEPB is a cyclic Quad. Proof: ATˆB  90  ( L in a semi circle) APˆ E  90 (Given) TEPB is a cyclic quad ( converse ExtL = int opp) 6.2 RTP: ATB  APE Aˆ  Aˆ ( common) ATˆB  APˆ E ( proved) ABˆ T  Eˆ ( L sum Triangle) ATB  APE ( A.A.A.) AT TB AB   AP PE AE 6.3 RTP: TP = PE ATˆB  BTˆP  ETˆP  180 ( L’s on a st line)  ˆ ˆ ˆ ATB  ABT  A  180 ( L’s on a st line) ATB  ATB But A  BTP( Alt segt thm) ABˆ T  ETˆP 6.4 RTP: ATPEPB Proof: ATˆB  EPˆ B (Proved) Aˆ  BTˆP Aˆ  BEˆ P ABˆ T  PBˆ E TABPEB ( AAA) AT AB BT   EP BE BP

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6.5

127

AT AB BT ( from d)   EP BE BP But AB = 2BP ( given) 2 BP BT  BE BP 2BP2 = BT.BE

7. 7.1

RTP: QR bisects PQT Proof: (co- int l’s SP//QR) SPˆ Q  Qˆ1  180   SPˆ Q  Rˆ  180 (opp ls of a parm) 3

Rˆ 3  Qˆ 2 Qˆ 1  Rˆ 3 Qˆ  Qˆ 1

(alt L’s QT//SR)

2

7.2 RTP; MN //PT Proof: Join RT ˆ TRN  RQˆ T (alt segt thm) PQˆ R  PTˆR (L’s in same segt) PQˆ R  RQˆ T (proved in a) PTˆR  TRˆ N MN // PT (alt l’s equal)

8.

RTP: CDF /// CED Proof: Cˆ 3  Cˆ 3 ( common) Eˆ  Bˆ ( Ls in same segt) 2

Bˆ 2  Dˆ 2 (L’s opp == sides)  Dˆ  Eˆ 2

CFˆD  CDˆ E ( L sum ) CDF /// CED (AAA) CD CF DF   CE CD DE 8.2 RTP: CDˆ E  CGˆ B Proof: CFˆD  Gˆ ( Ext L cyclic quad FCGB) CFˆD  CDˆ E ( proved in a) CDˆ E  CGˆ B

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Grade 12 Core Mathematics

128

8.3 RTP: CGB /// CDE Proof: CGˆ B  CDˆ E Bˆ 3  Dˆ 2 (Alt Segt Thm) Dˆ  Eˆ (proved) 2

 Bˆ 3  Eˆ  Cˆ1  Cˆ 3 ( L sum ) GBC /// DEC (AAA) GB BC GC   DE EC DC 8.4

CB = CD GB BC  DE EC GB CD  DE EC

(from c)

EC.GB = DE . CD

9.1

RTP: AC bisects BCˆ D Proof: Bˆ 2  Cˆ 2 (Alt segt thm) Bˆ  Cˆ ( L’s in same segt) 2

1

 Cˆ1  Cˆ 2

AC bisects BCˆ D 9.2

RTP ABC /// DEC Proof Cˆ1  Cˆ 2 (proved in a) Aˆ 2  Dˆ 1 ( L’s in same segt) ABˆ C  Eˆ 1

CAB /// CDE CA AB CB   CD DE CE 9.3 9.4

BC.DC  AC.EC RTP:

ADˆ C  BAˆ C  ACˆ D Cˆ  Cˆ (Proved) 1

2

Cˆ 2  Dˆ 2 ( L’s in same segt) Aˆ 2  Dˆ 1 ( L’s in same segt) ˆ C  BAˆ C  ACˆ D  AD

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Grade 12 Core Mathematics

9.5

129

RTP: ABE /// ACB Proof: Aˆ 2  Aˆ 2 ( common) Bˆ  Cˆ ( proved) 2

2

AEˆ B  ABˆ C ( L sum ) ABE /// ACB (AAA) AB BE AE   AC CB AB 9.6

AC 2  AE. AC

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