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Additional Mathematics SPM Chapter 3

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1 (b) (c) (d) (e) and (h)

2 f ( x) = 4 x2 ndash 8 x + 6

When x = ndash2

f (ndash2) = 4(ndash2)2 ndash 8(ndash2) + 6

= 16 + 16 + 6

= 38

3 f ( x) = x2 ndash 3 x + 2

When f ( x) = 0

x2 ndash 3 x + 2 = 0

( x ndash 1)( x ndash 2) = 0

x ndash 1 = 0 or x ndash 2 = 0

x = 1 or x = 2

4 f ( x) = ndash3 x2 + 5 x ndash 1

When f ( x) = 1

ndash3 x2 + 5 x ndash 1 = 1

ndash3 x2 + 5 x ndash 1 ndash 1 = 0

3 x2 ndash 5 x + 2 = 0 (3 x ndash 2)( x ndash 1) = 0

3 x ndash 2 = 0 or x ndash 1 = 0

x =2mdash3

or x = 1

5 (a) (b)

(c) (d)

(e) (f)

6 (a) Two different real roots

(b) One real root or two similar real roots

(c) No real roots

7 (a) The minimum value is 3

(b) The maximum value is 4

(c) The minimum value is ndash10

(d) f ( x) = 2983091 1mdash2

( x ndash 3)2 +1mdash4 983092

= ( x ndash 3)2 +1mdash2

Therefore the minimum value is1mdash2

(e) f ( x) =1mdash3

[6 ndash ( x + 1)2] + 5

= 2 ndash 1mdash3

( x + 1)2 + 5

= ndash1mdash3

( x + 1)2 + 7

Therefore the maximum value is 7

(f) The minimum value is 3

8 (a) f ( x) = x2 ndash 4 x + 2

= x2 ndash 4 x + 983089 4mdash2 983090

2

ndash 983089 4mdash2 983090

2

+ 2

= ( x ndash 2)2 ndash 4 + 2

= ( x ndash 2)2 ndash 2

Hence the minimum value is ndash2

(b) f ( x) = 2 x2 + 6 x ndash 5

= 2( x2 + 3 x) ndash 5

= 2983091 x2 + 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 ndash 5

= 2983091983089 x + 3mdash2 983090

2

ndash9mdash4 983092 ndash 5

= 2983089 x + 3mdash2 983090

2

ndash9mdash2

ndash 5

= 2983089 x + 3mdash2 983090

2

ndash19

ndashndashndash2

Hence the minimum value is ndash19

ndashndashndash2

(c) f ( x) = x2 + 5 x

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

= 983089 x +5mdash2 983090

2

ndash25

ndashndashndash4

Hence the minimum value is ndash25

ndashndashndash4

CHAPTER

3 Quadratic Functions

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(d) f ( x) = 6 x ndash x2

= ndash( x2 ndash 6 x)

= ndash983091 x2 ndash 6 x + 983089 6mdash2 983090

2

ndash 983089 6mdash2 983090

2

983092 = ndash[( x ndash 3)2 ndash 9]

= ndash ( x ndash 3)2 + 9

Hence the maximum value is 9

(e) f ( x) = 3 ndash 4 x ndash x2

= ndash x2 ndash 4 x + 3

= ndash( x2 + 4 x) + 3

= ndash983091 x2 + 4 x + 983089 4mdash2 983090

2

ndash 983089 4mdash2 983090

2

983092 + 3

= ndash[( x + 2)2 ndash 4] + 3

= ndash( x + 2)2 + 4 + 3

= ndash( x + 2)2 + 7

Hence the maximum value is 7

(f) f ( x) = 4 x ndash 2 x2

= ndash2 x2 + 4 x

= ndash2( x2

ndash 2 x) = ndash2983091 x2 ndash 2 x + 983089 2

mdash2 983090

2

ndash 983089 2mdash2 983090

2

983092 = ndash2[( x ndash 1)2 ndash 1]

= ndash2( x ndash 1)2 + 2

Hence the maximum value is 2

(g) f ( x) = 10 + 5 x ndash 3 x2

= ndash3 x2 + 5 x + 10

= ndash3983089 x2 ndash5mdash3 x983090 + 10

= ndash3983091 x2 ndash5mdash3 x + 983089 5

mdash6 983090

2

ndash 983089 5mdash6 983090

2

983092 + 10

= ndash3983091983089 x ndash5mdash6 983090

2

ndash25

ndashndashndash36 983092 + 10

= ndash3983089 x ndash5mdash6 983090

2

+25

ndashndashndash12

+ 10

= ndash3983089 x ndash5mdash6 983090

2

+145

ndashndashndashndash12

Hence the maximum value is145

ndashndashndashndash12

(h) f ( x) = (2 x ndash 1)( x + 3)

= 2 x2 + 5 x ndash 3

= 2983089 x2

+

5

mdash2 x983090 ndash 3

= 2983091 x2 +5mdash2 x + 983089 5

mdash4 983090

2

ndash 983089 5mdash4 983090

2

983092 ndash 3

= 2983091983089 x +5mdash4 983090

2

ndash25

ndashndashndash16 983092 ndash 3

= 2983089 x +5mdash4 983090

2

ndash25

ndashndashndash8

ndash 3

= 2983089 x +5mdash4 983090

2

ndash49

ndashndashndash8

Hence the minimum value is ndash49

ndashndashndash8

(i) f ( x) = (1 ndash 4 x)( x + 2)

= x + 2 ndash 4 x2 ndash 8 x

= ndash 4 x2 ndash 7 x + 2

= ndash 4983089 x2 +7mdash4 x983090 + 2

= ndash 4983091 x2 +7mdash4 x + 983089 7

mdash8 983090

2

ndash 983089 7mdash8 983090

2

983092 + 2

= ndash 4983091983089 x + 7mdash8 983090

2

ndash49

ndashndashndash64 983092 + 2

= ndash 4983089 x + 7mdash8 983090

2

+49

ndashndashndash16

+ 2

= ndash 4983089 x + 7mdash8 983090

2

+81

ndashndashndash16

Hence the maximum value is81

ndashndashndash16

9 (a) f ( x) = x2 ndash 4

Therefore the minimum point is (0 ndash 4)

x ndash2 2 4 f ( x) 0 0 12

f (x )

x

ndash4 ndash2 2 4

12

0

(b) f ( x) = 3 x2 + 5

Therefore the minimum point is (0 5)

x ndash1 3

f ( x) 8 32

f (x )

x

ndash1 3

5

8

32

0

(c) f ( x) = 8 ndash x2

Therefore the maximum point is (0 8)

x ndash3 plusmn9831059831068 3

f ( x) ndash1 0 ndash1

f (x )

x

ndash1

8

ndash3 3 ndash 8 8

0

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(d) f ( x) = 10 ndash 2 x2

Therefore the maximum point is (0 10)

x ndash3 plusmn9831059831065 4

f ( x) ndash8 0 ndash22

f (x )

x

ndash8

10

ndash3 4 ndash 5 5 0

ndash22

(e) f ( x) = x( x + 2)

= x2 + 2 x

= x2 + 2 x + 12 ndash 12

= ( x + 1)2 ndash 1

Therefore the minimum point is (ndash1 ndash1)

x ndash 4 ndash2 0 2 f ( x) 8 0 0 8

f (x )

x

(ndash1ndash1) ndash4

8

ndash2 20

(f) f ( x) = ( x ndash 1)(2 x + 1)

= 2 x2

ndash x ndash 1 = 2983089 x2 ndash

xmdash2 983090 ndash 1

= 2983091 x2 ndash xmdash2

+ 983089 1mdash4 983090

2

ndash 983089 1mdash4 983090

2

983092 ndash 1

= 2983091983089 x ndash 1mdash4 983090

2

ndash1

ndashndashndash16 983092 ndash 1

= 2983089 x ndash 1mdash4 983090

2

ndash1mdash8

ndash 1

= 2983089 x ndash 1mdash4 983090

2

ndash9mdash8

Therefore the minimum point is (

1

mdash4 ndash

9

mdash8 )

x ndash1 ndash1mdash2

0 1 2

f ( x) 2 0 ndash1 0 5

f (x )

x

1 2ndash1 1

2ndash ndash

98ndashndash)14 ( ndash

2

0

5

ndash1

(g) f ( x) = ndash( x ndash 3)2 + 5

Therefore the maximum point is (3 5)

x ndash2 0 3 ndash 9831059831065 3 + 9831059831065 6

f ( x) ndash20 ndash 4 0 0 ndash4

f (x )

x

ndash2 ndash4

ndash20

(3 5)

63 + 53 ndash 50

(h) f ( x) = x2 + 4 x + 5

= x2 + 4 x + 22 ndash 22 + 5

= ( x + 2)2 + 1

Therefore the minimum point is (ndash2 1)

x ndash3 0 1 f ( x) 2 5 10

f (x )

x

(ndash2 1)

ndash3 1

2

5

10

0

(i) f ( x) = 2 x2 + 6 x ndash 8

= 2( x2

+ 3 x) ndash 8

= 2983091 x2 + 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 ndash 8

= 2983091983089 x +3mdash2 983090

2

ndash9mdash4 983092 ndash 8

= 2983089 x + 3mdash2 983090

2

ndash9mdash2

ndash 8

= 2983089 x + 3mdash2 983090

2

ndash25

ndashndashndash2

Therefore the minimum point is (ndash3mdash2

ndash25

ndashndashndash2

)

x ndash3 0 1 2 f ( x) ndash8 ndash8 0 12

ndash25

2ndashndash)3

2( ndash

f (x )

x

ndash3

12

ndash8

0 1 2

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(j) f ( x) = ( x ndash 4)2

Therefore the minimum point is (4 0)

x 0 4 5

f ( x) 16 0 1

f (x )

x

4 501

16

(k) f ( x) = ndash x2 + 6 x ndash 9

= ndash( x2 ndash 6 x) ndash 9

= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9

= ndash[( x ndash 3)2 ndash 9] ndash 9

= ndash( x ndash 3)2

Therefore the maximum point is (3 0)

x 0 3 4

f ( x) ndash9 0 ndash1

f (x )

x 0

ndash1

ndash9

(3 0)

4

10 (a) x( x ndash 2) 983102 0

0 2x

f (x )

The range of values of x is x 983100 0 or x 983102 2

(b) ( x ndash 3)( x ndash 4) 983100 0

0 43

x

f (x )

The range of values of x is 3 983100 x 983100 4

(c) x2 ndash 3 x ndash 4 983086 0

( x ndash 4)( x + 1) 983086 0

4 ndash1 0 x

f (x )

The range of values of x is x 983084 ndash1 or x 983086 4

(d) 2 x2 + 5 x ndash 3 983084 0

(2 x ndash 1)( x + 3) 983084 0

1 ndash

2 ndash3 0

x

f (x )

The range of values of x is ndash3 983084 x 983084 1mdash2

(e) ( x ndash 3)( x + 2) 983100 ndash4

x2 ndash x ndash 6 983100 ndash4

x2 ndash x ndash 2 983100 0

( x ndash 2)( x + 1) 983100 0

ndash1 20 x

f (x )

The range of values of x is ndash1 983100 x 983100 2

(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)

2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12

2 x2 ndash 11 x + 15 983100 0

( x ndash 3)(2 x ndash 5) 983100 0

5 ndash

2

30 x

f (x )

The range of values of x is5mdash2

983100 x 983100 3

(g) x2 + 4

ndashndashndashndashndashndash5

983100 2 x ndash 1

x2 + 4 983100 5(2 x ndash 1)

x2 + 4 983100 10 x ndash 5

x2 ndash 10 x + 9 983100 0

( x ndash 1)( x ndash 9) 983100 0

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910 x

f (x )

The range of values of x is 1 983100 x 983100 9

(h) x (1 ndash 4 x) 983084 5 x ndash 8

x ndash 4 x2 ndash 5 x + 8 983084 0

ndash 4 x2 ndash 4 x + 8 983084 0

x2 + x ndash 2 983086 0

( x + 2)( x ndash 1) 983086 0

1 ndash2 0 x

f (x )

The range of values of x is x 983084 ndash2 or x 983086 1

1 (a) x-coordinate of the maximum part =1 + 7

ndashndashndashndashndash2

= 4

Therefore the equation of the axis of symmetry

is x = 4

(b) f ( x) = p ndash ( x + q)2

= 5 ndash ( x ndash 4)2

2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1

= 2( x2 ndash 8 x) + k 2 + 2k + 1

= 2983091 x2 ndash 8 x + 983089 8mdash2 983090

2

ndash 983089 8mdash2 983090

2

983092 + k 2 + 2k + 1

= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1

= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1

= 2( x ndash 4)2 + k 2 + 2k ndash 31

Given minimum value = ndash28

there4 k 2 + 2k ndash 31 = ndash28

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0 k = ndash3 1

3 (a) f ( x) = 2( x ndash 3)2 + k

p is the x-coordinate of the minimum point

Therefore p = 3

(b) k is the minimum value of f ( x)

Therefore k = ndash4

(c) The equation of the axis of symmetry is x = 3

4 f ( x) = 3 x2 ndash 2 x + p

a = 3 b = ndash2 c = p

Since the graph does not intersect the x-axis

b2 ndash 4ac lt 0

(ndash2)2 ndash 4(3)( p) lt 0

4 ndash 12 p lt 0

4 lt 12 p

13

lt p

p gt13

5 f ( x) = 2 x2 ndash 12 x + 5

= 2( x2 ndash 6 x) + 5

= 2[( x ndash 3)2 ndash 32] + 5

= 2( x ndash 3)2 ndash 18 + 5

= 2( x ndash 3)2 ndash 13

there4 p = 2 q = ndash3 ndashr + 1 = ndash13

r = 14

6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2

= ndash( x2 ndash 6 px) + 1 ndash 4 p2

= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash

2 983090

2

ndash 983089 6 p ndashndashndash

2 983090

2

983092 + 1 ndash 4 p2

= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 1 + 5 p2

The maximum value given is q2 ndash p

Therefore q2 ndash p = 1 + 5 p2

5 p2 + p + 1 = q2

(b) x = 3 is symmetrical axis 3 p = 3

p = 1

Substitute p = 1 into 5 p2 + p + 1 = q2

5(1)2 + 1 + 1 = q2

q2 = 7

q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067

7 4t (t + 1) ndash 3t 2 + 12 983086 0

4t 2 + 4t ndash 3t 2 + 12 983086 0

t 2 + 4t + 12 983086 0

(t + 2)(t + 6) 983086 0

0 ndash2 ndash6x

f (x )

The range of values of t is t 983084 ndash6 or t 983086 ndash2

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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash

2 = ndash2

Equation of the axis of symmetry is x = ndash2

2 Let x be the x-coordinate of A

0 + x

ndashndashndashndashndash2

= 3

x = 6

The coordinates of A are (6 4)

3 Let x be the x-coordinate of A

x + 6 ndashndashndashndashndash

2 = 2

x = 4 ndash 6

x = ndash2

The coordinates of A are (ndash2 0)

4 x-coordinate of A =0 + 8

ndashndashndashndashndash2

= 4

Let C be the centre of OB

4

5

A

C O

AC 2

= OA2

ndash OC 2

= 52 ndash 42

= 9

AC = 3

The coordinates of A are (4 3)

5 x-coordinate of minimum point =0 + 4

ndashndashndashndashndash2

= 2

x-coordinate of minimum point for the image is ndash2

6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)

Hence p = 2 and q = ndash1

(b) y = ( x ndash 2)2 ndash 1

When y = 0

( x ndash 2)2 ndash 1 = 0

( x ndash 2)2 = 1

x ndash 2 = plusmn1

x = plusmn1 + 2

= 1 3

Hence A is (1 0) and B is (3 0)

7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090

2

Given (ndash1 k ) is the maximum point

Therefore 2k + 1 = k

k = ndash1

x +1mdash2 p = 0 when x = ndash1

ndash1 + 1mdash2 p = 0

1mdash2 p = 1

p = 2

8 Given ( p 2q) is the minimum point of

y = 2 x2 ndash 4 x + 5

= 2( x2 ndash 2 x) + 5

= 2( x2 ndash 2 x + 12 ndash 12) + 5

= 2[( x ndash 1)2 ndash 1] + 5

= 2( x ndash 1)2 ndash 2 + 5

= 2( x ndash 1)2 + 3

2q = 3

q =3mdash2

p ndash 1 = 0

p = 1

9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1

substitute x = 1 y = 4 into the equation

4 = 12 ndash 2k (1) + 1

2k = ndash2

k = ndash1

(b) y = x2 ndash 2(ndash1) x + 1

= x2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0

10 f ( x) = ndash x2 ndash 8 x + k ndash 1

= ndash( x2 + 8 x) + k ndash 1

= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1

= ndash[( x + 4)2 ndash 16] + k ndash 1

= ndash( x + 4)2 + 16 + k ndash 1

= ndash( x + 4)2 + 15 + k

Since 13 is the maximum value

then 15 + k = 13

k = ndash2

11 f ( x) = 2 x2 ndash 6 x + 7

= 2( x2 ndash 3 x) + 7

= 2983091 x2 ndash 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 + 7

= 2983091983089 x ndash3mdash2 983090

2

ndash9mdash4 983092 + 7

= 2983089 x ndash3mdash2 983090

2

ndash9mdash2

+ 7

= 2983089 x ndash3mdash2 983090

2

+5mdash2

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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(d) f ( x) = 6 x ndash x2

= ndash( x2 ndash 6 x)

= ndash983091 x2 ndash 6 x + 983089 6mdash2 983090

2

ndash 983089 6mdash2 983090

2

983092 = ndash[( x ndash 3)2 ndash 9]

= ndash ( x ndash 3)2 + 9

Hence the maximum value is 9

(e) f ( x) = 3 ndash 4 x ndash x2

= ndash x2 ndash 4 x + 3

= ndash( x2 + 4 x) + 3

= ndash983091 x2 + 4 x + 983089 4mdash2 983090

2

ndash 983089 4mdash2 983090

2

983092 + 3

= ndash[( x + 2)2 ndash 4] + 3

= ndash( x + 2)2 + 4 + 3

= ndash( x + 2)2 + 7

Hence the maximum value is 7

(f) f ( x) = 4 x ndash 2 x2

= ndash2 x2 + 4 x

= ndash2( x2

ndash 2 x) = ndash2983091 x2 ndash 2 x + 983089 2

mdash2 983090

2

ndash 983089 2mdash2 983090

2

983092 = ndash2[( x ndash 1)2 ndash 1]

= ndash2( x ndash 1)2 + 2

Hence the maximum value is 2

(g) f ( x) = 10 + 5 x ndash 3 x2

= ndash3 x2 + 5 x + 10

= ndash3983089 x2 ndash5mdash3 x983090 + 10

= ndash3983091 x2 ndash5mdash3 x + 983089 5

mdash6 983090

2

ndash 983089 5mdash6 983090

2

983092 + 10

= ndash3983091983089 x ndash5mdash6 983090

2

ndash25

ndashndashndash36 983092 + 10

= ndash3983089 x ndash5mdash6 983090

2

+25

ndashndashndash12

+ 10

= ndash3983089 x ndash5mdash6 983090

2

+145

ndashndashndashndash12

Hence the maximum value is145

ndashndashndashndash12

(h) f ( x) = (2 x ndash 1)( x + 3)

= 2 x2 + 5 x ndash 3

= 2983089 x2

+

5

mdash2 x983090 ndash 3

= 2983091 x2 +5mdash2 x + 983089 5

mdash4 983090

2

ndash 983089 5mdash4 983090

2

983092 ndash 3

= 2983091983089 x +5mdash4 983090

2

ndash25

ndashndashndash16 983092 ndash 3

= 2983089 x +5mdash4 983090

2

ndash25

ndashndashndash8

ndash 3

= 2983089 x +5mdash4 983090

2

ndash49

ndashndashndash8

Hence the minimum value is ndash49

ndashndashndash8

(i) f ( x) = (1 ndash 4 x)( x + 2)

= x + 2 ndash 4 x2 ndash 8 x

= ndash 4 x2 ndash 7 x + 2

= ndash 4983089 x2 +7mdash4 x983090 + 2

= ndash 4983091 x2 +7mdash4 x + 983089 7

mdash8 983090

2

ndash 983089 7mdash8 983090

2

983092 + 2

= ndash 4983091983089 x + 7mdash8 983090

2

ndash49

ndashndashndash64 983092 + 2

= ndash 4983089 x + 7mdash8 983090

2

+49

ndashndashndash16

+ 2

= ndash 4983089 x + 7mdash8 983090

2

+81

ndashndashndash16

Hence the maximum value is81

ndashndashndash16

9 (a) f ( x) = x2 ndash 4

Therefore the minimum point is (0 ndash 4)

x ndash2 2 4 f ( x) 0 0 12

f (x )

x

ndash4 ndash2 2 4

12

0

(b) f ( x) = 3 x2 + 5

Therefore the minimum point is (0 5)

x ndash1 3

f ( x) 8 32

f (x )

x

ndash1 3

5

8

32

0

(c) f ( x) = 8 ndash x2

Therefore the maximum point is (0 8)

x ndash3 plusmn9831059831068 3

f ( x) ndash1 0 ndash1

f (x )

x

ndash1

8

ndash3 3 ndash 8 8

0

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(d) f ( x) = 10 ndash 2 x2

Therefore the maximum point is (0 10)

x ndash3 plusmn9831059831065 4

f ( x) ndash8 0 ndash22

f (x )

x

ndash8

10

ndash3 4 ndash 5 5 0

ndash22

(e) f ( x) = x( x + 2)

= x2 + 2 x

= x2 + 2 x + 12 ndash 12

= ( x + 1)2 ndash 1

Therefore the minimum point is (ndash1 ndash1)

x ndash 4 ndash2 0 2 f ( x) 8 0 0 8

f (x )

x

(ndash1ndash1) ndash4

8

ndash2 20

(f) f ( x) = ( x ndash 1)(2 x + 1)

= 2 x2

ndash x ndash 1 = 2983089 x2 ndash

xmdash2 983090 ndash 1

= 2983091 x2 ndash xmdash2

+ 983089 1mdash4 983090

2

ndash 983089 1mdash4 983090

2

983092 ndash 1

= 2983091983089 x ndash 1mdash4 983090

2

ndash1

ndashndashndash16 983092 ndash 1

= 2983089 x ndash 1mdash4 983090

2

ndash1mdash8

ndash 1

= 2983089 x ndash 1mdash4 983090

2

ndash9mdash8

Therefore the minimum point is (

1

mdash4 ndash

9

mdash8 )

x ndash1 ndash1mdash2

0 1 2

f ( x) 2 0 ndash1 0 5

f (x )

x

1 2ndash1 1

2ndash ndash

98ndashndash)14 ( ndash

2

0

5

ndash1

(g) f ( x) = ndash( x ndash 3)2 + 5

Therefore the maximum point is (3 5)

x ndash2 0 3 ndash 9831059831065 3 + 9831059831065 6

f ( x) ndash20 ndash 4 0 0 ndash4

f (x )

x

ndash2 ndash4

ndash20

(3 5)

63 + 53 ndash 50

(h) f ( x) = x2 + 4 x + 5

= x2 + 4 x + 22 ndash 22 + 5

= ( x + 2)2 + 1

Therefore the minimum point is (ndash2 1)

x ndash3 0 1 f ( x) 2 5 10

f (x )

x

(ndash2 1)

ndash3 1

2

5

10

0

(i) f ( x) = 2 x2 + 6 x ndash 8

= 2( x2

+ 3 x) ndash 8

= 2983091 x2 + 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 ndash 8

= 2983091983089 x +3mdash2 983090

2

ndash9mdash4 983092 ndash 8

= 2983089 x + 3mdash2 983090

2

ndash9mdash2

ndash 8

= 2983089 x + 3mdash2 983090

2

ndash25

ndashndashndash2

Therefore the minimum point is (ndash3mdash2

ndash25

ndashndashndash2

)

x ndash3 0 1 2 f ( x) ndash8 ndash8 0 12

ndash25

2ndashndash)3

2( ndash

f (x )

x

ndash3

12

ndash8

0 1 2

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(j) f ( x) = ( x ndash 4)2

Therefore the minimum point is (4 0)

x 0 4 5

f ( x) 16 0 1

f (x )

x

4 501

16

(k) f ( x) = ndash x2 + 6 x ndash 9

= ndash( x2 ndash 6 x) ndash 9

= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9

= ndash[( x ndash 3)2 ndash 9] ndash 9

= ndash( x ndash 3)2

Therefore the maximum point is (3 0)

x 0 3 4

f ( x) ndash9 0 ndash1

f (x )

x 0

ndash1

ndash9

(3 0)

4

10 (a) x( x ndash 2) 983102 0

0 2x

f (x )

The range of values of x is x 983100 0 or x 983102 2

(b) ( x ndash 3)( x ndash 4) 983100 0

0 43

x

f (x )

The range of values of x is 3 983100 x 983100 4

(c) x2 ndash 3 x ndash 4 983086 0

( x ndash 4)( x + 1) 983086 0

4 ndash1 0 x

f (x )

The range of values of x is x 983084 ndash1 or x 983086 4

(d) 2 x2 + 5 x ndash 3 983084 0

(2 x ndash 1)( x + 3) 983084 0

1 ndash

2 ndash3 0

x

f (x )

The range of values of x is ndash3 983084 x 983084 1mdash2

(e) ( x ndash 3)( x + 2) 983100 ndash4

x2 ndash x ndash 6 983100 ndash4

x2 ndash x ndash 2 983100 0

( x ndash 2)( x + 1) 983100 0

ndash1 20 x

f (x )

The range of values of x is ndash1 983100 x 983100 2

(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)

2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12

2 x2 ndash 11 x + 15 983100 0

( x ndash 3)(2 x ndash 5) 983100 0

5 ndash

2

30 x

f (x )

The range of values of x is5mdash2

983100 x 983100 3

(g) x2 + 4

ndashndashndashndashndashndash5

983100 2 x ndash 1

x2 + 4 983100 5(2 x ndash 1)

x2 + 4 983100 10 x ndash 5

x2 ndash 10 x + 9 983100 0

( x ndash 1)( x ndash 9) 983100 0

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910 x

f (x )

The range of values of x is 1 983100 x 983100 9

(h) x (1 ndash 4 x) 983084 5 x ndash 8

x ndash 4 x2 ndash 5 x + 8 983084 0

ndash 4 x2 ndash 4 x + 8 983084 0

x2 + x ndash 2 983086 0

( x + 2)( x ndash 1) 983086 0

1 ndash2 0 x

f (x )

The range of values of x is x 983084 ndash2 or x 983086 1

1 (a) x-coordinate of the maximum part =1 + 7

ndashndashndashndashndash2

= 4

Therefore the equation of the axis of symmetry

is x = 4

(b) f ( x) = p ndash ( x + q)2

= 5 ndash ( x ndash 4)2

2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1

= 2( x2 ndash 8 x) + k 2 + 2k + 1

= 2983091 x2 ndash 8 x + 983089 8mdash2 983090

2

ndash 983089 8mdash2 983090

2

983092 + k 2 + 2k + 1

= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1

= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1

= 2( x ndash 4)2 + k 2 + 2k ndash 31

Given minimum value = ndash28

there4 k 2 + 2k ndash 31 = ndash28

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0 k = ndash3 1

3 (a) f ( x) = 2( x ndash 3)2 + k

p is the x-coordinate of the minimum point

Therefore p = 3

(b) k is the minimum value of f ( x)

Therefore k = ndash4

(c) The equation of the axis of symmetry is x = 3

4 f ( x) = 3 x2 ndash 2 x + p

a = 3 b = ndash2 c = p

Since the graph does not intersect the x-axis

b2 ndash 4ac lt 0

(ndash2)2 ndash 4(3)( p) lt 0

4 ndash 12 p lt 0

4 lt 12 p

13

lt p

p gt13

5 f ( x) = 2 x2 ndash 12 x + 5

= 2( x2 ndash 6 x) + 5

= 2[( x ndash 3)2 ndash 32] + 5

= 2( x ndash 3)2 ndash 18 + 5

= 2( x ndash 3)2 ndash 13

there4 p = 2 q = ndash3 ndashr + 1 = ndash13

r = 14

6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2

= ndash( x2 ndash 6 px) + 1 ndash 4 p2

= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash

2 983090

2

ndash 983089 6 p ndashndashndash

2 983090

2

983092 + 1 ndash 4 p2

= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 1 + 5 p2

The maximum value given is q2 ndash p

Therefore q2 ndash p = 1 + 5 p2

5 p2 + p + 1 = q2

(b) x = 3 is symmetrical axis 3 p = 3

p = 1

Substitute p = 1 into 5 p2 + p + 1 = q2

5(1)2 + 1 + 1 = q2

q2 = 7

q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067

7 4t (t + 1) ndash 3t 2 + 12 983086 0

4t 2 + 4t ndash 3t 2 + 12 983086 0

t 2 + 4t + 12 983086 0

(t + 2)(t + 6) 983086 0

0 ndash2 ndash6x

f (x )

The range of values of t is t 983084 ndash6 or t 983086 ndash2

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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash

2 = ndash2

Equation of the axis of symmetry is x = ndash2

2 Let x be the x-coordinate of A

0 + x

ndashndashndashndashndash2

= 3

x = 6

The coordinates of A are (6 4)

3 Let x be the x-coordinate of A

x + 6 ndashndashndashndashndash

2 = 2

x = 4 ndash 6

x = ndash2

The coordinates of A are (ndash2 0)

4 x-coordinate of A =0 + 8

ndashndashndashndashndash2

= 4

Let C be the centre of OB

4

5

A

C O

AC 2

= OA2

ndash OC 2

= 52 ndash 42

= 9

AC = 3

The coordinates of A are (4 3)

5 x-coordinate of minimum point =0 + 4

ndashndashndashndashndash2

= 2

x-coordinate of minimum point for the image is ndash2

6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)

Hence p = 2 and q = ndash1

(b) y = ( x ndash 2)2 ndash 1

When y = 0

( x ndash 2)2 ndash 1 = 0

( x ndash 2)2 = 1

x ndash 2 = plusmn1

x = plusmn1 + 2

= 1 3

Hence A is (1 0) and B is (3 0)

7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090

2

Given (ndash1 k ) is the maximum point

Therefore 2k + 1 = k

k = ndash1

x +1mdash2 p = 0 when x = ndash1

ndash1 + 1mdash2 p = 0

1mdash2 p = 1

p = 2

8 Given ( p 2q) is the minimum point of

y = 2 x2 ndash 4 x + 5

= 2( x2 ndash 2 x) + 5

= 2( x2 ndash 2 x + 12 ndash 12) + 5

= 2[( x ndash 1)2 ndash 1] + 5

= 2( x ndash 1)2 ndash 2 + 5

= 2( x ndash 1)2 + 3

2q = 3

q =3mdash2

p ndash 1 = 0

p = 1

9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1

substitute x = 1 y = 4 into the equation

4 = 12 ndash 2k (1) + 1

2k = ndash2

k = ndash1

(b) y = x2 ndash 2(ndash1) x + 1

= x2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0

10 f ( x) = ndash x2 ndash 8 x + k ndash 1

= ndash( x2 + 8 x) + k ndash 1

= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1

= ndash[( x + 4)2 ndash 16] + k ndash 1

= ndash( x + 4)2 + 16 + k ndash 1

= ndash( x + 4)2 + 15 + k

Since 13 is the maximum value

then 15 + k = 13

k = ndash2

11 f ( x) = 2 x2 ndash 6 x + 7

= 2( x2 ndash 3 x) + 7

= 2983091 x2 ndash 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 + 7

= 2983091983089 x ndash3mdash2 983090

2

ndash9mdash4 983092 + 7

= 2983089 x ndash3mdash2 983090

2

ndash9mdash2

+ 7

= 2983089 x ndash3mdash2 983090

2

+5mdash2

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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(d) f ( x) = 10 ndash 2 x2

Therefore the maximum point is (0 10)

x ndash3 plusmn9831059831065 4

f ( x) ndash8 0 ndash22

f (x )

x

ndash8

10

ndash3 4 ndash 5 5 0

ndash22

(e) f ( x) = x( x + 2)

= x2 + 2 x

= x2 + 2 x + 12 ndash 12

= ( x + 1)2 ndash 1

Therefore the minimum point is (ndash1 ndash1)

x ndash 4 ndash2 0 2 f ( x) 8 0 0 8

f (x )

x

(ndash1ndash1) ndash4

8

ndash2 20

(f) f ( x) = ( x ndash 1)(2 x + 1)

= 2 x2

ndash x ndash 1 = 2983089 x2 ndash

xmdash2 983090 ndash 1

= 2983091 x2 ndash xmdash2

+ 983089 1mdash4 983090

2

ndash 983089 1mdash4 983090

2

983092 ndash 1

= 2983091983089 x ndash 1mdash4 983090

2

ndash1

ndashndashndash16 983092 ndash 1

= 2983089 x ndash 1mdash4 983090

2

ndash1mdash8

ndash 1

= 2983089 x ndash 1mdash4 983090

2

ndash9mdash8

Therefore the minimum point is (

1

mdash4 ndash

9

mdash8 )

x ndash1 ndash1mdash2

0 1 2

f ( x) 2 0 ndash1 0 5

f (x )

x

1 2ndash1 1

2ndash ndash

98ndashndash)14 ( ndash

2

0

5

ndash1

(g) f ( x) = ndash( x ndash 3)2 + 5

Therefore the maximum point is (3 5)

x ndash2 0 3 ndash 9831059831065 3 + 9831059831065 6

f ( x) ndash20 ndash 4 0 0 ndash4

f (x )

x

ndash2 ndash4

ndash20

(3 5)

63 + 53 ndash 50

(h) f ( x) = x2 + 4 x + 5

= x2 + 4 x + 22 ndash 22 + 5

= ( x + 2)2 + 1

Therefore the minimum point is (ndash2 1)

x ndash3 0 1 f ( x) 2 5 10

f (x )

x

(ndash2 1)

ndash3 1

2

5

10

0

(i) f ( x) = 2 x2 + 6 x ndash 8

= 2( x2

+ 3 x) ndash 8

= 2983091 x2 + 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 ndash 8

= 2983091983089 x +3mdash2 983090

2

ndash9mdash4 983092 ndash 8

= 2983089 x + 3mdash2 983090

2

ndash9mdash2

ndash 8

= 2983089 x + 3mdash2 983090

2

ndash25

ndashndashndash2

Therefore the minimum point is (ndash3mdash2

ndash25

ndashndashndash2

)

x ndash3 0 1 2 f ( x) ndash8 ndash8 0 12

ndash25

2ndashndash)3

2( ndash

f (x )

x

ndash3

12

ndash8

0 1 2

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(j) f ( x) = ( x ndash 4)2

Therefore the minimum point is (4 0)

x 0 4 5

f ( x) 16 0 1

f (x )

x

4 501

16

(k) f ( x) = ndash x2 + 6 x ndash 9

= ndash( x2 ndash 6 x) ndash 9

= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9

= ndash[( x ndash 3)2 ndash 9] ndash 9

= ndash( x ndash 3)2

Therefore the maximum point is (3 0)

x 0 3 4

f ( x) ndash9 0 ndash1

f (x )

x 0

ndash1

ndash9

(3 0)

4

10 (a) x( x ndash 2) 983102 0

0 2x

f (x )

The range of values of x is x 983100 0 or x 983102 2

(b) ( x ndash 3)( x ndash 4) 983100 0

0 43

x

f (x )

The range of values of x is 3 983100 x 983100 4

(c) x2 ndash 3 x ndash 4 983086 0

( x ndash 4)( x + 1) 983086 0

4 ndash1 0 x

f (x )

The range of values of x is x 983084 ndash1 or x 983086 4

(d) 2 x2 + 5 x ndash 3 983084 0

(2 x ndash 1)( x + 3) 983084 0

1 ndash

2 ndash3 0

x

f (x )

The range of values of x is ndash3 983084 x 983084 1mdash2

(e) ( x ndash 3)( x + 2) 983100 ndash4

x2 ndash x ndash 6 983100 ndash4

x2 ndash x ndash 2 983100 0

( x ndash 2)( x + 1) 983100 0

ndash1 20 x

f (x )

The range of values of x is ndash1 983100 x 983100 2

(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)

2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12

2 x2 ndash 11 x + 15 983100 0

( x ndash 3)(2 x ndash 5) 983100 0

5 ndash

2

30 x

f (x )

The range of values of x is5mdash2

983100 x 983100 3

(g) x2 + 4

ndashndashndashndashndashndash5

983100 2 x ndash 1

x2 + 4 983100 5(2 x ndash 1)

x2 + 4 983100 10 x ndash 5

x2 ndash 10 x + 9 983100 0

( x ndash 1)( x ndash 9) 983100 0

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Additional Mathematics SPM Chapter 3

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910 x

f (x )

The range of values of x is 1 983100 x 983100 9

(h) x (1 ndash 4 x) 983084 5 x ndash 8

x ndash 4 x2 ndash 5 x + 8 983084 0

ndash 4 x2 ndash 4 x + 8 983084 0

x2 + x ndash 2 983086 0

( x + 2)( x ndash 1) 983086 0

1 ndash2 0 x

f (x )

The range of values of x is x 983084 ndash2 or x 983086 1

1 (a) x-coordinate of the maximum part =1 + 7

ndashndashndashndashndash2

= 4

Therefore the equation of the axis of symmetry

is x = 4

(b) f ( x) = p ndash ( x + q)2

= 5 ndash ( x ndash 4)2

2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1

= 2( x2 ndash 8 x) + k 2 + 2k + 1

= 2983091 x2 ndash 8 x + 983089 8mdash2 983090

2

ndash 983089 8mdash2 983090

2

983092 + k 2 + 2k + 1

= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1

= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1

= 2( x ndash 4)2 + k 2 + 2k ndash 31

Given minimum value = ndash28

there4 k 2 + 2k ndash 31 = ndash28

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0 k = ndash3 1

3 (a) f ( x) = 2( x ndash 3)2 + k

p is the x-coordinate of the minimum point

Therefore p = 3

(b) k is the minimum value of f ( x)

Therefore k = ndash4

(c) The equation of the axis of symmetry is x = 3

4 f ( x) = 3 x2 ndash 2 x + p

a = 3 b = ndash2 c = p

Since the graph does not intersect the x-axis

b2 ndash 4ac lt 0

(ndash2)2 ndash 4(3)( p) lt 0

4 ndash 12 p lt 0

4 lt 12 p

13

lt p

p gt13

5 f ( x) = 2 x2 ndash 12 x + 5

= 2( x2 ndash 6 x) + 5

= 2[( x ndash 3)2 ndash 32] + 5

= 2( x ndash 3)2 ndash 18 + 5

= 2( x ndash 3)2 ndash 13

there4 p = 2 q = ndash3 ndashr + 1 = ndash13

r = 14

6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2

= ndash( x2 ndash 6 px) + 1 ndash 4 p2

= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash

2 983090

2

ndash 983089 6 p ndashndashndash

2 983090

2

983092 + 1 ndash 4 p2

= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 1 + 5 p2

The maximum value given is q2 ndash p

Therefore q2 ndash p = 1 + 5 p2

5 p2 + p + 1 = q2

(b) x = 3 is symmetrical axis 3 p = 3

p = 1

Substitute p = 1 into 5 p2 + p + 1 = q2

5(1)2 + 1 + 1 = q2

q2 = 7

q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067

7 4t (t + 1) ndash 3t 2 + 12 983086 0

4t 2 + 4t ndash 3t 2 + 12 983086 0

t 2 + 4t + 12 983086 0

(t + 2)(t + 6) 983086 0

0 ndash2 ndash6x

f (x )

The range of values of t is t 983084 ndash6 or t 983086 ndash2

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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash

2 = ndash2

Equation of the axis of symmetry is x = ndash2

2 Let x be the x-coordinate of A

0 + x

ndashndashndashndashndash2

= 3

x = 6

The coordinates of A are (6 4)

3 Let x be the x-coordinate of A

x + 6 ndashndashndashndashndash

2 = 2

x = 4 ndash 6

x = ndash2

The coordinates of A are (ndash2 0)

4 x-coordinate of A =0 + 8

ndashndashndashndashndash2

= 4

Let C be the centre of OB

4

5

A

C O

AC 2

= OA2

ndash OC 2

= 52 ndash 42

= 9

AC = 3

The coordinates of A are (4 3)

5 x-coordinate of minimum point =0 + 4

ndashndashndashndashndash2

= 2

x-coordinate of minimum point for the image is ndash2

6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)

Hence p = 2 and q = ndash1

(b) y = ( x ndash 2)2 ndash 1

When y = 0

( x ndash 2)2 ndash 1 = 0

( x ndash 2)2 = 1

x ndash 2 = plusmn1

x = plusmn1 + 2

= 1 3

Hence A is (1 0) and B is (3 0)

7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090

2

Given (ndash1 k ) is the maximum point

Therefore 2k + 1 = k

k = ndash1

x +1mdash2 p = 0 when x = ndash1

ndash1 + 1mdash2 p = 0

1mdash2 p = 1

p = 2

8 Given ( p 2q) is the minimum point of

y = 2 x2 ndash 4 x + 5

= 2( x2 ndash 2 x) + 5

= 2( x2 ndash 2 x + 12 ndash 12) + 5

= 2[( x ndash 1)2 ndash 1] + 5

= 2( x ndash 1)2 ndash 2 + 5

= 2( x ndash 1)2 + 3

2q = 3

q =3mdash2

p ndash 1 = 0

p = 1

9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1

substitute x = 1 y = 4 into the equation

4 = 12 ndash 2k (1) + 1

2k = ndash2

k = ndash1

(b) y = x2 ndash 2(ndash1) x + 1

= x2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0

10 f ( x) = ndash x2 ndash 8 x + k ndash 1

= ndash( x2 + 8 x) + k ndash 1

= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1

= ndash[( x + 4)2 ndash 16] + k ndash 1

= ndash( x + 4)2 + 16 + k ndash 1

= ndash( x + 4)2 + 15 + k

Since 13 is the maximum value

then 15 + k = 13

k = ndash2

11 f ( x) = 2 x2 ndash 6 x + 7

= 2( x2 ndash 3 x) + 7

= 2983091 x2 ndash 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 + 7

= 2983091983089 x ndash3mdash2 983090

2

ndash9mdash4 983092 + 7

= 2983089 x ndash3mdash2 983090

2

ndash9mdash2

+ 7

= 2983089 x ndash3mdash2 983090

2

+5mdash2

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Additional Mathematics SPM Chapter 3

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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(j) f ( x) = ( x ndash 4)2

Therefore the minimum point is (4 0)

x 0 4 5

f ( x) 16 0 1

f (x )

x

4 501

16

(k) f ( x) = ndash x2 + 6 x ndash 9

= ndash( x2 ndash 6 x) ndash 9

= ndash( x2 ndash 6 x + 32 ndash 32) ndash 9

= ndash[( x ndash 3)2 ndash 9] ndash 9

= ndash( x ndash 3)2

Therefore the maximum point is (3 0)

x 0 3 4

f ( x) ndash9 0 ndash1

f (x )

x 0

ndash1

ndash9

(3 0)

4

10 (a) x( x ndash 2) 983102 0

0 2x

f (x )

The range of values of x is x 983100 0 or x 983102 2

(b) ( x ndash 3)( x ndash 4) 983100 0

0 43

x

f (x )

The range of values of x is 3 983100 x 983100 4

(c) x2 ndash 3 x ndash 4 983086 0

( x ndash 4)( x + 1) 983086 0

4 ndash1 0 x

f (x )

The range of values of x is x 983084 ndash1 or x 983086 4

(d) 2 x2 + 5 x ndash 3 983084 0

(2 x ndash 1)( x + 3) 983084 0

1 ndash

2 ndash3 0

x

f (x )

The range of values of x is ndash3 983084 x 983084 1mdash2

(e) ( x ndash 3)( x + 2) 983100 ndash4

x2 ndash x ndash 6 983100 ndash4

x2 ndash x ndash 2 983100 0

( x ndash 2)( x + 1) 983100 0

ndash1 20 x

f (x )

The range of values of x is ndash1 983100 x 983100 2

(f) (2 x ndash 1)( x ndash 3) 983100 4( x ndash 3)

2 x2 ndash 6 x ndash x + 3 983100 4 x ndash 12

2 x2 ndash 11 x + 15 983100 0

( x ndash 3)(2 x ndash 5) 983100 0

5 ndash

2

30 x

f (x )

The range of values of x is5mdash2

983100 x 983100 3

(g) x2 + 4

ndashndashndashndashndashndash5

983100 2 x ndash 1

x2 + 4 983100 5(2 x ndash 1)

x2 + 4 983100 10 x ndash 5

x2 ndash 10 x + 9 983100 0

( x ndash 1)( x ndash 9) 983100 0

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Additional Mathematics SPM Chapter 3

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910 x

f (x )

The range of values of x is 1 983100 x 983100 9

(h) x (1 ndash 4 x) 983084 5 x ndash 8

x ndash 4 x2 ndash 5 x + 8 983084 0

ndash 4 x2 ndash 4 x + 8 983084 0

x2 + x ndash 2 983086 0

( x + 2)( x ndash 1) 983086 0

1 ndash2 0 x

f (x )

The range of values of x is x 983084 ndash2 or x 983086 1

1 (a) x-coordinate of the maximum part =1 + 7

ndashndashndashndashndash2

= 4

Therefore the equation of the axis of symmetry

is x = 4

(b) f ( x) = p ndash ( x + q)2

= 5 ndash ( x ndash 4)2

2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1

= 2( x2 ndash 8 x) + k 2 + 2k + 1

= 2983091 x2 ndash 8 x + 983089 8mdash2 983090

2

ndash 983089 8mdash2 983090

2

983092 + k 2 + 2k + 1

= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1

= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1

= 2( x ndash 4)2 + k 2 + 2k ndash 31

Given minimum value = ndash28

there4 k 2 + 2k ndash 31 = ndash28

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0 k = ndash3 1

3 (a) f ( x) = 2( x ndash 3)2 + k

p is the x-coordinate of the minimum point

Therefore p = 3

(b) k is the minimum value of f ( x)

Therefore k = ndash4

(c) The equation of the axis of symmetry is x = 3

4 f ( x) = 3 x2 ndash 2 x + p

a = 3 b = ndash2 c = p

Since the graph does not intersect the x-axis

b2 ndash 4ac lt 0

(ndash2)2 ndash 4(3)( p) lt 0

4 ndash 12 p lt 0

4 lt 12 p

13

lt p

p gt13

5 f ( x) = 2 x2 ndash 12 x + 5

= 2( x2 ndash 6 x) + 5

= 2[( x ndash 3)2 ndash 32] + 5

= 2( x ndash 3)2 ndash 18 + 5

= 2( x ndash 3)2 ndash 13

there4 p = 2 q = ndash3 ndashr + 1 = ndash13

r = 14

6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2

= ndash( x2 ndash 6 px) + 1 ndash 4 p2

= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash

2 983090

2

ndash 983089 6 p ndashndashndash

2 983090

2

983092 + 1 ndash 4 p2

= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 1 + 5 p2

The maximum value given is q2 ndash p

Therefore q2 ndash p = 1 + 5 p2

5 p2 + p + 1 = q2

(b) x = 3 is symmetrical axis 3 p = 3

p = 1

Substitute p = 1 into 5 p2 + p + 1 = q2

5(1)2 + 1 + 1 = q2

q2 = 7

q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067

7 4t (t + 1) ndash 3t 2 + 12 983086 0

4t 2 + 4t ndash 3t 2 + 12 983086 0

t 2 + 4t + 12 983086 0

(t + 2)(t + 6) 983086 0

0 ndash2 ndash6x

f (x )

The range of values of t is t 983084 ndash6 or t 983086 ndash2

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Additional Mathematics SPM Chapter 3

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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash

2 = ndash2

Equation of the axis of symmetry is x = ndash2

2 Let x be the x-coordinate of A

0 + x

ndashndashndashndashndash2

= 3

x = 6

The coordinates of A are (6 4)

3 Let x be the x-coordinate of A

x + 6 ndashndashndashndashndash

2 = 2

x = 4 ndash 6

x = ndash2

The coordinates of A are (ndash2 0)

4 x-coordinate of A =0 + 8

ndashndashndashndashndash2

= 4

Let C be the centre of OB

4

5

A

C O

AC 2

= OA2

ndash OC 2

= 52 ndash 42

= 9

AC = 3

The coordinates of A are (4 3)

5 x-coordinate of minimum point =0 + 4

ndashndashndashndashndash2

= 2

x-coordinate of minimum point for the image is ndash2

6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)

Hence p = 2 and q = ndash1

(b) y = ( x ndash 2)2 ndash 1

When y = 0

( x ndash 2)2 ndash 1 = 0

( x ndash 2)2 = 1

x ndash 2 = plusmn1

x = plusmn1 + 2

= 1 3

Hence A is (1 0) and B is (3 0)

7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090

2

Given (ndash1 k ) is the maximum point

Therefore 2k + 1 = k

k = ndash1

x +1mdash2 p = 0 when x = ndash1

ndash1 + 1mdash2 p = 0

1mdash2 p = 1

p = 2

8 Given ( p 2q) is the minimum point of

y = 2 x2 ndash 4 x + 5

= 2( x2 ndash 2 x) + 5

= 2( x2 ndash 2 x + 12 ndash 12) + 5

= 2[( x ndash 1)2 ndash 1] + 5

= 2( x ndash 1)2 ndash 2 + 5

= 2( x ndash 1)2 + 3

2q = 3

q =3mdash2

p ndash 1 = 0

p = 1

9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1

substitute x = 1 y = 4 into the equation

4 = 12 ndash 2k (1) + 1

2k = ndash2

k = ndash1

(b) y = x2 ndash 2(ndash1) x + 1

= x2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0

10 f ( x) = ndash x2 ndash 8 x + k ndash 1

= ndash( x2 + 8 x) + k ndash 1

= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1

= ndash[( x + 4)2 ndash 16] + k ndash 1

= ndash( x + 4)2 + 16 + k ndash 1

= ndash( x + 4)2 + 15 + k

Since 13 is the maximum value

then 15 + k = 13

k = ndash2

11 f ( x) = 2 x2 ndash 6 x + 7

= 2( x2 ndash 3 x) + 7

= 2983091 x2 ndash 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 + 7

= 2983091983089 x ndash3mdash2 983090

2

ndash9mdash4 983092 + 7

= 2983089 x ndash3mdash2 983090

2

ndash9mdash2

+ 7

= 2983089 x ndash3mdash2 983090

2

+5mdash2

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Additional Mathematics SPM Chapter 3

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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Additional Mathematics SPM Chapter 3

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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910 x

f (x )

The range of values of x is 1 983100 x 983100 9

(h) x (1 ndash 4 x) 983084 5 x ndash 8

x ndash 4 x2 ndash 5 x + 8 983084 0

ndash 4 x2 ndash 4 x + 8 983084 0

x2 + x ndash 2 983086 0

( x + 2)( x ndash 1) 983086 0

1 ndash2 0 x

f (x )

The range of values of x is x 983084 ndash2 or x 983086 1

1 (a) x-coordinate of the maximum part =1 + 7

ndashndashndashndashndash2

= 4

Therefore the equation of the axis of symmetry

is x = 4

(b) f ( x) = p ndash ( x + q)2

= 5 ndash ( x ndash 4)2

2 f ( x) = 2 x2 ndash 16 x + k 2 + 2k + 1

= 2( x2 ndash 8 x) + k 2 + 2k + 1

= 2983091 x2 ndash 8 x + 983089 8mdash2 983090

2

ndash 983089 8mdash2 983090

2

983092 + k 2 + 2k + 1

= 2[( x ndash 4)2 ndash 16] + k 2 + 2k + 1

= 2( x ndash 4)2 ndash 32 + k 2 + 2k + 1

= 2( x ndash 4)2 + k 2 + 2k ndash 31

Given minimum value = ndash28

there4 k 2 + 2k ndash 31 = ndash28

k 2 + 2k ndash 3 = 0

(k + 3)(k ndash 1) = 0 k = ndash3 1

3 (a) f ( x) = 2( x ndash 3)2 + k

p is the x-coordinate of the minimum point

Therefore p = 3

(b) k is the minimum value of f ( x)

Therefore k = ndash4

(c) The equation of the axis of symmetry is x = 3

4 f ( x) = 3 x2 ndash 2 x + p

a = 3 b = ndash2 c = p

Since the graph does not intersect the x-axis

b2 ndash 4ac lt 0

(ndash2)2 ndash 4(3)( p) lt 0

4 ndash 12 p lt 0

4 lt 12 p

13

lt p

p gt13

5 f ( x) = 2 x2 ndash 12 x + 5

= 2( x2 ndash 6 x) + 5

= 2[( x ndash 3)2 ndash 32] + 5

= 2( x ndash 3)2 ndash 18 + 5

= 2( x ndash 3)2 ndash 13

there4 p = 2 q = ndash3 ndashr + 1 = ndash13

r = 14

6 (a) f ( x) = ndash x2 + 6 px + 1 ndash 4 p2

= ndash( x2 ndash 6 px) + 1 ndash 4 p2

= ndash983091 x2 ndash 6 px + 983089 6 p ndashndashndash

2 983090

2

ndash 983089 6 p ndashndashndash

2 983090

2

983092 + 1 ndash 4 p2

= ndash[( x ndash 3 p)2 ndash 9 p2] + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 9 p2 + 1 ndash 4 p2

= ndash ( x ndash 3 p)2 + 1 + 5 p2

The maximum value given is q2 ndash p

Therefore q2 ndash p = 1 + 5 p2

5 p2 + p + 1 = q2

(b) x = 3 is symmetrical axis 3 p = 3

p = 1

Substitute p = 1 into 5 p2 + p + 1 = q2

5(1)2 + 1 + 1 = q2

q2 = 7

q = plusmn9831059831067 Hence p = 1 q = plusmn9831059831067

7 4t (t + 1) ndash 3t 2 + 12 983086 0

4t 2 + 4t ndash 3t 2 + 12 983086 0

t 2 + 4t + 12 983086 0

(t + 2)(t + 6) 983086 0

0 ndash2 ndash6x

f (x )

The range of values of t is t 983084 ndash6 or t 983086 ndash2

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Additional Mathematics SPM Chapter 3

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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash

2 = ndash2

Equation of the axis of symmetry is x = ndash2

2 Let x be the x-coordinate of A

0 + x

ndashndashndashndashndash2

= 3

x = 6

The coordinates of A are (6 4)

3 Let x be the x-coordinate of A

x + 6 ndashndashndashndashndash

2 = 2

x = 4 ndash 6

x = ndash2

The coordinates of A are (ndash2 0)

4 x-coordinate of A =0 + 8

ndashndashndashndashndash2

= 4

Let C be the centre of OB

4

5

A

C O

AC 2

= OA2

ndash OC 2

= 52 ndash 42

= 9

AC = 3

The coordinates of A are (4 3)

5 x-coordinate of minimum point =0 + 4

ndashndashndashndashndash2

= 2

x-coordinate of minimum point for the image is ndash2

6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)

Hence p = 2 and q = ndash1

(b) y = ( x ndash 2)2 ndash 1

When y = 0

( x ndash 2)2 ndash 1 = 0

( x ndash 2)2 = 1

x ndash 2 = plusmn1

x = plusmn1 + 2

= 1 3

Hence A is (1 0) and B is (3 0)

7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090

2

Given (ndash1 k ) is the maximum point

Therefore 2k + 1 = k

k = ndash1

x +1mdash2 p = 0 when x = ndash1

ndash1 + 1mdash2 p = 0

1mdash2 p = 1

p = 2

8 Given ( p 2q) is the minimum point of

y = 2 x2 ndash 4 x + 5

= 2( x2 ndash 2 x) + 5

= 2( x2 ndash 2 x + 12 ndash 12) + 5

= 2[( x ndash 1)2 ndash 1] + 5

= 2( x ndash 1)2 ndash 2 + 5

= 2( x ndash 1)2 + 3

2q = 3

q =3mdash2

p ndash 1 = 0

p = 1

9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1

substitute x = 1 y = 4 into the equation

4 = 12 ndash 2k (1) + 1

2k = ndash2

k = ndash1

(b) y = x2 ndash 2(ndash1) x + 1

= x2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0

10 f ( x) = ndash x2 ndash 8 x + k ndash 1

= ndash( x2 + 8 x) + k ndash 1

= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1

= ndash[( x + 4)2 ndash 16] + k ndash 1

= ndash( x + 4)2 + 16 + k ndash 1

= ndash( x + 4)2 + 15 + k

Since 13 is the maximum value

then 15 + k = 13

k = ndash2

11 f ( x) = 2 x2 ndash 6 x + 7

= 2( x2 ndash 3 x) + 7

= 2983091 x2 ndash 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 + 7

= 2983091983089 x ndash3mdash2 983090

2

ndash9mdash4 983092 + 7

= 2983089 x ndash3mdash2 983090

2

ndash9mdash2

+ 7

= 2983089 x ndash3mdash2 983090

2

+5mdash2

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Additional Mathematics SPM Chapter 3

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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Additional Mathematics SPM Chapter 3

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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1 x-coordinate of maximum point = ndash 4 + 0 ndashndashndashndashndashndash

2 = ndash2

Equation of the axis of symmetry is x = ndash2

2 Let x be the x-coordinate of A

0 + x

ndashndashndashndashndash2

= 3

x = 6

The coordinates of A are (6 4)

3 Let x be the x-coordinate of A

x + 6 ndashndashndashndashndash

2 = 2

x = 4 ndash 6

x = ndash2

The coordinates of A are (ndash2 0)

4 x-coordinate of A =0 + 8

ndashndashndashndashndash2

= 4

Let C be the centre of OB

4

5

A

C O

AC 2

= OA2

ndash OC 2

= 52 ndash 42

= 9

AC = 3

The coordinates of A are (4 3)

5 x-coordinate of minimum point =0 + 4

ndashndashndashndashndash2

= 2

x-coordinate of minimum point for the image is ndash2

6 (a) y = ( x ndash p)2 + q and minimum point is (2 ndash1)

Hence p = 2 and q = ndash1

(b) y = ( x ndash 2)2 ndash 1

When y = 0

( x ndash 2)2 ndash 1 = 0

( x ndash 2)2 = 1

x ndash 2 = plusmn1

x = plusmn1 + 2

= 1 3

Hence A is (1 0) and B is (3 0)

7 f ( x) = 2k + 1 ndash 983089 x + 1mdash2 p983090

2

Given (ndash1 k ) is the maximum point

Therefore 2k + 1 = k

k = ndash1

x +1mdash2 p = 0 when x = ndash1

ndash1 + 1mdash2 p = 0

1mdash2 p = 1

p = 2

8 Given ( p 2q) is the minimum point of

y = 2 x2 ndash 4 x + 5

= 2( x2 ndash 2 x) + 5

= 2( x2 ndash 2 x + 12 ndash 12) + 5

= 2[( x ndash 1)2 ndash 1] + 5

= 2( x ndash 1)2 ndash 2 + 5

= 2( x ndash 1)2 + 3

2q = 3

q =3mdash2

p ndash 1 = 0

p = 1

9 (a) Since (1 4) is the point on y = x2 ndash 2kx + 1

substitute x = 1 y = 4 into the equation

4 = 12 ndash 2k (1) + 1

2k = ndash2

k = ndash1

(b) y = x2 ndash 2(ndash1) x + 1

= x2 + 2 x + 1

= ( x + 1)2

Minimum value of y is 0

10 f ( x) = ndash x2 ndash 8 x + k ndash 1

= ndash( x2 + 8 x) + k ndash 1

= ndash( x2 + 8 x + 42 ndash 42) + k ndash 1

= ndash[( x + 4)2 ndash 16] + k ndash 1

= ndash( x + 4)2 + 16 + k ndash 1

= ndash( x + 4)2 + 15 + k

Since 13 is the maximum value

then 15 + k = 13

k = ndash2

11 f ( x) = 2 x2 ndash 6 x + 7

= 2( x2 ndash 3 x) + 7

= 2983091 x2 ndash 3 x + 983089 3mdash2 983090

2

ndash 983089 3mdash2 983090

2

983092 + 7

= 2983091983089 x ndash3mdash2 983090

2

ndash9mdash4 983092 + 7

= 2983089 x ndash3mdash2 983090

2

ndash9mdash2

+ 7

= 2983089 x ndash3mdash2 983090

2

+5mdash2

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Additional Mathematics SPM Chapter 3

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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Additional Mathematics SPM Chapter 3

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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The minimum point is (3mdash2

5mdash2

)

x ndash1 0 3

f ( x) 15 7 7

ndash ndash

3

5322

0

7

15

x

f (x )

ndash1

The range is5mdash2

983100 f ( x) 983100 15

12 f ( x) = 5 ndash 4 x ndash 2 x2

= ndash2 x2 ndash 4 x + 5

= ndash2( x2 + 2 x) + 5

= ndash2( x2 + 2 x + 12 ndash 12) + 5

= ndash2[( x + 1)2 ndash 1] + 5

= ndash2( x + 1)2 + 2 + 5

= ndash2( x + 1)2 + 7

When f ( x) = ndash1

ndash2( x + 1)2 + 7 = ndash1

ndash2( x + 1)2 = ndash8

( x + 1)2 = 4

x + 1 = plusmn2

x = plusmn2 ndash 1

= ndash3 or 1

13 y = ( x ndash 3)2 ndash 4

Minimum point is (3 ndash4)

x ndash1 0 1 5 6

y 12 5 0 0 5

x 0 1 5 6

5

12

ndash1

(3 ndash4)

y

The range is ndash 4983100 y 983100 12

14 y = ndash x2 + 4 x ndash 5

= ndash( x2 ndash 4 x) ndash 5

= ndash( x2 ndash 4 x + 22 ndash 22) ndash 5

= ndash [( x ndash 2)2 ndash 4] ndash 5

= ndash( x ndash 2)2 + 4 ndash 5

= ndash( x ndash 2)2 ndash 1

Maximum point is (2 ndash1)

x ndash1 0 3

y ndash10 ndash5 ndash2

x

0 3(2 ndash1)

ndash1

ndash5

ndash10

y

The range is ndash10 983100 y 983100 ndash1

15 y = 9 ndash ( x ndash 3)2Maximum point is (3 9)

x ndash1 0 6 7

y 7 0 0 7

x

y

0 ndash1 6

(3 9)

7

7

The range is 0 983100 y 983100 9

16 3 x2 983084 x

3 x2

ndash x 983084 0 x(3 x ndash 1)983084 0

ndash

13

0 x

f (x )

The range is 0 983084 x 983084 1mdash3

17 3 x ndash x

2

ndashndashndashndashndashndash2 983084 1

3 x ndash x2 983084 2

ndash x2 + 3 x ndash 2 983084 0

x2 ndash 3 x + 2 983086 0

( x ndash 1)( x ndash 2) 983086 0

20 x

f (x )

1

The range is x 983084

1 or x 983086

2

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Additional Mathematics SPM Chapter 3

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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18 Given that x ndash 2 y = 1

there4 x = 1 + 2 y 983089

Substitute 983089 into y + 3983102 2 xy

y + 3983102 2(1 + 2 y) y

y + 3983102 2 y + 4 y2

0983102 4 y2 + y ndash 3

0983102 (4 y ndash 3)( y + 1)

3 ndash

4

0 y

f (y )

ndash1

The range is ndash1 983100 y 983100 3mdash4

19 f ( x) 983084 0

5 x2 ndash 4 x ndash 1 983084 0

(5 x + 1)( x ndash 1) 983084 0

1 ndash ndash

5

0 x

f (x )

1

The range is ndash1mdash5

983084 x 983084 1

20 g( x) 983086 0

4 x2 ndash 9 983086 0(2 x + 3)(2 x ndash 3) 983086 0

3 ndash ndash

23 ndash

2

0 x

g (x )

The range is x 983084 ndash3mdash2

or x 983086 3mdash2

21 (a) Since y = 3 x

2

ndash 9 x + t 983086

0 for all values of x andit does not have root when y = 0

Then b2 ndash 4ac 983084 0 for 3 x2 ndash 9 x + t = 0

(ndash9)2 ndash 4(3)(t ) 983084 0

81 ndash 12t 983084 0

ndash12t 983084 ndash81

t 983086 ndash81 ndashndashndashndash ndash12

t 983086 27

ndashndashndash4

(b) Let f ( x) = a( x ndash b)2 + c

f ( x) = a( x ndash 2)2 + 0

f ( x) = a( x ndash 2)2

Substitute x = 0 f ( x) = ndash3 into the equation

ndash3 = a(0 ndash 2)2

= 4a

a = ndash3mdash4

Hence the quadratic function is

f ( x) = ndash3mdash4

( x ndash 2)2

22 (a) Given 2 x2 ndash 3 y + 2 = 0

3 y = 2 x2 + 2

y =2 x2

ndashndashndash3

+2mdash3

983089

Substitute 983089 into y 983084 10

2 x2

ndashndashndash

3

+2mdash

3

983084 10

2 x2 + 2 983084 30

2 x2 ndash 28 983084 0

x2 ndash 14 983084 0

( x + 98310598310698310614 )( x ndash 98310598310698310614 ) 983084 0

14 ndash14

0 x

f (x )

The range is ndash98310598310698310614 983084 x 983084 98310598310698310614

(b) 2 x2 ndash 8 x ndash 10 = 2( x2 ndash 4 x) ndash 10

= 2( x2 ndash 4 x + 22 ndash 22) ndash 10

= 2[( x ndash 2)2 ndash 4] ndash 10

= 2( x ndash 2)2 ndash 8 ndash 10

= 2( x ndash 2)2 ndash 18

Therefore a = 2 b = ndash2 and c = ndash18

Hence the minimum value of 2 x2 ndash 8 x ndash 10 is

ndash18

23 5 ndash 2 x 983100 0 3 x2 ndash 4 x 983086 ndash1

3 x2

ndash 4 x + 1983086

0(3 x ndash 1)( x ndash 1) 983086 0

11 ndash

3

0 x

f (x )

x 983084 1mdash3

x 983086 1

ndash2 x 983100

ndash5 x 983102

ndash5 ndashndashndash ndash2

x 983102 5mdash2

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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5ndash

2

1ndash

3

1

x 5mdash2

x x 11mdash3

The range is x 983102 5mdash2

24 5 983084 f ( x) 983084 9

5 983084 5 ndash 3 x + x2 983084 9

5 983084 5 ndash 3 x + x2 5 ndash 3 x + x2 983084 9

5 ndash 3 x + x2 ndash 9 983084 0

x2 ndash 3 x ndash 4 983084 0

( x ndash 4)( x + 1) 983084 0

4 ndash1 0 x

f (x )

ndash1 983084 x 983084 4

0 983084 x2 ndash 3 x

0 983084 x( x ndash 3)

30 x

f (x )

x 983084 0 x 983086 3

0 ndash1 43

x lt 0 x gt 3

ndash1 lt x lt 4

The range is ndash1983084

x 983084

0 or 3983084

x 983084

4

25 1 983102 x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x ndash 3 983086 ndash3

x2 + 3 x 983086 0

x( x + 3) 983086 0

ndash3 0 x

f (x )

x 983084 ndash3 x 983086 0

1 983102 x2 + 3 x ndash 3

0 983102 x2 + 3 x ndash 4

0 983102 ( x + 4)( x ndash 1)

1 ndash4 0 x

f (x )

ndash 4 983100 x 983100 1

ndash4

x lt ndash3 x gt 0

ndash4 x 1

ndash3 0 1

The range is ndash 4983100 x 983084 ndash3 or 0 983084 x 983100 1

26 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 10 983084 ndash 4 x2 + 5 x ndash 10 983086 4

x2 + 5 x ndash 14 983086 0

( x + 7)( x ndash 2) 983086 0

ndash7 20

x

f (x )

x 983084 ndash7 x 983086 2

x2 + 5 x ndash 6 983084 0

( x + 6)( x ndash 1) 983084 0

ndash6 10

x

f (x )

ndash6 983084 x 983084 1

The ranges are ndash6 lt x lt 1 x lt ndash7 x gt 2

27 f ( x) = (r + 1) x2 + 2rx + r ndash 3

Given that f ( x) does not intersect the x-axis

therefore b2 ndash 4ac 983084 0

(2r)2 ndash 4(r + 1)(r ndash 3) 983084 0

4r2 ndash 4(r2 ndash 2r ndash 3) 983084 0

4r2 ndash 4r2 + 8r + 12 983084 0

2r + 3 983084 0

r 983084 ndash3mdash2

28 Given that f ( x) = 9 ndash 6 x + 2 x2 does not have real root

when f ( x) = k

there4 9 ndash 6 x + 2 x2 = k

2 x2 ndash 6 x + 9 ndash k = 0

Use b2 ndash 4ac 983084 0

(ndash 6)2 ndash 4(2)(9 ndash k ) 983084 0

36 ndash 72 + 8k 983084 0

ndash36 + 8k 983084 0 8k 983084 36

k 983084 36

ndashndashndash8

k 983084 9mdash2

29 2 x2 + 10 x ndash 20 983100 8

ndash8 983100 2 x2 + 10 x ndash 20 983100 8

ndash8983100 2 x2 + 10 x ndash 20 2 x2 + 10 x ndash 20 983100 8

2 x2 + 10 x ndash 28 983100 0

x2 + 5 x ndash 14 983100 0

( x + 7)( x ndash 2) 983100 0

ndash7 20

x

f (x )

ndash7 983100 x 983100 2

0983100 2 x2 + 10 x ndash 12

0983100 x2 + 5 x ndash 6

0983100 ( x + 6)( x ndash 1)

ndash6 10

x

f (x )

x 983100 ndash6 x 983102 1

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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x ndash6 x 1

ndash7 x 2

ndash6 ndash7 1 2

The range is ndash7 983100 x 983100 ndash6 or 1 983100 x 983100 2

30 y = x2 + 5 x ndash 6

= x2 + 5 x + 983089 5mdash2 983090

2

ndash 983089 5mdash2 983090

2

ndash 6

= 983089 x + 5mdash2 983090

2

ndash25

ndashndashndash4

ndash 6

= 983089 x + 5mdash2 983090

2

ndash49

ndashndashndash4

The minimum point is (ndash5mdash2

ndash49

ndashndashndash4

)

x ndash13 ndash 6 0 1 3

y 98 0 ndash 6 0 18

x 0

ndash6 ndash13 ndash6

18

98

1 3

) ) ndash ndash ndash ndash

5

2

49

4

y

The range is ndash49

ndashndashndash4

983084 y 983084 98

31 y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + (2 ndash n) x + 16

= ndash[ x2 ndash (2 ndash n) x] + 16

= ndash983091 x2 ndash (2 ndash n) x + 983089 2 ndash n ndashndashndashndashndash

2 983090

2

ndash 983089 2 ndash n ndashndashndashndashndash

2 983090

2

983092 + 16

= ndash

983091983089 x ndash

2 ndash n ndashndashndashndashndash

2 983090

2

ndash

9830892 ndash n

ndashndashndashndashndash2 983090

2

983092 + 16

= ndash983089 x ndash 2 ndash n ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash n ndashndashndashndashndash

2 983090

2

+ 16

Since y = x2 ndash 3k and y = ndash x2 + 2 x ndash nx + 16 have the

same axis of symmetry that is x = 0

then ndash2 ndash n

ndashndashndashndashndash2

= 0

2 ndash n = 0

n = 2

The equation y = ndash x2 + 2 x ndash nx + 16

= ndash x2 + 2 x ndash 2 x + 16

= ndash983089 x ndash 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 983089 2 ndash 2 ndashndashndashndashndash

2 983090

2

+ 16

= ndash x2 + 16

When y = 0

ndash x2

+ 16 = 0 x2 = 16

x = plusmn98310598310698310616

= plusmn4

Therefore B = (4 0)

Substitute x = 4 y = 0 into y = x2 ndash 3k

0 = 42 ndash 3k

0 = 16 ndash 3k

k =16

ndashndashndash3

32 (a) y = ndash2[(3k ndash x)2 + n] ndash 10 has a maximum point

(4 11)

y = ndash2(3k ndash x)2 ndash 2n ndash 10

there4 3k ndash 4 = 0 and ndash2n ndash 10 = 11

k =4mdash3

ndash2n = 21

n = ndash21

ndashndashndash2

(b) Substitute k =4mdash3

and n = ndash21

ndashndashndash2

into

y = ndash2[(3k ndash x)2 + n] ndash 10

y = ndash2983091(4 ndash x)2 ndash21

ndashndashndash2 983092 ndash 10

= ndash2(4 ndash x)2 + 21 ndash 10

= ndash2(4 ndash x)2

+ 11 When y = 0

ndash2(4 ndash x)2 + 11= 0

2(4 ndash x)2 = 11

(4 ndash x)2 =11

ndashndashndash2

4 ndash x = plusmn98310598310698310698310611 ndashndashndash

2

x = 4 plusmn 98310598310698310698310611 ndashndashndash

2

= 4 ndash 98310598310698310698310611 ndashndashndash

2 4 + 98310598310698310698310611

ndashndashndash2

= 1655 6345

(c) y = ndash2(4 ndash x)2 + 11

The maximum point is (4 11)

x ndash1 0 5

y ndash39 ndash21 9

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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1 y = 3(2 x ndash 1)( x + 1) ndash x(4 x ndash 5) + 2

= 3(2 x2 + x ndash 1) ndash 4 x2 + 5 x + 2

= 6 x2 + 3 x ndash 3 ndash 4 x2 + 5 x + 2

= 2 x2 + 8 x ndash 1

= 2( x2 + 4 x) ndash 1

= 2( x2 + 4 x + 22 ndash 22) ndash 1

= 2[( x + 2)2 ndash 4] ndash 1

= 2( x + 2)2 ndash 8 ndash 1

= 2( x + 2)2 ndash 9

Since a = 2 983086 0 therefore the minimum value of

y is ndash9

When y = 0 2( x + 2)2 ndash 9 = 0

( x + 2)2 =9mdash2

x + 2 = plusmn9831059831069831069mdash2 x = plusmn9831059831069831069mdash2 ndash 2

= 9831059831069831069mdash2 ndash 2 or ndash9831059831069831069mdash2 ndash 2

= 01213 or ndash 4121

When x = 0 y = 2(2)2 ndash 9

= ndash1

The minimum point is (ndash2 ndash9)

ndash1 ndash4121

(ndash2 ndash9)

012130

y

x

2 5 983084 Area of rectangle ABCD 983084 21

5 983084 ( x + 3)( x ndash 1) 983084 21

5 983084 ( x + 3)( x ndash 1)

5 983084 x2 + 2 x ndash 3

0 983084 x2 + 2 x ndash 8

0 983084 ( x + 4)( x ndash 2)

0 ndash4 2

f (x )

x

x 983084 ndash 4 x gt 2

( x + 3)( x ndash 1) 983084 21

x2 + 2 x ndash 3 983084 21

x2 + 2 x ndash 24 983084 0

( x ndash 4)( x + 6) 983084 0

0 ndash6 4

f (x )

x

ndash 6 983084 x 983084 4

ndash4

ndash6 lt x lt 4

x lt ndash4 x gt 2

2

x

ndash6 4

The range is ndash 6983084 x 983084 ndash 4 or 2 983084 x 983084 4

3 (a) p =1 + 5

ndashndashndashndashndash2

= 3

(b) y = ( x ndash 1)( x ndash 5) = x2 ndash 6 x + 5

4 y = a( x ndash 2)2 + 1

Substitute x = 0 y = 9 into the equation

9 = a(ndash2)2 + 1

8 = 4a

a = 2

Therefore the quadratic function is

f ( x) = 2( x ndash 2)2 + 1

5 x2

+ (1 + k ) x ndash k 2

+ 1 = 0For quadratic equation to have real roots

b2 ndash 4ac 983102 0

(1 + k )2 ndash 4(1)(1 ndash k 2) 983102 0

1 + 2k + k 2 ndash 4 + 4k 2 983102 0

5k 2 + 2k ndash 3 983102 0

(5k ndash 3)(k + 1) 983102 0

0 ndash1 3 ndash

5

f (k )

k

The range of values of k is k 983100 ndash1 or k 983102 3mdash5

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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6 y = x2 + 7 x ndash 8 ndash 2k

For y to be positive for all real values of x there is

no roots for y = 0

Therefore b2 ndash 4ac 983084 0

72 ndash 4(1)(ndash8 ndash 2k ) 983084 0

49 + 32 + 8k 983084 0

8k 983084 ndash81

k 983084 ndash81

ndashndashndash8

Alternative

y = x2 + 7 x ndash 8 ndash 2k

= x2 + 7 x + 983089 7mdash2 983090

2

ndash 983089 7mdash2 983090

2

ndash 8 ndash 2k

= 983089 x +7mdash2 983090

2

ndash49

ndashndashndash4

ndash 8 ndash 2k

For y to be positive for all real values of x

ndash49

ndashndashndash4

ndash 8 ndash 2k 983086 0

ndash2k 983086 49

ndashndashndash4

+ 8

ndash2k 983086 81

ndashndashndash4

k 983084 ndash81

ndashndashndash8

7 Substitute x = 6 y = 0 into y = px2 + qx

0 = p(6)2 + q(6)

0 = 36 p + 6q

q + 6 p = 0 983089

y = px2 + qx

= p983089 x2 +q mdash p x983090

= p983091 x2 +q mdash p x + 983089

q ndashndashndash2 p 983090

2

ndash 983089q

ndashndashndash2 p 983090

2

983092

= p983091983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndashndash4 p2 983092

= p983089 x +q

ndashndashndash2 p 983090

2

ndashq2

ndashndashndash4 p

ndash q2

ndashndashndash4 p

= ndash12

q2 = 48 p

p =q2

ndashndashndash48

983090

Substitute 983090 into 983089

q + 6983089 q2

ndashndashndash48

983090 = 0

q +q2

mdash8

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = ndash8

When q = 0 p =02

ndashndashndash48

= 0

When q = ndash8 p =(ndash8)2

ndashndashndashndashndash48

=64

ndashndashndash48

=4mdash3

Therefore the values of p =

4

mdash3 and q = ndash8

8 (2 ndash 3k ) x2 + x +3mdash4k = 0

b2 ndash 4ac = 12 ndash 4(2 ndash 3k )983089 3mdash4k 983090

= 1 ndash 6k + 9k 2

= 9k 2 ndash 6k + 1

= (3k ndash 1)2

Since (3k ndash 1)2 983102 0 for all values of k

therefore (2 ndash 3k ) x2 + x +3mdash4k = 0 has real roots for

all values of k

9 f ( x) = 3( x2 + 2mx + m2 + n)

= 3[( x + m)2 + n]

= 3( x + m)2 + 3n

The minimum point is (ndashm 3n)

Compare to A(t 3t 2)

there4 ndashm = t and 3n = 3t 2

m = ndasht n = t 2

10 (a) y = px2 + 8 x + 10 ndash p

When the graph does not intercept the x-axis

there are no roots for px2 + 8 x + 10 ndash p = 0 Therefore b2 ndash 4ac 983084 0

82 ndash 4 p(10 ndash p) 983084 0

64 ndash 40 p + 4 p2 983084 0

p2 ndash 10 p + 16 983084 0

( p ndash 2)( p ndash 8) 983084 0

2 8

Hence r = 2 and t = 8

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Additional Mathematics SPM Chapter 3

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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(b) When p = 2

y = 2 x2 + 8 x + 8

= 2( x2 + 4 x) + 8

= 2( x2 + 4 x + 22 ndash 22) + 8

= 2[( x + 2)2 ndash 4] + 8

= 2( x + 2)2 ndash 8 + 8

= 2( x + 2)2

Therefore the minimum point is (ndash2 0)

When x = 0 y = 8

When y = 0 2( x + 2)2 = 0

x = ndash2

When p = 8

y = 8 x2 + 8 x + 2

= 8( x2 + x) + 2

= 8983091 x2 + x + 983089 1mdash2 983090

2

ndash 983089 1mdash2 983090

2

983092 + 2

= 8983091983089 x +1mdash2 983090

2

ndash1mdash4 983092 + 2

= 8983089 x +1mdash2 983090

2

ndash 2 + 2

= 8983089 x +1mdash2 983090

2

Therefore the minimum point is (ndash1mdash2

0)

When x = 0 y = 2

When y = 0 0 = 8983089 x + 1mdash2 983090

2

x = ndash1mdash2

0

2

8

1 ndash ndash

2

ndash2

y p = 2 p = 8

x

11 (a) f ( x) = 24 x ndash 4 x2 + r

= ndash 4 x2 + 24 x + r

= ndash 4( x2 ndash 6 x) + r

= ndash 4( x2 ndash 6 x + 32 ndash 32) + r

= ndash 4[( x ndash 3)2 ndash 9] + r

= ndash 4( x ndash 3)2 + 36 + r

Compare to f ( x) = p( x ndash q)2 + 16

Therefore p = ndash 4 q = 3 and 36 + r = 16

r = ndash20

(b) The turning point is (3 16)

(c) f ( x) = 24 x ndash 4 x2 ndash 20

When x = 0 f ( x) = ndash20

When f ( x) = 0 ndash 4( x ndash 3)2 + 16 = 0

4( x ndash 3)2 = 16

( x ndash 3)2 = 4

x ndash 3 = plusmn2

x = plusmn2 + 3

= ndash2 + 3 or 2 + 3

= 1 or 5

0 1

(3 16)

5

ndash20

y

x

12 (a) y = ndash| p( x ndash 3)2 + q|

Substitute x = 3 y = ndash5 into the equation

ndash5 = ndash| p(3 ndash 3)2 + q| 5 = |q| q = plusmn5

Substitute x = 4 y = 0 into the equation

0 = ndash| p(4 ndash 3)2 plusmn 5| p plusmn 5 = 0

p = 5

Therefore p = 5 q = ndash5 or p = ndash5 q = 5

(b) When x = 3 y = ndash5

For p = 5 q = ndash5

When x = 6 y = ndash|5(6 ndash 3)2 ndash 5| = ndash|40| = ndash40

Based on the graph the range of values of y is

ndash 40 983100 y 983100 0

For p = ndash5 q = 5

When x = 6 y = ndash| ndash5(6 ndash 3)2 + 5| = ndash| ndash 40| = ndash 40

Therefore the range of values of y is

ndash 40 983100 y 983100 0

13 (a) y = ndash2( x ndash 3)2 + 2k

= ndash x2 + 2 x + px ndash 8

= ndash x2 + (2 + p) x ndash 8

= ndash[ x2 ndash (2 + p) x] ndash 8

= ndash983091 x2 ndash (2 + p) x + 983089 2 + p ndashndashndashndashndash

2 9830902

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983091983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

ndash 983089 2 + p ndashndashndashndashndash

2 9830902

983092 ndash 8

= ndash983089 x ndash2 + p

ndashndashndashndashndash2 983090

2

+(2 + p)2

ndashndashndashndashndashndashndash4

ndash 8

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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Since the x-coordinate of the maximum point for

both the graphs are same

therefore2 + p

ndashndashndashndashndash2

= 3

p = 4

y = ndash x2 + 2 x + px ndash 8 becomes

y = ndash x2 + 2 x + 4 x ndash 8 y = ndash x2 + 6 x ndash 8

When y = 0

ndash x2 + 6 x ndash 8 = 0

x2 ndash 6 x + 8 = 0

( x ndash 2)( x ndash 4) = 0

x = 2 or 4

Hence A(2 0) and B(4 0)

Substitute x = 2 y = 0 into y = ndash2( x ndash 3)2 + 2k

0 = ndash2(2 ndash 3)2 + 2k

2k = 2

k = 1 Hence k = 1 and p = 4

(b) For y = ndash2( x ndash 3)2 + 2k

= ndash2( x ndash 3)2 + 2(1)

= ndash2( x ndash 3)2 + 2

Maximum value of the curve is 2

For y = ndash x2 + 2 x + px ndash 8

= ndash x2 + 2 x + 4 x ndash 8

= ndash x2 + 6 x ndash 8

When x = 3

y = ndash9 + 18 ndash 8

= 1

Maximum value of the curve is 1

14 Since 3 x2 983102 0 for all values of x

therefore3 x2

ndashndashndashndashndashndashndashndashndashndashndashndashndash(2 x ndash 1)( x + 4)

983100 0

(2 x ndash 1)( x + 4) 983100 0

0 ndash4 1 ndash

2

f (x )

x

Hence ndash 4983100 x 983100 1mdash2

15 Since x2 + 1 983086 0

therefore x2 + 3 x + 2 ndashndashndashndashndashndashndashndashndashndash

x2 + 1 983086 0

x2 + 3 x + 2 983086 0

( x + 1)( x + 2) 983086 0

0 ndash2 ndash1

f (x )

x

Hence x 983084 ndash2 x gt ndash1

16 ndash4

ndashndashndashndashndashndash1 ndash 3 x

983100 x

0 983100 x +4

ndashndashndashndashndashndash1 ndash 3 x

0 983100 x(1 ndash 3 x) + 4

ndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0983100

x ndash 3 x2 + 4

ndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

0 983100 ndash3 x2 + x + 4 ndashndashndashndashndashndashndashndashndashndashndash

1 ndash 3 x

0 983100 (ndash3 x + 4)( x + 1)

ndashndashndashndashndashndashndashndashndashndashndashndashndashndash1 ndash 3 x

For ndash3 x + 4 983102 0

4 983102 3 x

x 983100 4mdash3

For x + 1 983102 0

x 983102 ndash1

For 1 ndash 3 x 983086 0

ndash3 x 983086 ndash1

x 983084 1mdash3

ndash1 ndash + ndash +

1 4 ndash ndash

3

4x ndash

3

1x lt ndash

3

x ndash1

3

x

Therefore the range is ndash1 983100 x 983084 1mdash3

x 983102 4mdash3

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4

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Additional Mathematics SPM Chapter 3

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17 y2 ndash 9 = x

x = y2 ndash 9

When x = 0

y2 = 9

y = plusmn3

When y = 0

x = ndash9

When x = 7

7 = y2 ndash 9

y2 = 16

y = plusmn4

x ndash9 0 7

y 0 plusmn3 plusmn4

0 ndash9 7

ndash4 ndash3

34

y

x

The range of values of y is ndash 4 983100 y 983100 4