Addmath Chapter 3

7/30/2019 Addmath Chapter 3

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Additional Mathematics SPM Chapter 3

Penerbitan Pelangi Sdn. Bhd.

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

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

Whenx = 2,

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

= 16 + 16 + 6

= 38

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

Whenf(x) = 0,

x2 3x + 2 = 0

(x 1)(x 2) = 0

x 1 = 0 or x 2 = 0

x = 1 or x = 2

4. f(x) = 3x2 + 5x 1

Whenf(x) = 1,

3x2 + 5x 1 = 1

3x2 + 5x 1 1 = 0

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

3x 2 = 0 or x 1 = 0

x =23

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 10.

(d) f(x) = 23 12 (x 3)2 +

14 4

= (x 3)2 +12

Therefore, the minimum value is12

.

(e) f(x) =13

[6 (x + 1)2] + 5

= 2 13

(x + 1)2 + 5

= 13

(x + 1)2 + 7

Therefore, the maximum value is 7.

(f) The minimum value is 3.

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

=x2 4x + 1 42 22

1 42 22

+ 2

= (x 2)2 4 + 2

= (x 2)2 2

Hence, the minimum value is 2.

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

= 2(x2 + 3x) 5

= 23x2 + 3x + 1 32 22

1 32 22

4 5

= 231x + 32 22

94 4 5

= 21x + 32 22

92

5

= 21x + 32 22

19

2

Hence, the minimum value is 19

2

.

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

=x2 + 5x + 1 52 22

1 52 22

= 1x + 52 22

25

4


4

.

3 Quadratic Functions


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(d) f(x) = 6x x2

= (x2 6x)

= 3x2 6x + 1 62 22

1 62 22

4= [(x 3)2 9]

= (x 3)2 + 9

Hence, the maximum value is 9.

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

= x2 4x + 3

= (x2 + 4x) + 3

= 3x2 + 4x + 1 42 22

1 42 22

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

= (x + 2)2 + 4 + 3

= (x + 2)2 + 7


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

= 2x2 + 4x

= 2(x2

2x)= 23x2 2x + 1 22 2

2

1 22 22

4= 2[(x 1)2 1]

= 2(x 1)2 + 2


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

= 3x2 + 5x + 10

= 31x2 53x2 + 10

= 33x2 53x + 156 2

2

1 56 22

4 + 10

= 331x 56 22

25

36 4 + 10

= 31x 56 22

+25

12

+ 10

= 31x 56 22

+145

12

Hence, the maximum value is145

12

.

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

= 2x2 + 5x 3

= 21x2

+

5

2x2 3= 23x2 + 52x + 1

54 2

2

1 54 22

4 3

= 231x + 54 22

25

16 4 3

= 21x + 54 22

25

8

3

= 21x + 54 22

49

8


8

.

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

=x + 2 4x2 8x

= 4x2 7x + 2

= 41x2 + 74x2 + 2

= 43x2 + 74x + 178 2

2

1 78 22

4 + 2

= 431x + 78 22

49

64 4 + 2

= 41x + 78 22

+49

16

+ 2

= 41x + 78 22

+81

16

Hence, the maximum value is81

16

.

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

Therefore, the minimum point is (0, 4).

x 2 2 4f(x) 0 0 12

f(x)

x

42 2 4

12

0

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


x 1 3

f(x) 8 32

f(x)

x

1 3

5

8

32

0

(c) f(x) = 8 x2

Therefore, the maximum point is (0, 8).

x 3 AB8 3

f(x) 1 0 1

f(x)

x

1

8

3 3 8 8

0


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(d) f(x) = 10 2x2


x 3 AB5 4

f(x) 8 0 22

f(x)

x

8

10

3 4 5 50

22

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

=x2 + 2x

=x2 + 2x + 12 12

= (x + 1)2 1


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

f(x)

x

(1,1)4

8

2 20

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

= 2x2

x 1= 21x2 x2 2 1= 23x2 x2 + 1

14 2

2

1 14 22

4 1

= 231x 14 22

1

16 4 1

= 21x 14 22

18

1

= 21x 14 22

98

Therefore, the minimum point is (

1

4 ,

9

8 ).

x 1 12

0 1 2

f(x) 2 0 1 0 5

f(x)

x

1 21 1

2

98)14 (,

2

0

5

1

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


x 2 0 3 AB5 3 + AB5 6

f(x) 20 4 0 0 4

f(x)

x

24

20

(3, 5)

63 + 53 50

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

=x2 + 4x + 22 22 + 5

= (x + 2)2 + 1


x 3 0 1f(x) 2 5 10

f(x)

x

(2, 1)

3 1

2

5

10

0

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

= 2(x2

+ 3x) 8

= 23x2 + 3x + 1 32 22

1 32 22

4 8

= 231x + 32 22

94 4 8

= 21x + 32 22

92

8

= 21x + 32 22

25

2

Therefore, the minimum point is (32

, 25

2

).

x 3 0 1 2f(x) 8 8 0 12

25

2)32( ,

f(x)

x

3

12

8

0 1 2


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


x 0 4 5

f(x) 16 0 1

f(x)

x

4 501

16

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

= (x2 6x) 9

= (x2 6x + 32 32) 9

= [(x 3)2 9] 9

= (x 3)2


x 0 3 4

f(x) 9 0 1

f(x)

x0

1

9

(3, 0)

4

10. (a) x(x 2) > 0

0 2x

f(x)

The range of values ofx isx< 0 or x> 2.

(b) (x 3)(x 4) < 0

0 43

x

f(x)

The range of values ofx is 3


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910x

f(x)

The range of values ofx is 1 3

x2 + 2x 3 > 0

(x + 3)(x 1) > 0

13 0x

f(x)

The range of values ofx isx< 3 orx> 1.

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

= 2(x2 6x) + 5

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

= 2(x 3)2 18 + 5

= 2(x 3)2 13

\p = 2, q = 3 , r + 1 = 13

r = 14

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

= (x2 6px) + 1 4p2

= 3x2 6px + 16p

2

22

1 6p2

22

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

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

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

The maximum value given is q2 p.

Therefore, q2 p = 1 + 5p2

5p2 +p + 1 = q2

(b) x = 3 is symmetrical axis

3p = 3

p = 1

Substitutep = 1 into 5p2 +p + 1 = q2,

5(1)2 + 1 + 1 = q2

q2 = 7

q = AB7Hence,p = 1, q = AB7

7. 4t(t+ 1) 3t2 + 12 . 0

4t2 + 4t 3t2 + 12 . 0

t2 + 4t+ 12 . 0(t+ 2)(t+ 6) . 0

026x

f(x)

The range of values oftis t, 6 or t. 2.


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1. x-coordinate of maximum point = 4 + 0

2= 2

Equation of the axis of symmetry isx = 2

2. Letx be thex-coordinate ofA

0 + x

2

= 3

x = 6

The coordinates ofA are (6, 4).

3. Letx be thex-coordinate ofA

x + 6

2= 2

x = 4 6

x = 2


4. x-coordinate ofA =0 + 8

2

= 4

Let Cbe the centre ofOB,

4

5

A

CO

AC2

= OA2

OC2

= 52 42

= 9

AC= 3


5. x-coordinate of minimum point =0 + 4

2

= 2

x-coordinate of minimum point for the image is 2.

6. (a) y = (x p)2 + q and minimum point is (2, 1)

Hence,p = 2 and q = 1

(b) y = (x 2)2 1

Wheny = 0,

(x 2)2 1 = 0

(x 2)2 = 1

x 2 = 1

x = 1 + 2

= 1, 3

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

7. f(x) = 2k+ 1 1x + 12p22

Given (1, k) is the maximum point.

Therefore, 2k+ 1 = k

k = 1

x +12p = 0 whenx = 1,

1 + 12p = 0

12p = 1

p = 2

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

y = 2x2 4x + 5

= 2(x2 2x) + 5

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

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

= 2(x 1)2 2 + 5

= 2(x 1)2 + 3

2q = 3

q =32

p 1 = 0

p = 1

9. (a) Since (1, 4) is the point ony =x2 2kx + 1,

substitutex = 1, y = 4 into the equation,

4 = 12 2k(1) + 1

2k= 2

k= 1

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

=x2 + 2x + 1

= (x + 1)2

Minimum value ofy is 0.

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

= (x2 + 8x) + k 1

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

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

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

= (x + 4)2 + 15 + k

Since 13 is the maximum value,

then 15 + k= 13

k= 2

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

= 2(x2 3x) + 7

= 23x2 3x + 1 32 22

1 32 22

4 + 7

= 231x 32 22

94 4 + 7

= 21x 32 22

92

+ 7

= 21x 32 22

+52


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The minimum point is (32

,52

).

x 1 0 3

f(x) 15 7 7

,

3

5322

0

7

15

x

f(x)

1

The range is52


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18. Given thatx 2y = 1,

\x = 1 + 2y .......................................1

Substitute 1 into y + 3 > 2xy, y + 3 > 2(1 + 2y)y

y + 3 > 2y + 4y2

0> 4y2 +y 3

0> (4y 3)(y + 1)

3

4

0y

f(y)

1

The range is 1 52


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5

2

1

3

1

x52

x x 113

The range isx>52

.

24. 5 , f(x) , 9

5 , 5 3x +x2, 9

5 , 5 3x +x2 , 5 3x +x2, 9

5 3x +x2 9 , 0

x2 3x 4 , 0

(x 4)(x + 1) , 0

41 0x

f(x)

1 ,x, 4

0 ,x2 3x

0 ,x(x 3)

30x

f(x)

x, 0,x. 3

01 43

x< 0 x> 3

1 < x< 4

The range is 1,

x,

0 or 3,

x,

4.

25. 1 >x2 + 3x 3 . 3

x2 + 3x 3 . 3

x2 + 3x. 0

x(x + 3) . 0

3 0x

f(x)

x, 3,x. 0

1 >x2 + 3x 3 ,

0 >x2 + 3x 4

0 > (x + 4)(x 1)

14 0x

f(x)

4 0

4x 1

3 0 1

The range is 4


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10



x 6 x 1

7x 2

67 1 2

The range is 7


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11



(1, 39)

21

(5, 9)

(4, 11)

0x

y

The range is 39


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12



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

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

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

= 2x2 + 8x 1

= 2(x2 + 4x) 1

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

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

= 2(x + 2)2 8 1

= 2(x + 2)2 9

Since a = 2 . 0, therefore the minimum value of

y is 9.

Wheny = 0, 2(x + 2)2 9 = 0

(x + 2)2 =92

x + 2 = ABB92 x = ABB92 2

= ABB92 2 or ABB92

2

= 0.1213 or 4.121

Whenx = 0, y = 2(2)2 9

= 1

The minimum point is (2, 9).

14.121

(2, 9)

0.12130

y

x

2. 5 , Area of rectangleABCD, 21

5 , (x + 3)(x 1) , 21

5 , (x + 3)(x 1)

5 ,x2 + 2x 3

0 ,x2 + 2x 8

0 , (x + 4)(x 2)

04 2

f(x)

x

x, 4,x > 2

(x + 3)(x 1) , 21

x2 + 2x 3 , 21

x2 + 2x 24 , 0

(x 4)(x + 6) , 0

06 4

f(x)

x

6 ,x, 4

4

6 < x < 4

x < 4 x > 2

2

x

6 4

The range is 6,x, 4 or 2 ,x, 4.

3. (a) p =1 + 5

2

= 3

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

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

Substitutex = 0, y = 9 into the equation,

9 = a(2)2 + 1

8 = 4a

a = 2

Therefore, the quadratic function is

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

5. x2

+ (1 + k)x k2

+ 1 = 0For quadratic equation to have real roots,

b2 4ac> 0

(1 + k)2 4(1)(1 k2) > 0

1 + 2k+ k2 4 + 4k2> 0

5k2 + 2k 3 > 0

(5k 3)(k+ 1) > 0

01 3

5

f(k)

k

The range of values ofkis k< 1 or k>35

.


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6. y =x2 + 7x 8 2k

For y to be positive for all real values ofx, there is

no roots fory = 0.

Therefore, b2 4ac, 0

72 4(1)(8 2k) , 0

49 + 32 + 8k, 0

8k, 81

k, 81

8

Alternative

y =x2 + 7x 8 2k

=x2 + 7x + 1 72 22

1 72 22

8 2k

= 1x + 72 22

49

4

8 2k

Fory to be positive for all real values ofx,

49

4

8 2k. 0

2k.49

4

+ 8

2k.81

4

k, 81

8

7. Substitutex = 6, y = 0 into y =px2 + qx,

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

0 = 36p + 6q

q + 6p = 0 .........................1

y =px2 + qx

=p1x2 +qpx2

=p3x2 +qpx + 1

q2p 2

2

1q

2p 2

2

4

=p31x +q

2p 2

2

q2

4p2 4

=p1x +q

2p 2

2

q2

4p

q2

4p

= 12

q2 = 48p

p =q2

48

..........................................................2

Substitute 2 into 1,

q + 61 q2

48

2 = 0 q +

q28

= 0

8q + q2 = 0

q(8 + q) = 0

q = 0 or q = 8

When q = 0, p =02

48

= 0

When q = 8, p =(8)2

48

=64

48

=43

Therefore, the values ofp =

4

3 and q = 8.

8. (2 3k)x2 +x +34k= 0

b2 4ac = 12 4(2 3k)1 34 k2= 1 6k+ 9k2

= 9k2 6k+ 1

= (3k 1)2

Since (3k 1)2> 0 for all values ofk,

therefore, (2 3k)x2 +x +34k= 0 has real roots for

all values ofk.

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

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

= 3(x + m)2 + 3n

The minimum point is (m, 3n)

Compare toA(t, 3t2),

\ m = t and 3n = 3t2

m = t n = t2

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

When the graph does not intercept the x-axis,

there are no roots forpx2 + 8x + 10 p = 0.Therefore, b2 4ac, 0

82 4p(10 p) , 0

64 40p + 4p2, 0

p2 10p + 16 , 0

(p 2)(p 8) , 0

2 8

Hence, r = 2 and t= 8


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(b) Whenp = 2,

y = 2x2 + 8x + 8

= 2(x2 + 4x) + 8

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

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

= 2(x + 2)2 8 + 8

= 2(x + 2)2


Whenx = 0, y = 8

Wheny = 0, 2(x + 2)2 = 0

x = 2

Whenp = 8,

y = 8x2 + 8x + 2

= 8(x2 +x) + 2

= 83x2 +x + 1 12

22

1 12

22

4 + 2

= 831x + 12

22

14

4 + 2

= 81x + 12

22

2 + 2

= 81x + 12

22

Therefore, the minimum point is (12

, 0).

Whenx = 0, y = 2

Wheny = 0, 0 = 81x + 12

22

x = 12

0

2

8

1

2

2

yp= 2 p= 8

x

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

= 4x2 + 24x + r

= 4(x2 6x) + r

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

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

= 4(x 3)2 + 36 + r

Compare tof(x) = p(x q)2 + 16

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

r = 20

(b) The turning point is (3, 16).

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

Whenx = 0, f(x) = 20

Whenf(x) = 0, 4(x 3)2 + 16 = 0

4(x 3)2 = 16

(x 3)2 = 4

x 3 = 2

x = 2 + 3

= 2 + 3 or 2 + 3

= 1 or 5

0 1

(3, 16)

5

20

y

x

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


5 = |p(3 3)2 + q|5 = |q|

q = 5


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

p = 5

Therefore,p = 5, q = 5 orp = 5, q = 5.

(b) Whenx = 3, y = 5

Forp = 5, q = 5,

Whenx = 6, y = |5(6 3)2 5|= |40|= 40

Based on the graph, the range of values ofy is

40


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15



Since thex-coordinate of the maximum point for

both the graphs are same,

therefore,2 + p

2

= 3

p = 4

y = x2 + 2x +px 8 becomes

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

Wheny = 0,

x2 + 6x 8 = 0

x2 6x + 8 = 0

(x 2)(x 4) = 0

x = 2 or 4

Hence,A(2, 0) andB(4, 0).

Substitutex = 2, y = 0 intoy = 2(x 3)2 + 2k,

0 = 2(2 3)2 + 2k

2k = 2

k = 1Hence, k= 1 andp = 4.

(b) Fory = 2(x 3)2 + 2k

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

= 2(x 3)2 + 2

Maximum value of the curve is 2.

Fory = x2 + 2x +px 8

= x2 + 2x + 4x 8

= x2 + 6x 8

Whenx = 3,

y = 9 + 18 8

= 1

Maximum value of the curve is 1.

14. Since 3x2> 0 for all values ofx,

therefore3x2

(2x 1)(x + 4)

< 0

(2x 1)(x + 4) < 0

04 1

2

f(x)

x

Hence, 4


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16



17. y2 9 =x

x =y2 9

Whenx = 0,

y2 = 9

y = 3

When y = 0,

x = 9

When x = 7,

7 =y2 9

y2 = 16

y = 4

x 9 0 7

y 0 3 4

09 7

43

34

y

x

The range of values ofy is 4

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