Remainder & Factor Theorems



Simultaneous Equations

Exercise 1A

2 (a) 2y = 3x ( 1 ( y = [pic] ----- (1)

[pic] + [pic] = 15 ----- (2)

Subs. (1) into (2): 4x ÷ [pic] + [pic] ÷ x = 15

[pic] + [pic] = 15

[pic] = 15

[pic] = 15

97x2 ( 54x + 9 = 15(6x2 ( 2x)

7x2 ( 24x + 9 = 0

(7x ( 3)(x ( 3) = 0

x = [pic] or 3

When x = [pic] in (1): y = [3([pic]) ( 1] ÷ 2 = [pic]

When x = 3 in (1): y = [pic] = 4

(x = [pic], y = [pic] and x = 3, y = 4.

7 Let the numbers be x and y.

x ( y = [pic] ( x = [pic] ----- (1)

and x2 + y2 = [pic] ----- (2)

Subs. (1) into (2): ([pic])2 + y2 = [pic]

[pic] + y2 = [pic]

× 8: 2(4y2 + 12y + 9) + 8y2 = 73

16y2 + 24y ( 55 = 0

(4y ( 5)(4y + 11) = 0

y = [pic] or ([pic] (rejected, since y > 0 is given)

When y = [pic] in (1): x = [2([pic]) + 3] ÷ 2 = [pic]

(the numbers are 1[pic], and 2[pic].

8 2x + 3y = 8 ( x = [pic] ----- (1)

2x2 + 3y2 = 110 ----- (2)

Subs. (1) into (2): 2([pic])2 + 3y2 = 110

[pic] + 3y2 = 110

× 3: 64 ( 48y + 9y2 + 6y2 = 220

15y2 ( 48y ( 156 = 0

5y2 (16y ( 52 = 0

(y + 2)(5y ( 26) = 0 ( y = (2 or [pic]

When y = (2 in (1): x = [8 ( 3((2)] ÷ 2 = 7

When y = [pic] in (1): x = [8 ( 3([pic])] ÷ 2 = ([pic]

(required coordinates are (7, (2) and ((3[pic], 5[pic]).

Exercise 1B

5 (g) 5(j)

Quotient = 6x2 + 3x + 2 Quotient = 2x ( 1

Remainder = (1 Remainder = 11x + 5

6 5(k)

Quotient = 5x2 + 25x + 112

Remainder = 492x ( 340

((x3 + 7x2 ( 4x ( 12 = (2x ( 1)(([pic]x2 + [pic]x ( [pic]) ( 12[pic]

9 3x2 + 5x + C = A(x + 1)2 + B(x + 1) + 4

Let x = (1: 3 ( 5 + C = 0 + 0 + 4 ( C = 6

Equating coeff. of x2: 3 = A

Let x = 0: C = A + B + 4 ( B = (1

(A = 3, B = (1 and C = 6.

10 Given: (ax + b)(x ( 1) + c(x2 + 5) = 18

Let x = 1: 0 + 6c = 18 ( c = 3

Equating coeff. of x2: a + c = 0 ( a = (3

Equating constants: (b + 5c = 18 ( b = (3

(a = (3, b = (3 and c = 3.

Exercise 1C

6 Let f(x) = x6 + 5x3 ( px ( q and g(x) = px2 ( qx ( 1

Given: f((1) = 7 g((1) = (6

1 ( 5 + p ( q = 7 p + q ( 1 = (6

p ( q = 11 ----- (1) p + q = (5 ----- (2)

(1) + (2): 2p = 6 ( p = 3

From (2): q = (8

(p = 3 and q = (8.

7 Given: 2x4 + px3 + 5x2 + 7 = (x + 1)(x ( 2)Q(x) + qx + 18

Let x = (1: 2 ( p + 5 + 7 = 0 ( q + 18

p ( q = (4 ----- (1)

Let x = 2: 32 + 8p + 20 + 7 = 2q + 18

8p ( 2q = (41 ----- (2)

(2) ( 2(1): 6p = (33 ( p = (5[pic]

From (1): q = (1[pic]

(p = (5[pic] and q = (1[pic].

8 Let f(x) = (3x + k)3 + (4x ( 7)2

Given: f(3) = 33 ( (9 + k)3 + 52 = 33

(9 + k)3 = 8

9 + k = 2 ( k = (7.

Exercise 1D

6 Let f(x) = x3 + 3x2 + hx + k

Given: f(2) = 0 f((1) = 30

8 + 12 + 2h + k = 0 (1 + 3 ( h + k = 30

2h + k = (20 ----- (1) h ( k = (28 ----- (2)

(1) + (2): 3h = (48 ( h = (16

From (1): k = 12

(h = (16 and k = 12.

Now, f(x) = x3 + 3x2 ( 16x + 12

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

= (x ( 2)(x ( 1)(x + 6)

7 Let f(x) = kx3 + 5x2 ( 7kx ( 8

Given: f(k) = 0 ( k4 + 5k2 ( 7k2 ( 8 = 0

( (k2)2 ( 2k2 ( 8 = 0

( (k2 + 2)(k2 ( 4) = 0

( (k2 + 2)(k + 2)(k ( 2) = 0

( k = 2 (given k > 0)

So, f(x) = 2x3 + 5x2 ( 14x ( 8

= (x ( 2)(2x2 + 9x + 4) = (x ( 2)(x + 4)(2x + 1)

The other 2 factors are (x + 4) and (2x + 1).

9 f(x) = a(x ( 1)2 + b(x ( 1) + c

Given: f(1) = 9 ( c = 9

f((1) = (11 ( 4a ( 2b + c = (11

( 4a ( 2b = (20

( 2a ( b = (10 ----- (1)

f((2) = 0 ( 9a ( 3b + c = 0

( 9a ( 3b = (9

( 3a ( b = (3 ----- (2)

(2) ( (1): a = 7

From (2): b = 3(7) + 3 = 24

(a = 7, b = 24 and c = 9.

Exercise 1E

1 (g) Let f(x) = 12x3 ( 8x2 ( 3x + 2

By trial & error, f([pic]) = 12([pic])3 ( 8([pic])2 ( 3([pic]) + 2 = 0

So, (2x ( 1) is a factor of f(x).

By inspection, f(x) = (2x ( 1)(6x2 + ax ( 2)

Equating coeff. of x: (3 = (a ( 4 ( a = (1

Now, f(x) = (2x ( 1)(6x2 ( x ( 2) = (2x ( 1)(2x + 1)(3x ( 2)

[Alternatively, f(x) = 12x3 ( 8x2 ( 3x + 2

= 4x2(3x ( 2) ( (3x ( 2)

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

2 (e) 216 ( (3x ( 4)3 = 63 ( (3x ( 4)3

= [6 ( (3x ( 4)] [62 + 6(3x ( 4) + (3x ( 4)2]

= (10 ( 3x) (36 + 18x ( 24 + 9x2 ( 24x + 16)

= (10 ( 3x) (9x2 ( 6x + 28)

3 (d) 3x3 ( 16x2 ( 37x + 14 = 0

Let f(x) = 3x3 ( 16x2 ( 37x + 14

By trial & error, f((2) = 3((2)3 ( 16((2)2 ( 37((2) + 14 = 0

So, (x + 2) is a factor of f(x).

By inspection, f(x) = (x + 2)(3x2 + ax + 7)

Equating coeff. of x2: (16 = 6 + a ( a = (22

Now, f(x) = (x + 2)(3x2 ( 22x + 7) = (x + 2)(x ( 7)(3x ( 1)

When f(x) = 0, (x + 2)(x ( 7)(3x ( 1) = 0

x = (2, 7 or [pic]

(e) 5x3 ( 26x2 + 35x ( 6 = 0

Let f(x) = 5x3 ( 26x2 + 35x ( 6

By trial & error, f(2) = 5(2)3 ( 26(2)2 + 35(2) ( 6 = 0

So, (x ( 2) is a factor of f(x).

By inspection, f(x) = (x ( 2)(5x2 + ax + 3)

Equating coeff. of x: 35 = (2a + 3 ( a = (16

Now, f(x) = (x ( 2)(5x2 ( 16x + 3) = (x ( 2)(x ( 3)(5x ( 1)

When f(x) = 0, (x ( 2)(x ( 3)(5x ( 1) = 0

x = 2, 3 or [pic]

4 (f) 4x3 ( 16x2 + 15x ( 4 = 0

Let f(x) = 4x3 ( 16x2 + 15x ( 4

By trial & error, f([pic]) = 4([pic])3 ( 16([pic])2 + 15([pic]) ( 4 = 0

So, (2x ( 1) is a factor of f(x).

By inspection, f(x) = (2x ( 1)(2x2 + ax + 4)

Equating coeff. of x: 15 = 8 ( a ( a = (7

Now, f(x) = (2x (12)(2x2 ( 7x + 4)

When f(x) = 0, (2x ( 1)(2x2 ( 7x + 4) = 0

x = [pic] or [pic] = 2.78 or 0.72

4 (h) 2x4 ( 19x3 + 61x2 ( 74x + 24 = 0

Let f(x) = 2x4 ( 19x3 + 61x2 ( 74x + 24

By trial & error, f(2) = 2(2)4 ( 19(2)3 + 61(2)2 ( 74(2) + 24 = 0

f(3) = 2(3)4 ( 19(3)3 + 61(3)2 ( 74(3) + 24 = 0

So, (x ( 2) and (x ( 3) are factors of f(x).

By inspection, f(x) = (x ( 2)(x ( 3)(2x2 + ax + 4)

Equating coeff. of x3: (19 = a ( 6 ( 4 ( a = (9

Now, f(x) = (x ( 2)(x ( 3)(2x2 ( 9x + 4) = (x ( 2)(x ( 3)(x ( 4)(2x ( 1)

When f(x) = 0, (x ( 2)(x ( 3)(x ( 4)(2x ( 1) = 0

x = 2, 3, 4, [pic]

7 Let the radii of the balls be x and y (using ( = [pic]).

x + y = 22 ( y = 22 ( x ----- (1)

[pic](x3 ( [pic](y3 = 3050[pic]

×[pic]: (x3 ( (y3 = 2288

×[pic]: x3 ( y3 = 728 ----- (2)

Subs. (1) into (2): x3 ( (22 ( x)3 = 728

x3 ( (10648 ( 1452x + 66x2 ( x3) = 728

2x3 ( 66x2 + 1452x ( 11376 = 0

x3 ( 33x2 + 726x ( 5688 = 0

Let f(x) = x3 ( 33x2 + 726x ( 5688

By trial & error, f(12) = 103 ( 33(10)2 + 726(10) ( 5688 = 0

So, (x ( 12) is a factor of f(x).

By inspection, f(x) = (x ( 12)(x2 + ax + 474)

Equating coeff. of x2: (33 = a ( 12 ( a = (21

Now, f(x) = (x ( 12)(x2 ( 21x + 474)

When f(x) = 0, (x ( 12)(x2 ( 21x + 474) = 0

x = 12 or [pic] (no solution)

When x = 12 in (1): y = 10.

(the radii are 10 cm and 12 cm.

Exercise 1F

5 [pic]

= 2 + [pic]

(a = 2, b = 4, c = (3 and d = (4.

6 [pic]

= 6x + 7 + [pic]

(a = 6, b = 7, c = 3 and d = (17.

7 [pic]

= 3x + 4 + [pic]

(a = 3, b = 4, c = 1 and d = (1.

Exercise 1G

4 (d) [pic]

= 3x + 2 + [pic]

Let [pic] = [pic] + [pic]

×(2x ( 3)(x2 + 5): (9x2 + 21x + 35 = A(x2 + 5) + (Bx + C)(2x ( 3)

Let x = [pic]: (9([pic]) + 21([pic]) + 35 = A([pic] + 5) + 0

A = [pic]

Equating coeff. of x2: (9 = A + 2B ( B = ([pic]

Equating constants: 35 = 5A ( 3C ( C = ([pic]

([pic] = 3x + 2 + [pic] ( [pic]

6 [pic]

= 3x ( 2 + [pic]

Let [pic] = [pic] + [pic]

×(2x + 3)(3x ( 5): x ( 27 = A(3x ( 5) + B(2x + 3)

Let x = ([pic]: ([pic] ( 27 = A(([pic] ( 5) + 0 ( A = 3

Let x = [pic]: [pic] ( 27 = 0 + B([pic] + 3) ( B = (4

([pic] = 3x ( 2 + [pic] ( [pic]

7 Let f(x) = 2x3 + 3x2 ( 50x ( 75

By trial & error, f(5) = 2(5)3 + 3(5)2 ( 50(5) ( 75 = 0

So, (x ( 5) is a factor of f(x).

By inspection, f(x) = (x ( 5)(2x2 + ax + 15)

Equating coeff. of x2: 3 = (10 + a ( a = 13

Now, f(x) = (x ( 5)(2x2 + 13x + 15) = (x ( 5)(x + 5)(2x + 3)

Let [pic] = [pic] + [pic] + [pic]

×(x ( 5)(x + 5)(2x + 3):

3x2 + 19x + 90 = A(x + 5)(2x + 3) + B(x ( 5)(2x + 3) + C(x ( 5)(x + 5)

Let x = 5: 3(5)2 + 19(5) + 90 = A(10)(13) ( A = 2

Let x = (5: 3((5)2 + 19((5) + 90 = B((10)( (7) ( B = 1

Let x = ([pic]: 3(([pic])2 + 19(([pic]) + 90 = C(([pic])([pic]) ( C = (3

([pic] = [pic] + [pic] ( [pic]

8 Let f(x) = 8x3 ( 22x2 ( 9x + 9

By trial & error, f([pic]) = 8([pic])3 ( 22([pic])2 ( 9([pic]) + 9 = 0

So, (2x ( 1) is a factor of f(x).

By inspection, f(x) = (2x ( 1)(4x2 + ax ( 9)

Equating coeff. of x: (9 = (a ( 18 ( a = (9

Now, f(x) = (2x ( 1)(4x2 ( 9x ( 9) = (2x ( 1)(x ( 3)(4x + 3)

Let [pic] = [pic] + [pic] + [pic]

×(2x ( 1)(x ( 3)(4x + 3):

25x + 75 = A(x ( 3)(4x + 3) + B(2x ( 1)(4x + 3) + C(2x ( 1)(x ( 3)

Let x = [pic]: 25([pic]) + 75 = A(([pic])(5) ( A = (7

Let x = 3: 25(3) + 75 = B(5)(15) ( B = 2

Let x = ([pic]: 25(([pic]) + 75 = C(([pic])(([pic]) ( C = 6

([pic] = ([pic] + [pic] + [pic]

Exercise 2A

3 For 2x2 + 5x ( 7 = 0, ( + ( = ([pic] and (( = ([pic]

(iii) New sum of roots = (2 + ( 2

= (( + ()2 ( 2((

= (([pic])2 ( 2(([pic]) = [pic]

New product of roots = (2(( 2)

= ((()2

= (([pic])2 = [pic]

Required new equation is x2 ( [pic]x + [pic] = 0

4x2 ( 53x + 49 = 0

(iv) New sum of roots = (3 + ( 3

= (( + ()((2 ( (( + ( 2)

= (([pic])[[pic] ( (([pic])] = ([pic]

New product of roots = (3(( 3)

= ((()3

= (([pic])3 = ([pic]

Required new equation is x2 ( (([pic])x ( [pic] = 0

8x2 + 335x ( 343 = 0

4 For 3x2 ( 8x + 2 = 0, ( + ( = ((([pic]) = [pic] and (( = [pic]

(iv) [pic] + [pic] = [pic] = [pic]

= [([pic])2 ( 2([pic])] ÷ [pic]

= [pic] = 8[pic]

(v) (4 + ( 4 = ((2)2 + (( 2)2

= ((2 + ( 2)2 ( 2(2( 2

= [(( + ()2 ( 2((]2 ( 2([pic])2

= [([pic])2 ( 2([pic])]2 ( 2([pic])2 = 32[pic]

5 For 3x2 + 2x ( 3 = 0, ( + ( = ([pic] and (( = ([pic] = (1

New sum of roots = [pic] + [pic] = [pic] = [pic]

= [(([pic])2 ( 2((1)] ÷ ((1)

= ([pic]

New product of roots = [pic]([pic]) = 1

Required new equation is x2 ( (([pic])x + 1 = 0

9x2 + 22x + 9 = 0

9 For 3x2 + 5x + 1 = 0, ( + ( = ([pic] and (( = [pic]

For hx2 ( 4x + k = 0, (( + 3) + (( + 3) = ((([pic])

( + ( + 6 = [pic]

([pic] + 6 = [pic]

[pic] = [pic] ( h = [pic]

And (( + 3)(( + 3) = [pic]

[pic] = (( + 3(( + () + 9

[pic] = [pic] + 3(([pic]) + 9 = [pic]

( k = [pic]([pic]) = 4

(h = [pic] and k = 4.

10 For 2x2 ( 4x ( 13 = 0, ( + ( = ((([pic]) = 2 and (( = ([pic]

(i) [pic] + [pic] = [pic]

= [pic]

= [22 ( 2(([pic]) + 2(2)] ÷ [([pic] + 2(2) + 4] = 14

(ii) New product of roots = [pic] × [pic] = [pic]

= (([pic]) ÷ [([pic] + 2(2) + 4] = ([pic]

Required new equation is x2 ( 14x ( [pic] = 0

3x2 ( 42x ( 13 = 0

12 If ( is a root of x2 ( 7x + k = 0, then

(2 ( 7( + k = 0

k = 7( ( (2 ----- (1)

If ( is a root of x2 + 8x ( 2k = 0, then

(2 + 8( ( 2k = 0 ----- (2)

Subs. (1) into (2): (2 + 8( ( 2(7( ( (2) = 0

3(2 ( 6( = 0

3((( ( 2) = 0

( = 0 or 2

But when ( = 0 in (1): k = 0 (rejected as given k ( 0)

When ( = 2 in (1): k = 7(2) ( 22 = 10

(( = 2 and k = 10.

15 If ( is a root of x2 ( 2x + 5 = 0, then

(2 ( 2( + 5 = 0 ----- (1)

(i) Show: 2(3 + (2 + 25 = 0

(1) × 2(: 2(3 ( 4(2 + 10( = 0

2(3 ( 4(2 + 5(2() = 0

2(3 ( 4(2 + 5((2 + 5) = 0 [from (1)]

2(3 ( 4(2 + 5(2 + 25 = 0

2(3 + (2 + 25 = 0 (Shown)

(ii) Show: (4 = 5 ( 12(

From (1): (2 = 2( ( 5

Square both sides: (4 = (2( ( 5)2

= 4(2 ( 20( + 25

= 4(2( ( 5) ( 20( + 25 [from (1)]

= 8( ( 20 ( 20( + 25

= 5 ( 12( (Shown)

Exercise 2B

9 5x2 + 7x ( 3k = 6 has no real roots.

Discriminant < 0

72 ( 4(5)((3k ( 6) < 0

49 + 60k + 120 < 0

60k < (169

k < (2[pic]

11 (p + 2)x2 ( 2px = 5 ( p has no real roots.

Discriminant < 0

((2p)2 ( 4(p + 2)(p ( 5) < 0

4p2 ( 4p2 + 12p + 40 < 0

12p < (40

p < (3[pic]

12 Show: x + 5 = [pic] has coincident roots when p = ([pic]

For x + 5 = [pic] ( x2 + 5x = p ( 3

( x2 + 5x + (3 ( p) = 0

Discriminant = 52 ( 4(1)(3 ( p)

= 25 ( 12 + 4p = 13 + 4p

For coincident roots, discriminant = 0

( 13 + 4p = 0

( p = ([pic] (Shown)

13 Show: 2x2 + 5kx = 3k has real and distinct roots for all k > 0.

Discriminant = (5k)2 ( 4(2)((3k)

= 25k2 + 24k

Since k > 0, then 25k2 > 0 and 24k > 0

( discriminant > 0

( the equation has real and distinct roots.

Exercise 2C

6 y = 3x2 ( 4x + 5 = 3(x2 ( [pic]x) + 5

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

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

For minimum value, y = [pic], so lowest portfolio value = [pic]($1000) = $3666.67

It occurs when x = [pic] years (or 8 months)

7 y = 2x2 + 4x + 17 = 2(x2 + 2x) + 17

= 2[(x + 1)2 ( 1] + 17

= 2(x + 1)2 + 15

Minimum value of y = 15 when x = (1.

8 y = 14 ( 3x ( x2 = 14 ( (x2 + 3x)

= 14 ( [(x + [pic])2 ( [pic]]

= 14 ( (x + [pic])2 + [pic]

= 16[pic] ( (x + [pic])2

Maximum value of y = 16[pic] when x = (1[pic].

9 y = (x2 + 4x ( 1 = ((x2 ( 4x) ( 1

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

= 3 ( (x ( 2)2

Greatest height = 3 m and horizontal distance = 2 m.

Exercise 2D

5 (3x ( 5)2 ( ([pic])2 ( 0

[(3x ( 5) ( ([pic])] [(3x ( 5) + ([pic])] ( 0

(3x ( [pic])(3x ( [pic]) ( 0

x ( [pic] or x ( [pic]

(x ( 1[pic] or x ( 2[pic]

7 1 ( x < (x ( 1)(5 ( x) < 3

1 ( x < (x2 + 6x ( 5 and (x2 + 6x ( 5 < 3

x2 ( 7x + 6 < 0 x2 ( 6x + 8 > 0

(x ( 1)(x ( 6) < 0 (x ( 2)(x ( 4) > 0

1 < x < 6 x < 2 or x > 4

For intersections, 1 < x < 2 or 4 < x < 6

9 x2 + 3kx + 2k = x ( 10 has real and distinct roots.

x2 + (3k ( 1)x + 2k + 10 = 0

Discriminant > 0

(3k ( 1)2 ( 4(1)(2k + 10) > 0

9k2 ( 6k + 1 ( 8k ( 40 > 0

9k2 ( 14k ( 39 > 0

(9k + 13)(k ( 3) > 0

k < (1[pic] or k > 3

For the equation to have equal roots, discriminant = 0 ( k = (1[pic] or 3

12 (7x2 + 9x + k < 0 (always negative)

Discriminant < 0

92 ( 4((7)(k) < 0

81 + 28k < 0

k < (2[pic]

So largest integer value = (3

14 3 ( 4k ( (k + 3)x ( x2 < 0

Discriminant < 0

[((k + 3)]2 ( 4((1)(3 ( 4k) < 0

k2 + 6k + 9 + 12 ( 16k < 0

k2 ( 10k + 21 < 0

(k ( 3)(k ( 7) < 0

3 < k < 7

Exercise 2E

7 y = [pic]x + k ----- (1)

x2 + y2 = 8k ----- (2)

Subs. (1) into (2): x2 + ([pic]x + k)2 ( 8k = 0

x2 + [pic]x2 + kx + k2 ( 8k = 0

[pic]x2 + kx + k2 ( 8k = 0

For line to be tangent to curve, discriminant = 0

k2 ( 4([pic])(k2 ( 8k) = 0

k2 ( 5k2 + 40k = 0

(4k(k ( 10) = 0

k = 0 or 10

8 y = 2x + k ----- (1)

x2 + 2y2 = 8 ----- (2)

Subs. (1) into (2): x2 + 2(2x + k)2 ( 8 = 0

x2 + 8x2 + 8kx + 2k2 ( 8 = 0

9x2 + 8kx + 2k2 ( 8 = 0

Since line does not intersect curve, discriminant < 0

(8k)2 ( 4(9)(2k2 ( 8) < 0

64k2 ( 72k2 + 288 < 0

288 ( 8k2 < 0

36 ( k2 < 0

(6 + k)(6 ( k) < 0

k < (6 or k > 6

9 x + 2y = 3 ( x = 3 ( 2y ----- (1)

x2 + y2 = p ----- (2)

Subs. (1) into (2): (3 ( 2y)2 + y2 ( p = 0

4y2 ( 12y + 9 + y2 ( p = 0

5y2 ( 12y + 9 ( p = 0

Since line intersecting curve at 2 real & distinct points, discriminant > 0

((12)2 ( 4(5)(9 ( p) > 0

144 ( 180 + 20p > 0

20p > 36

p > 1[pic]

10 y = mx + c ----- (1)

x2 + y2 = 4 ----- (2)

Subs. (1) into (2): x2 + (mx + c)2 ( 4 = 0

x2 + m2x2 + 2cmx + c2 ( 4 = 0

(1 + m2)x2 + 2cmx + c2 ( 4 = 0

For line to be tangent to curve, discriminant = 0

(2cm)2 ( 4(1 + m2)(c2 ( 4) = 0

4c2m2 ( 4(c2 ( 4 + c2m2 ( 4m2) = 0

4c2m2 ( 4c2 + 16 ( 4c2m2 + 16m2 = 0

16m2 ( 4c2 + 16 = 0

16m2 = 4c2 ( 16

4m2 = c2 ( 4 (Proved)

Exercise 2F

3 (b) [pic] = 13 ( [pic] = 13 and [pic] = (13

3x ( 8 = 13x + 26 and 3x ( 8 = (13x ( 26

10x = (34 and 16x = (18

x = (3[pic] and (1[pic]

3 (d) | 3x ( 7 | = 5x + 6 ( 3x ( 7 = 5x + 6 and 3x ( 7 = (5x ( 6

2x = (13 and 8x = 1

Rejected as | 3x ( 7 | > 0 x = [pic]

4 (d) | 2x2 + 3 | = 7x ( 2x2 + 3 = 7x and 2x2 + 3 = (7x

2x2 ( 7x + 3 = 0 and 2x2 + 7x + 3 = 0

(2x ( 1)(x ( 3) = 0 and (2x + 1)(x + 3) = 0

x = [pic] or 3 and x = ([pic] or (3

But when x = ([pic] or (3, | 2x2 + 3 | ( 7x

( x = [pic] or 3

4 (e) | 15x2 + 13x | = 20 ( 15x2 + 13x = 20 and 15x2 + 13x = (20

15x2 + 13x ( 20 = 0 and 15x2 + 13x + 20 = 0

(3x + 5)(5x ( 4) = 0 and since discriminant < 0

x = (1[pic] or [pic] no solution

5 (b) 2| 8 ( 3( | + 3| ( ( 13 | = 2(3( ( 8) + 3(13 ( ()

= 6( ( 16 + 39 ( 3(

= 3( + 23

5 (c) 3| 15 ( 8e | ( 2| 9 ( 4e | = 3(8e ( 15) ( 2(4e ( 9)

= 24e ( 45 ( 8e + 18

= 16e ( 27

Exercise 2G

2 (e) y = | x2 + 5x ( 6 |

= | (x + 6)(x ( 1) |

3 (b) y = | 11 ( 3x |, 2 ( x ( 6

Consider y = 11 ( 3x

When x = 2, y = 5

When x = 6, y = (7

4 (b) y = | x2 ( x ( 6 |, (4 ( x ( 4

= | (x + 2)(x ( 3) |

Consider y = x2 ( x ( 6

When x = (4, y = 14

When x = 4, y = 6

Exercise 4A

4 (c) 3x × 243y = 1 8x ÷ 2y ( 12 = [pic]

3x × 35y = 30 23x ÷ 2y ( 12 = 2[pic]

3x + 5y = 30 23x ( (y ( 12) = 2[pic]

x + 5y = 0 3x ( y + 12 = ([pic] ----- (2)

x = (5y ----- (1)

Subs. (1) into (2): 3((5y) ( y + 12 = ([pic]

((16y + 12)y = (4

4y2 ( 3y ( 1 = 0

(4y + 1)(y ( 1) = 0

y = ([pic] or 1

When y = ([pic] in (1): x = 1[pic]

When y = 1 in (1): x = (5

(x = ([pic], y = 1[pic] or x = 1, y = (5

5 y = abx

At (1, 21), 21 = ab ( a = [pic] ----- (1)

At (2, 63), 63 = ab2 ----- (2)

Subs. (1) into (2): ([pic])b2 = 63 ( b = 3

When b = 3 in (1): a = 7

At (k, 189), 189 = 7(3k) ( 3k = 27 = 33

( k = 3

(a = 7, b = 3 and k = 3.

6 y = pxq + 7

When x = 3, y =25, 25 = p(3q) + 7 ( p(3q) = 18 ----- (1)

When x = 5, y =57, 57 = p(5q) + 7 ( p(5q) = 50

( p = [pic] ----- (2)

Subs. (2) into (1): ([pic])3q = 18 ( [pic] = [pic] = [pic] = [pic]

( [pic] = [pic]

( q = 2

When q = 2 in (2): p = 2

(p = 2 and q = 2.

Exercise 4B

3 (d) (2[pic] + 3[pic])(2[pic] ( 3[pic]) = (2[pic])2 ( (3[pic])2

= 28 ( 45 = (17

4 (d) [pic] = [pic] = [pic] × [pic]

= [pic] = [pic]

6 Length = [pic] = [pic] × [pic]

= [pic] = (2[pic] + 4) cm

7 (c) [pic]([pic] + [pic]) = [pic] + [pic]

= [pic] + [pic] = 3[pic]

7 (d) [pic]([pic] ( [pic]) = [pic] ( 2[pic]

= [pic] ( 2(8) = (11[pic]

10 (c) [pic] = [pic] × [pic]

= [pic]

= [pic] = 2[pic] ( 3

10 (d) [pic] = [pic] × [pic]

= [pic]

= [pic]

13 Height = [pic] = [pic] = [pic] × [pic]

= [pic] = (3 ( [pic]) cm

Exercise 4C

4 (d) [pic] = [pic] = (2

5 (g) log2 ( 3k 125 = 3

125 = (2 ( 3k)3

( 2 ( 3k = [pic] = 5

( 3k = 2 ( 5 = (3

k = (1

5 (h) logy [pic] = 3

[pic] = y3

2y3 = y ( y2

2y3 + y2 ( y = 0

y(2y2 + y ( 1) = 0

y(2y ( 1)(y + 1) = 0

y = 0, [pic] or (1

But when y = 0 or (1, logy [pic] is not defined

( y = [pic]

5 (i) (log3 x)2 = 4log3 x

(log3 x)2 ( 4log3 x = 0

log3 x (log3 x ( 4) = 0

log3 x = 0 or log3 x = 4

x = 30 or x = 34

x = 1 or 81

7 (c) [pic](ln y) = 3

ln y = ([pic])3

y = e[pic] = 5.41 × 1013

7 (d) [pic](logy [pic]) = 3

logy [pic] = ([pic])3 = 2

[pic] = y2 ( y = [pic] (base y > 0)

Exercise 4D

5 (c) log6 [pic] + log6 [pic] ( log6 [pic]

= log6 [pic]

= log6 6 = 1

(d) 4log2 0.6 + 2log2 [pic] ( 2log2 0.15

= log2 [pic] + log2 [pic] ( log2 [pic]

= log2 [[pic] × [pic] ÷ [pic]]

= log2 16 = log2 24 = 4

6 (d) logc [pic] + log3 [pic]

= [pic] + log3 [pic]

= [pic] + log3 2[pic]

= [pic] + log3 2 + log3 [pic]

= [pic] + log3 2 + [pic]

= 1 + log3 2 = log3 3 + log2 2 = log3 6

7 (iv) [pic] = log3 [pic] ÷ log3 5[pic]

= log3 2(3 ÷ [pic]log3 5

= (3(0.6309) ÷ [pic](1.465) = (0.861

9 (e) log7 10 × lg [pic] = [pic] × lg (72)[pic]

= [pic] × [pic]lg 7 = [pic]

(f) logx 16 × log64 x = logx 24 × [pic]

= 4 × [pic] = [pic]

12 Given: log6 2 = a and log5 3 = b

log5 2 = [pic] = log6 2 ÷ ([pic])

= a × log5 (2 × 3)

log5 2 = a(log5 2 + log5 3)

log5 2 ( a log5 2 = a log5 3

log5 2(1 ( a) = ab

log5 2 = [pic]

Exercise 4E

3 (d) logx 32 = 3 ( logx 2

logx 32 + logx 2 = 3

logx 64 = 3

( 64 = x3

( x = 4

4 (c) 2p × 4q = 4 log4 (3q + 11) ( log4 p = 0.5

2p × 22q = 22 log4 [pic] = [pic]

p + 2q = 2 [pic] = 4[pic] = 2

p = 2 ( 2q ----- (1) 3q + 11 = 2p ----- (2)

Subs. (1) into (2): 3q + 11 = 2(2 ( 2q)

7q = (7 ( q = (1

From (1): p = 2 ( 2((1) = 4

(p = 4 and q = (1.

6 (d) 3ln p ( 2logp e = 4

3ln p ( [pic] = 4

× ln p: 3(ln p)2 ( 2 = 4ln p

Let a = ln p: 3a2 ( 4a ( 2 = 0

a = [pic] = 1.7208 or (0.3873

ln p = 1.7208 or (0.3873

p = e1.7208 or e(0.3873 = 5.59 or 0.679

7 (f) 2(7y) = 3 ( 5[pic]

2(7[pic])2 + 5(7[pic]) ( 3 = 0

Let a = 7[pic]: 2a2 + 5a ( 3 = 0

(2a ( 1)(a + 3) = 0

a = [pic] or a = (3

7[pic] = [pic] or 7[pic] = (3 (rejected as 7[pic] > 0)

[pic] = [pic] ( y = (0.712

8 Given: 7x = 2 and 2y = 7

( x = log7 2 and y = log2 7

(xy = log7 2 × log2 7

= [pic] × log2 7 = 1

10 2log2 y = 4 + log2 [pic]

log2 y2 ( log2 [pic] = 4

log2 [pic] = 4

[pic] = 24 = 16

y2 = 16x[pic]

( y = 4x[pic] (reject y = (4x[pic], as log2 y is not defined)

Exercise 4F

5 (i) (ii) The 2 graphs are a reflection about

the line y = x.

6 (i) (ii) eex = xe4

ln eex = ln xe4

ex = ln x + 4ln e

ln x = ex ( 4

Draw y = ex ( 4

From the sketch, there are 2 solutions.

Exercise 4G

6 A = P(1 + [pic])n , A = Pe[pic]

(i) Bank X Bank Y

A = 1000e[pic] = $1061.84 A = 1000 + [pic] = $1090

Betty should put her money in Bank Y as she would earn more interest.

(ii) The interest rate has the biggest effect when the period is the same, because higher

interest rate would earn more interest.

Exercise 3A

3 (c) [pic] = [pic] = 7 × 13 = 91

(d) [pic] = [pic] = 3 × 2 × 7 × 6 = 252

6 (c) (1 ( ky)4 = 1 + 4((ky) + 6((ky)2 + 4((ky)3 + ((ky)4

= 1 ( 4ky + 6k2y2 ( 4k3y3 + k4y4

7 (d) (1 ( x2)10 = 1 + 10((x2) + [pic]((x2)2 + [pic]((x2)3 + ...

= 1 ( 10x2 + 45x4 ( 120x6 + ...

10 (1 + kx)9 = 1 + 9(kx) + [pic](kx)2 + [pic](kx)3 + [pic](kx)4 + [pic](kx)5 + ...

Given: coeff. of x5 = 3(coeff. of x4)

[pic]k5 = 3[pic]k4

[pic]k5 = 3([pic])k4

( k5 = 3k4

( k5 ( 3k4 = 0

( k4(k ( 3) = 0

( k = 3 ([pic]k ( 0)

Exercise 3B

3 (d) (2 ( [pic])8

(r + 1)th term = [pic](2)8 ( r(([pic])r

= [pic](2)8 ( r((1)r x (2r

For the term [pic], we have (2r = (10 ( r = 5

Required coeff. of [pic] = [pic](2)8 ( 5((1)5 = (448

6 For (3 + 2x)6 = 36 + 6(3)5(2x) + [pic](3)4(2x)2 + [pic](3)3(2x)3 + [pic](3)2(2x)4 + ...

Coeff. of x4 = [pic](9)(16) = 2160

For (k + 3x)6 = k6 + 6(k)5(3x) + [pic](k)4(3x)2 + [pic](k)3(3x)3 + [pic](k)2(3x)4 + ...

Coeff. of x4 = [pic](k2)(81) = 1215k2

Given: 1215k2 = 2160 ( k2 = [pic]

( k = (1[pic]

7 (3 + kx)10 = 310 + 10(3)9(kx) + [pic](3)8(kx)2 + [pic](3)7(kx)3 + [pic](3)6(kx)4

+ [pic](3)5(kx)5 + ...

Given: coeff. of x5 = coeff. of x4

[pic]243k5 = [pic]729k4

[pic](k5) = [pic](3k4)

( [pic] = 3k4

( 2k5 = 5k4

( 2k5 ( 5k4 = 0

( k4(2k ( 5) = 0

( k = 2[pic] ([pic]k ( 0)

9 (a) (2 + [pic])6 = 26 + 6(2)5([pic]) + [pic](2)4([pic])2 + ...

= 64 + 96x + 60x2 + ...

(b) (3 ( 2x)6 = 36 + 6(3)5((2x) + [pic](3)4((2x)2 + ...

= 729 ( 2916x + 4860x2 + ...

(6 ( [pic]x ( x2)6 = [ (2 + [pic])(3 ( 2x) ]6

= (64 + 96x + 60x2 + ...) (729 ( 2916x + 4860x2 + ... )

Coeff. of x2 = 64(4860) + 96((2916) + 60(729) = 74844

11 For the term containing x5y5, then r = 5 in (x + [pic])10.

So, T6 = [pic](x)5 ([pic])5 = [pic]

15 (1 + 2x ( 3x2)8 = [1 + (2x ( 3x2)]8

= 1 + 8(2x ( 3x2) + [pic](2x ( 3x2)2 + [pic](2x ( 3x2)3 + ...

= 1 + 16x ( 24x2 + 28(4x2 ( 12x3 + ...) + 56(8x3 + ...) + ...

= 1 + 16x + 88x2 + 112x3 + ...

Let x = 0.01, then 1 + 2x ( 3x2 = 1 + 2(0.01) ( 3(0.01)2 = 1.0197

(1.01978 = 1 + 16(0.01) + 88(0.01)2 + 112(0.01)3 + ...

= 1.168912 ( 1.1689

17 Given: (1 + px + qx2)8 = 1 + 8x + 52x2 + kx3 + ...

L.H.S. = [1 + (px + qx2)]8

= 1 + 8(px + qx2) + [pic](px + qx2)2 + [pic](px + qx2)3 + ...

= 1 + 8px + 8qx2 + 28(p2x2 + 2pqx3 + ...) + 56(p3x3 + ...) + ...

= 1 + 8px + (8q + 28p2)x2 + (56pq + 56p3)x3 + ...

Equating coeff. of x: 8p = 8 ( p = 1

Equating coeff. of x2: 8q + 28(1)2 = 52 ( q = 3

Equating coeff. of x3: k = 56(1)(3) + 56(1)3 = 224

(p = 1, q = 3 and k = 224.

-----------------------

2x ( 1

x2 + 3x ( 4 2x3 + 5x2 + 9

((2x3 + 6x2 ( 8x)

(x2 + 8x + 9

(((x2 ( 3x + 4)

11x + 5

6x2 + 3x + 2

2x + 1 12x3 + 12x2 + 7x + 1

((12x3 + 6x2)

6x2 + 7x

((6x2 + 3x) 4x + 1

((4x + 2)

(1

([pic]x2 + [pic]x ( [pic]

2x ( 1 (x3 + 7x2 ( 4x ( 12

(((x3 + [pic]x2)

[pic]x2 ( 4x

(([pic]x2 ( [pic]x)

([pic]x ( 12

((([pic]x + [pic])

(12[pic]

5x2 + 25x + 112

x2 ( 5x + 3 5x4 + 2x2 + 7x ( 4

((5x4 ( 25x3 + 15x2)

25x3 ( 13x2 + 7x

((25x3 ( 125x2 + 75x) 112x2 ( 68x ( 4

((112x2 ( 560x + 336

492x ( 340

x2 + 5x ( 6

x ( 2 x3 + 3x2 ( 16x + 12

((x3 ( 2x2)

5x2 ( 16x

((5x2 ( 10x) (6x + 12

(((6x + 12)

0

2x2 + 9x + 4

x ( 2 2x3 + 5x2 ( 14x ( 8

((2x3 ( 4x2)

9x2 ( 14x

((9x2 ( 18x) 4x ( 8

((4x ( 8)

0

2

3x3 + 2 6x3 + 4x2 ( 3x

((6x3 + 4)

4x2 ( 3x ( 4

6x + 7

x2 + x ( 2 6x3 + 13x2 ( 2x ( 31

((6x3 + 6x2 ( 12x)

7x2 + 10x ( 31

((7x2 + 7x ( 14) 3x ( 17

3x + 4

3x3 + x 9x4 + 12x3 + 3x2 + 5x ( 1

((9x4 + 3x2)

12x3 + 5x ( 1

((12x3 + 4x) x ( 1

3x + 2

2x3 ( 3x2 + 10x ( 15 6x4 ( 5x3 + 15x2 ( 4x + 5

((6x4 ( 9x3 + 30x2 ( 45x)

4x3 ( 15x2 + 41x + 5

((4x3 ( 6x2 + 20x ( 30)

(9x2 + 21x + 35

3x ( 2

6x2 ( x ( 15 18x3 ( 15x2 ( 42x + 3

((18x3 ( 3x2 ( 45x)

(12x2 + 3x + 3

(((12x2 + 2x + 30) x ( 27

(1

y

x

y = 2x2 + 4x + 17

0

15

17

(1[pic]

y

x

y = 14 ( 3x ( x2

0

14

16[pic]

(6

x

y = x2 + 5x ( 6

1

y

0

6

(6

y = | x2 + 5x ( 6 |

y

x

y = 11 ( 3x

0

7

5

2

6

(7

y = | 11 ( 3x |

(4

y

x

y = x2 ( x ( 6

0

14

6

(2

3

4

(6

y = | x2 ( x ( 6 |

x

y

y = 10x

y = lg x

1

1

0

y = x

x

y

y = ln x

1

0

y = ex ( 4

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download