maths t (chap 1)

© Oxford Fajar Sdn. Bhd. (008974-T) 2013

cHAPTER 1 FuncTions

Focus on Exam 1

1 (a) g(x) = 16 – x2

For g(x) to be defined, 16 – x2 0(4 + x)(4 – x) 0

x−4 4

Hence, the domain of g is {x | – 4 x 4, x P }.

(b)

x

y

O 4

4

-4

16 - x2y =

The graph of g(x) isactually part of a circlewith the equationy 2 = 16 - x 2 ⇒ x 2 + y 2 = 42.

(c) The range is { y | 0 y 4, y P }.

2 (a) f : x x2 – 9 f (x) = x2 – 9 For f (x) to be defined, x2 – 9 0

(x + 3)(x – 3) 0

x−3 3

Hence, the domain of f is {x | x –3 or x 3, x P }.

(b)

xO

y

−3 3

x 2 − 9 y =

(c) The range is { y | y 0, y P }.

3 First, consider only g(x) = 12

x – 2. The

graph of g(x) is as shown below.

x

y

4

2

−2O

12

y = x − 2

12

y = − x + 2

– 12

x + 2, x < 4,

Thus, g(x) = 5 12

x – 2, x 4.

Next, consider only h(x) = 12 x + 2. The

graph of h(x) is as shown below.

x

y

−4

2

O

12

y = x + 2

12

y = − x − 2


Fully Worked Solution 3ACE AHEAD Mathematics (T) First Term Updated Edition2

– 12

x – 2, x < – 4,

Thus, h(x) = 5 12

x + 2, x – 4.

Therefore,

• for x < – 4, f (x) = – 12

x + 2 – 1– 12

x – 22= 4

• for – 4 x < 4, f (x) = – 12

x + 2 – 112x + 22= –x

• for x 4, f (x) = 12

x – 2 – 112

x + 22= – 4

4, x < – 4,

Hence, f (x) = 5– x, – 4 x < 4,– 4, x 4.

(a) The graph of f (x) is as shown below.

x

y

O

y = 4

y = −4

y = −x

4

−4

−4

4

(b) Thel range of f (x) is {y | – 4 y 4, y P }.

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

x

y

3

y = (x + 1)2 + 2

(−1, 2)

O

The domain of f (x) is {x | x P }.The range of f (x) is { y | y 2, y P }.

(ii) g(x) = 1

x – 2

x

y

O 2

y = 1x − 2

12

−

The domain of g(x) is {x | x P , x ≠ 2}.The range of g(x) is { y | y P , y ≠ 0}.

(b) g ° f = g[ f (x)]

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

= 1

(x + 1)2 + 2 – 2

= 1

(x + 1)2 , x ≠ –1

The domain of g ° f is {x | x P , x ≠ –1}.

x

y

O−1

y = g f (x)

(x + 1)2= 1

5 (a) (i) f (x) = x – 2

x

y

O 2

x − 2y =

The domain of f (x) is {x | x 2, x P }.The range of f (x) is { y | y 0, y P }.



(ii) g(x) = x2 – 3

x

y

Oy = x2 − 3

−3

The domain of g(x) is {x | x P }.The range of g(x) is { y | y –3, y P }.

(b) g ° f exists because Rf # Dg.

g ° f = g[ f(x)]

= g1 x – 2 2

= 1 x – 2 22 – 3= x – 5

(c) For f ° g to be defined, Rg # Df .Rg Df

x2 – 3 2

x2 – 5 0

1x + 521x – 5 2 0Hence, the required set of values of x is

{x | x – 5 or x 5, x P }.

6 (a) (i) f (x) = 25 – x2

x

5

−5 5O

y

25 − x 2y =

The domain of f is {x | –5 x 5, x P }.The range of f is { y | 0 y 5, y P }.

(ii) g (x) = x2 – 5

x

y

O

−5

y = x2 − 5

{ {

The domain of g is {x | x P }.The range of g is {y | y –5, y P }.

(b) f ° g does not exist because Rg # Df .

(c) For f ° g to be defined, Rg # Df .

Df Rg Df

–5 x2 – 5 5 0 x2 10Hence, the required set of values of x is{x | – 10 x 10, x P }.

7 (a) (i) f (x) = 1x – 2

x

y

O 212

−

The domain of f is {x | x P , x ≠ 2}.The range of f is { y | y P , y ≠ 0}.

(ii) g(x) = 2

x + 4

xO

y

12

−4

The domain of g(x) is {x | x P , x ≠ – 4}. …jThe range of g is { y | y P , y ≠ 0}.

(b) f ° g = f [g(x)]

= f 1 2x + 4 2

= 1

1 2

x + 42 – 2

= x + 4

2 – 2(x + 4)

} } }



= x + 4–6 – 2x

, x ≠ –3 …k

Combining j and k, the domain of f ° g is {x | x P , x ≠ – 4, x ≠ –3}.

8 (a) For f : x a xx + 1

, the domain is

{x | x P , x ≠ –1}.

For g : x a x + 2x

, the domain is

{x | x P , x ≠ 0}.

(b) g ° f = g f (x) = g1 xx – 12

=

xx + 1

+ 2

xx + 1

= x + 2(x + 1)

x

= 3x + 2x

= 3 + 2x, x ≠ 0

Other than x ≠ 0, the domain of g ° f also has to follow the domain of f, i.e. x P , x ≠ –1.Hence, the domain of g ° f is {x | x P , x ≠ 0, x ≠ –1}.

If x ≠ –1, then g f (x) ≠ 3 + 2(–1)

, i.e.

g f(x) ≠ 1.Thus, the range of g ° f cannot take the value 1. Other than that, based on the graph in (c), the range of g ° f also cannot take the value 3. Hence, the range of g ° f is {y | y P , y ≠ 1, y ≠ 3}.

(c) For h : x ! 3 + 2x,

the domain is {x | x P , x ≠ 0} and the range is { y | y P , y ≠ 3}.

3

O

y

x23

–

2xh(x) = 3 +

(d) h ≠ g ° f because the domain and the range of g ° f are not the same as the domain and the range of h.

9 f : x a 1x, x P \{0}

g : x a 2x – 1, x P f ° g = fg(x)

= f (2x – 1)

= 12x – 1

, x ≠ 12

The domain of f ° g is 5x | x P , x ≠ 126.

10 (a) Let y = f –1(x)f (y) = x

2 + y – 1 = xy – 1 = x – 2y – 1 = (x – 2)2

y – 1 = x2 – 4x + 4y = x2 – 4x + 5

[ f –1(x) = x2 – 4x + 5The domain of f –1 is the same as the range of f, i.e. {x | x 2, x P }.The range of f –1 is the same as the domain of f, i.e. {y | y 1, y P }.

(b) The graphs of y = f (x) and y = f –1(x) are as shown below.

x

y

O 1 2 3 4 5

12345

y = f −1(x)

y = f (x)

y = x

The graph of y = f −1(x) isthe reflection of the graph ofy = f (x) in the straight line y = x.

A\B means A – B or A ∩ B9.



The point of intersection of the graphs of y = f (x) and y = f –1(x) is the same as the point of intersection of the curve y = f –1(x) = x2 – 4x + 5 and the straight line y = x.y = x2 – 4x + 5 …jy = x …kx2 – 4x + 5 = xx2 – 5x + 5 = 0

x = –(–5) ± (–5)2 – 4(1)(5)

2(1)

x = 5 ± 5

2x = 1.38 or 3.62x = 1.38 is not accepted[ x = 3.62[ y = x = 3.62Hence, the required point of intersection is (3.62, 3.62).

11 (a) The graph of y = f (x) = x2 – 3x is as shown below.

x

y

O 3

12( 1

4, −2 )1

y = f (x) = x 2 − 3x

f –1 does not exist because f is not a one-to-one function.

(b) In order for f –1 to exist, the domain of f must be restricted to only

5x | x 112 , x P 6.

Let y = f –1(x)f (y) = x

y2 – 3y = xy2 – 3y – x = 0

y = –(–3) + (–3)2 – 4(1)(–x) 2(1)

y = 3 + 9 + 4x

2

[ f –1(x) = 3 + 9 + 4x

2The domain of f –1 is the same as the

range of f, i.e. 5x | x –214, x P 6.

12 Since (x – 2) is a factor of p(x) = qx3 – rx2 + x – 2,

p(2) = 0 q(2)3 – r(2)2 + 2 – 2 = 0

8q – 4r = 02q – r = 0 …j

p(x) has a remainder of –12 when it is divided by (x + 1).

p(–1) = –12 q(–1)3 – r(–1)2 – 1 – 2 = 0

–q – r = 3 …kj – k:

2q – r = 0–q – r = 3

3q = –3 q = –1

From j, 2(–1) – r = 0r = –2

[ p(x) = –x3 + 2x2 + x – 2–x2 + 1

x – 2 2 –x3 + 2x2 + x – 2–x3 + 2x2

x – 2x – 2

0[ p(x) = (x – 2)(–x2 + 1)

= (x – 2)(1 + x)(1 – x)Hence, the zeroes of p(x) are 2, –1 and 1.

13 Since q(x) is divisible by x2 + x – 6 = (x –2)(x + 3), then it is also divisible by (x – 2) and (x + 3).

q(2) = 0 m(2)3 – 5(2)2 + k(2) + 54 = 0

8m + 2k = –344m + k = –17 …j

q(–3) = 0 m(–3)3 – 5(–3)2 + k(–3) + 54 = 0

–27m – 3k = –99m + k = 3 …k

k – j: 5m = 20 ⇒ m = 4From j, 4(4) + k = –17 ⇒ k = –33

14 Since (x + 2) is a factor of p(x), thenp(–2) = 0

(–2)3 + 4(–2)2 –h(–2) + k = 02h + k = –8

k = –2h – 8 …j

–



When p(x) is divided by (x – h), the remainder is h3.

p(h) = h3

h3 + 4h2 – h2 + k = h3

3h2 + k = 0 …k Substituting j into k,

3h2 – 2h – 8 = 0(3h + 4)(h – 2) = 0

h = – 43

or 2

When h = – 43

, k = –21– 43 2 – 8 = – 16

3 When h = 2, k = –2(2) – 8 = –12

15 When a polynomial p(x) of degree n 2 is divided by 2x2 + 3x – 2 = (2x – 1)(x + 2), the remainder is an expression in the form ax + b, where a and b are constants. i.e.

p(x) = (2x – 1)(x + 2) q(x) + (ax + b) When p(x) is divided by (2x – 1), the

remainder is 32

.

p1 12 2 = (0)1 1

2 + 22 q(x) + 1

2 a + b = 3

2a + 2b = 3 …j

When p(x) is divided by (x + 2), the remainder is –1.

p(–2) = [2 × (–2) – 1](0) q(x) + (–2a + b) = –1

–2a + b = –1 …kSolving j and k, a = 1, b = 1Hence, the remainder when p(x) is divided by 2x2 + 3x – 2 is ax + b = x + 1.

16 x3 + x – 2x2 – 4 2 x5 – 3x3 – 2x2 – 4x + 8

x5 – 4x3

x3 – 2x2 – 4x + 8x3 – 4x

– 2x2 + 8– 2x2 + 8

0

The remainder is 0. Since the remainder is 0, (x2 – 4) is a factor

of p(x).

[ p(x) = (x2 – 4)(x3 + x – 2) Let q(x) = x3 + x – 2. If x = 1, q(x) = 13 + 1 – 2 = 0 [ (x – 1) is a factor of q(x). [ p(x) = (x2 – 4)(x – 1)(x2 + x + 2) When p(x) = 0, x2 – 4 = 0 or x – 1 = 0 or x2 + x + 2 = 0 x = ± 2, x = 1,

No real solutions because b2 – 4ac = 12 – 4(1)(2) = –7 (< 0)

The roots of p(x) = 0 are ±2 and 1.

17 (a) x2 – 1 = (x – 1)(x + 1) p(1) = 12n – (m + 2)(1)2 + m + 1

= 1 – m – 2 + m + 1 = 0 Thus, (x – 1) is a factor of p(x). p(–1) = (–1)2n – (m + 2)(–1)2 + m + 1

= 1 – m – 2 + m + 1 = 0 Thus, (x + 1) is a factor of p(x). Since (x – 1) and (x + 1) are factors of

p(x), then (x – 1)(x + 1) = x2 – 1 is a factor of p(x).

(b) When m = 8, p(x) = x2n – (8 + 2)x2 + 8 + 1= x2n – 10x2 + 9

Since (x – 3) is a factor, thenp(3) = 0

32n – 10(3)2 + 9 = 032n = 8132n = 34

2n = 4n = 2

Hence, p(x) = x4 – 10x2 + 9 = (x2 – 9) (x2 – 1)

= (x + 3)(x – 3)(x + 1) (x – 1)

18 (a) p(x) = x4 + ax3 – 7x2 – 4ax + bSince (x + 3) is a factor of p(x), p(–3) = 0.

(–3)4 + a(–3)3 – 7(–3)2 – 4a(–3) + b = 081 – 27a – 63 + 12a + b = 0

–15a + b = –18 … jWhen p(x) is divided by (x – 3), the remainder is 60.Therefore, p(3) = 6034 + a(3)3 – 7(3)2 – 4a(3) + b = 60

81 + 27a – 63 – 12a + b = 6015a + b = 42 … k



j + k, 2b = 24b = 12

From j, –15a + 12 = –18a = 2

[ p(x) = x4 + 2x3 – 7x2 – 8x + 12

(b) x3 – x2 – 4x + 4x + 3 2 x4 + 2x3 – 7x2 – 8x + 12

–(x4 + 3x3)–x3 – 7x2

–(–x3 – 3x2)– 4x2 – 8x

–(– 4x2 – 12x)4x + 12

–(4x + 12)0

Let f (x) = x3 – x2 – 4x + 4f (1) = 13 – 12 – 4(1) + 4 = 0Therefore, (x – 1) is another factor of p(x).

x2 – 4x – 1 2 x3 – x2 – 4x + 4

–(x3 – x2)– 4x + 4

–(– 4x + 4)0

Hence, p (x) = (x + 3)(x – 1)(x2 – 4)= (x + 3)(x – 1)(x + 2)(x – 2)

12y4 – 8y3 – 7y2 + 2y + 1 = 0

1211x2

4

– 811x2

3

– 711x2

2

+ 211x2 + 1 = 0

12 – 8x – 7x2 + 2x3 + x4 = 0(x + 3)(x – 1)(x + 2)(x – 2) = 0

x = –3, 1, –2 or 21y = –3, 1, –2 or 2

y = – 13

, 1, – 12

or 12

19 (a) p(x) = 2x 3 + 4x 2 + 12

x – k

Since (x + 1) is a factor of p(x), then p(–1) = 0

2(–1)3 + 4(–1)2 + 12

(–1) – k = 0

–2 + 4 – 12

– k = 0

32

– k = 0

k = 32

(b) p(x) = 2x3 + 4x2 + 12

x – 32

2x2 + 2x – 32

x + 1 2 2x3 + 4x2 + 12

x – 32

–(2x3 + 2x2)

2x2 + 12

x

–(2x2 + 2x)

– 32

x – 32

–1– 32

x – 3220

Hence, p(x) = (x + 1)12x2 + 2x – 322

= (x + 1)14x2 + 4x – 32 2

= 12

(x + 1)(2x + 3)(2x – 1)

20 (a) Since (x + 2) is a factor, thenp(–2) = 0

6(–2)4 – a(–2)3 – b(–2)2 + 28(–2) + 12 = 096 + 8a – 4b – 56 + 12 = 0

8a – 4b = –522a – b = –13 …j

Since (x – 2) is a factor, then p(2) = 06(2)4 – a(2)3 – b(2)2 + 28(2) + 12 = 096 – 8a – 4b + 56 + 12 = 0 –8a – 4b = –164 2a + b = 41 …k

j + k: 4a = 28a = 7

From j, 2(7) – b = –13b = 27

p(x) = (x + 2)(x – 2)g(x)

Letting y = 1x



6x2 – 7x – 3x2 – 4 2 6x4 – 7x3 – 27x2 + 28x + 12

(–) 6x4 – 24x2

–7x3 – 3x2 + 28x + 12(–) –7x3 + 28x

–3x2 + 12(–) –3x2 + 12

0

p(x) = (x + 2)(x – 2)(6x2 – 7x – 3)= (x + 2)(x – 2)(2x – 3)(3x + 1)

(b) p(x) = (x + 2)(x – 2)(2x – 3)(3x + 1)= (2x – 3)[(x + 2)(x – 2)(3x + 1)]= (2x – 3)[(x2 – 4)(3x + 1)]= (2x – 3)(3x3 + x2 – 12x – 4)= (2x – 3)(3x3 – 41 + 37 + x2 – 12x)

q(x)q(x) = x2 – 12x + 37

= x2 – 12x + (–6)2 – (–6)2 + 37= (x – 6)2 + 1

The minimum point is (6, 1).When x = –2, y = q(–2)

= (–2)2 – 12(–2) + 37 = 65

When x = 10, y = q(10)= 102 – 12(10) + 37 = 17

The graph of y = q(x) for x P [–2, 10]is as shown below.

y

x

(6, 1)

(10, 17)

(–2, 65)

O

Hence, the corresponding range for x P [–2, 10] is [1, 65].

21 4x2 – x + 3

x3 – 1 = 4x2 – x + 3

(x – 1)(x2 + x + 1)

≡ A

x – 1 +

Bx + Cx2 + x + 1

4x2 – x + 3 ≡ A(x2 + x + 1) + (Bx + C)(x – 1) Letting x = 1, 6 = 3A ⇒ A = 2 Letting x = 0, 3 = A + C(–1)

5

3 = 2 – CC = –1

Letting x = –1, 8 = A + (–B + C)(–2)8 = 2 + (–B – 1)(–2)8 = 2 + 2B + 2

2B = 4B = 2

[ 4x2 – x + 3x3 – 1

= 2x – 1

+ 2x – 1x2 + x + 1

22 Since the remainders when p(x) is divided by (x + 1) is 0, p(–1) = 0.

p(–1) = 0 (–1)3 + m(–1)2 + 15(–1) + k = 0

–1 + m – 15 + k = 0m + k = 16 …j

Since the remainders when p(x) is divided by (x + 2) is (– 4), p(–2) = – 4.

p(–2) = – 4 (–2)3 + m(–2)2 + 15(–2) + k = – 4

–8 + 4m – 30 + k = – 44m + k = 34 …k

k – j: 3m = 18 ⇒ m = 6 From j: 6 + k = 16 ⇒ k = 10

[ p(x) = x3 + 6x2 + 15x + 10Since the remainders when p(x) is divided by (x + 1) is 0, (x + 1) is a factor of p(x).

x2 + 5x + 10x + 12 x3 + 6x2 + 15x + 10

x3 + x2

5x2 + 15x5x2 + 5x

10x + 1010x + 10

0[ p(x) = (x + 1)(x2 + 5x + 10)

x + 7p(x)

= x + 7(x + 1)(x2 + 5x + 10)

≡ Ax + 1

+ Bx + Cx2 + 5x + 10

x + 7 ≡ A(x2 + 5x + 10) + (Bx + C)(x + 1)Letting x = –1, 6 = 6A ⇒ A = 1Letting x = 0, 7 = 10A + C

7 = 10(1) + CC = –3

Letting x = 1, 8 = 16A + 2B + 2C8 = 16(1) + 2B + 2(–3)



2B = –2B = –1

[ x + 7(x + 1)(x2 + 5x + 10)

= 1x + 1

+ –x – 3x2 + 5x + 10

= 1x + 1

– x + 3x2 + 5x + 10

23 –16 x3 – 4x2 + 4x – 16 0 When –16 x3 – 4x2 + 4x – 16,

x3 – 4x2 + 4x 0 x(x2 – 4x + 4) 0

x (x – 2)2 0Since (x – 2)2 0,

in order that x(x – 2)2 0, then x 0 … j When x3 – 4x2 + 4x – 16 0, we let f (x) = x3 – 4x2 + 4x – 16. f (4) = 43 – 4(4)2 + 4(4) – 16 = 0 Thus, (x – 4) is a factor of f (x).

x2 + 4x – 42 x3 – 4x2 + 4x – 16

x3 – 4x2

4x – 164x – 16

0 x3 – 4x2 + 4x – 16 0 (x – 4)(x2 + 4) 0 Since x2 + 4 > 0, in order that (x – 4)(x2 + 4) 0, then

x – 4 0 ⇒ x 4 …k Combining j and k, the required set of values of x is

{x | 0 x 4}.

24 3x – 5

x x – 3

3x – 5x

– x + 3 0

3x – 5 – x2 + 3xx

0

–x2 + 6x – 5x

0

x2 – 6x + 5x

0

(x – 1)(x – 5)x

0

−

−

−

x − 1 � 0

x − 5 � 0

x > 0−

−

+

+

+

+

+

+

−

+

+

x0 1 5 − −

The required set of values of x is{x | x < 0 or 1 x 5}.

25 xx – 3 < 4

|x||x – 3|

< 4

|x| < 4|x – 3|x2 < 16(x – 3)2

x2 < 16x2 – 96x + 1440 < 15x2 – 96x + 1440 < 5x2 – 32x + 480 < (x – 4)(5x – 12)

− + + 5x − 12 � 0

+−− x − 4 � 0

4125+ − +

x

Hence, the required set of values of x is

5x | x < 125 or x > 46.

Alternative method

– 4 < xx – 3

< 4

For the left-end For the right-end inequality, inequality,

– 4 < x

x – 3 x

x – 3 < 4

xx – 3 + 4 > 0

xx – 3

– 4 < 0

x + 4(x – 3)

x – 3 > 0 x – 4(x – 3)

x – 3 < 0

x + 4x – 12

x – 3 > 0 –3x + 12

x – 3 < 0

5x – 12x – 3

> 0 3(–x + 4)

x – 3 < 0

We write ‘<’ and not‘’ because x ≠ 0.



3125+ − +

− + +

x − 3 � 0

5x − 12 � 0

− − +−x + 4 � 0

x − 3 � 0

43− + −

− + +

+ + −

xx

[ x < 125 or x > 3 …j [ x < 3 or x > 4 …k

Combining j and k:

3

x4

x < 3 or x > 4

125

125

or x > 3x <

The required set of values of x is

5x | x < 125 or x > 46.

26 The graphs of y = |x + 2| and y = 1

x + 1 is as shown below.

x

y

O−2

2

−1

y = 1x + 1

y = x + 2

y = −x − 2

A

y = x + 2 …j

y = 1

x + 1 …k

Substituting j into k,

x + 2 = 1

x + 1 x2 + 3x + 2 = 1 x2 + 3x + 1 = 0

x = –3 ± 32 – 4(1)(1)2(1)

x = –3 ± 5

2 The x-coordinate of point A is

x = –3 + 5

2.

Based on the graphs, the solution set of x

for which |x + 2| > 1

x + 1 is

5x < –1 or x > –3 + 5

2 6.

27

xO−1

−1

1

1 3

y = x − 1

y

A y = −x − 1 x + 1y =

To determine the x-coordinate of point A, solve

y = x – 1 …j y = x + 1 …k Substituting j into k,

x – 1 = x + 1(x – 1)2 = x + 1

x2 – 2x + 1 = x + 1x2 – 3x = 0

x(x – 3) = 0 Thus, the x-coordinate of point A is x = 3. The part of the x-axis where the graph of

y = x + 1 is above the graph of y = |x| – 1 is –1 x 3.

Hence, the required set of values of x is {x | –1 x 3}.

28 p(x) = 2x3 + hx2 + kx + 36 Since (x – 3) is a factor, then

p(3) = 0 2(3)3 + h(3)2 + k(3) + 36 = 0

9h + 3k = –903h + k = –30 …j

p(x) = (x + 2) f (x) – 30 means that the remainder when p(x) is divided by (x + 2) is –30.

p(–2) = –30 2(–2)3 + h(–2)2 + k(–2) + 36 = –30

4h – 2k = –502h – k = –25 …k

This is the set of values of x where the graph of y = |x + 2| is above the graph of

y = 1x + 1

.



j + k: 5h = –55 ⇒ h = –11 From j: 3(–11) + k = –30

k = 3 Therefore, p(x) = 2x3 – 11x2 + 3x + 36. 2x2 – 5x – 12

x – 32 2x3 – 11x2 + 3x + 362x3 – 6x2

–5x2 + 3x–5x2 + 15x

–12x + 36–12x + 36

0 Therefore, p(x) = (x – 3)(2x2 – 5x – 12)

= (x – 3)(2x + 3)(x – 4)

3 4x

−32

The sets of values of x such that p(x) 0

is 5x | – 32 x 3 or x 46.

29 p(x) = 2x3 + px2 + qx + 6 Since (2x + 1) is a factor of p(x), then

p1– 122 = 0

21– 122

3

+ p1– 122

2

+ q1– 122 + 6 = 0

– 14

+ 14

p – 12

q + 6 = 0

– 1 + p – 2q + 24 = 0 p – 2q = –23 …j

When p(x) is divided by (x + 3), the remainder is –15.

p(–3) = –15 2(–3)3 + p(–3)2 + q(–3) + 6 = –15

9p – 3q = 333p – q = 11 …k

p – 2q = –23 …j –6p – 2q = 22 …k × 2

–5p = – 45 p = 9 From j: 9 – 2q = –23 ⇒ q = 16

[ p(x) = 2x3 + 9x2 + 16x + 6

x2 + 4x + 62x + 12 2x3 + 9x2 + 16x + 6

2x3 + x2

8x2 + 16x8x2 + 4x

12x + 612x + 6

0 Let q(x) = x2 + 4x + 6

= x2 + 4x + 14222

– 14222

+ 6

= (x + 2)2 + 2 [> 0] [Shown]

p(x) = (2x + 1)(x2 + 4x + 6)

Since x2 + 4x + 6 is positive for all real values of x, then p(x) < 0 only if

2x + 1 < 0 ⇒ x < – 12

.

Hence, the solution set is 5x | x < – 126.

30 Sketch the graphs of y = |x – 2| and y = 1x .

2

O 1 2

P

Q

1x

y =

1x

y =

y = −x + 2

y = x − 2

21 +

y

x

To determine the x-coordinates of the points of intersection of the graphs of

y = |x – 2| and y = 1x , solve the following

simultaneous equations.

Case 1 (for point P)

y = 1x …j

y = –x + 2 …k



Substituting j into k:1x = –x + 2

1 = –x2 + 2x x2 – 2x + 1 = 0

(x – 1)2 = 0x = 1

Case 2 (for point Q)

y = 1x … j

y = x – 2 … l

Substituting j into l:1x = x – 2

1 = x2 – 2x x2 – 2x – 1 = 0

x = – (–2) ± (–2)2 – 4(1)(–1)

2(1)

= 2 ± 82

= 2 ± 2 22

= 1 ± 2

x = 1 – 2 is not accepted because x must be positive.

[ x = 1 + 2

Hence, the solution set for the inequality

|x – 2| < 1x

is {x | 0 < x < 1 + 2 , x ≠ 1}.

31 y = 4

x – 1

4x – 1

, x > 1,y = 5– 1 4

x – 12, x < 1.

As y ! ± `, x – 1 ! 0x ! 1

Thus, x = 1 is the asymptote.

As x ! ± ̀ , y ! 0.

y = 3 – 3x

As y ! ± ̀ , x ! 0.

Thus, x = 0 (the y-axis) is the asymptote.

As x ! ± ̀ , y ! 3.Thus, y = 3 is the asymptote.

1O

y

x3

3

4

3x

y = 3 −

4x − 1

y =

A 3x

y = 3 −4x − 1

y = −

The x-coordinate of point A is obtained by solving the following equations simultaneously.

y = 4x – 1

…j

y = 3 – 3x …k

4x – 1

= 3 – 3x

4x – 1

= 3x – 3x

(3x – 3)(x – 1) = 4x3x2 – 6x + 3 – 4x = 0

3x2 – 10x + 3 = 0(3x – 1)(x – 3) = 0

x = 13

or 3

x = 13

is not accepted.

Thus, x = 3

The solution set for which * 4x – 1 * > 3 – 3

xis given by the part of the graph where the

curve y = *4

x – 1 * is above the curve

y = 3 – 3x, that is, {x | 0 < x < 1 or 1 < x < 3}.

This is the range of values of x where the graph of y = |x – 2| is below

the graph of y = 1x .



32 xx + 1

> 1x + 1

xx + 1

– 1x + 1

> 0

x – 1x + 1

> 0

Hence, the required set of values of x is{x | x < –1 or x > 1}.

33

x

y

O−1

−2

12y = e −x

y = 2e −

y = −e −x − 1

y = −e −x

x

34 (a)

xO

y

1

y = |ln x |

(b)

−1 1O

y

x

y = ln (−x) y = ln x

(c)

−1 O

y

x

y = −ln (−x)

35 (a) The graph of y = f (x) = ln (x + 1) is as shown below.

−1 O

y

x

y = f (x) = ln (x + 1)

f –1 exists because f is a one-to-one and an onto function.

(b) Let y = f –1(x)f (y) = x

ln (y + 1) = xy + 1 = ex

y = ex – 1f –1(x) = ex – 1

The domain of f –1 is the same as the range of f, i.e. {x | x P }.

The range of f –1 is the same as the domain of f, i.e. {y | y –1, y P }.

(c) g ° f –1 = g[ f –1(x)]

= g(ex – 1)= ex – 1 + 1

= e 12 x

The domain of g ° f –1 is the same as the domain of f –1, i.e. {x | x P }.

The range of g ° f –1 is { y | y > 0, y P }.

x

y

y = e

O

1

12 x

36 (a) f ° g = f [g(x)]

= f 3ln 1 x – 12 24

= 1 + 2eln 1

x – 12 2

= 1 + 21 x – 12 2

= x Since it is known that f f –1(x) = x, by

−

+

x − 1 � 0

x + 1 � 0−

− +

+

x−1

+1−

+



comparison f –1(x) = g(x) = ln x – 12 .

(b) The domain of f –1 is the same as the range of f, i.e. {x | x > 1, x P }.The range of f –1 is the same as the domain of f, i.e. {y | y P }.

(c) The graphs of y = f (x) and y = f –1(x) are as shown below.

1

−1

O

y

x

y = f (x) = 1 + 2ex

3

y = x

3

y = f −1(x) = ln x − 12 )(

37 x–

12 + 2x–1 = 15

x–

12 + 2 1x

– 122

2

= 15

Let x–

12 = u

u + 2u2 = 152u2 + u – 15 = 0

(2u – 5)(u + 3) = 0

u = 52 or u = –3

When u = 52

, When u = –3,

x–

12 =

52

x–

12 = –3

x–1 = 152 2

2

1x =

254

x = 425

38 8x + 6(8–x) = 5

8x + 68x

= 5

[Not possible because

x– 1

2 > 0 for all real

values of x.]

Let 8x = u

u + 6u

= 5

u2 + 6 = 5uu2 – 5u + 6 = 0

(u – 2)(u – 3) = 0u = 2 or u = 3

8x = 2 8x = 323x = 21 x lg 8 = lg 3

3x = 1 x = lg 3lg 8

x = 13

x = 0.528

39 log2 x – logx 8 + 2log2h + h logx 4 = 0

log2 x – log2 8

log2 x + h + h 1

log2 4

log2 x2 = 0

y – log2 2

3

y + h + h 1log2 2

2

y 2 = 0

y – 3y

+ h + h12y 2 = 0

y2 – 3 + hy + 2h = 0

y2 + hy + 2h – 3 = 0 [Shown]

When h = – 14

, y2 – 14

y + 21– 142 – 3 = 0

4y2 – y – 14 = 0(4y + 7)(y – 2) = 0

y = – 74

or 2

Given y = log2 x, then x = 2y.

When y = – 74

, x = 2–

74 = 0.297.

When y = 2, x = 22 = 4.

40 2 logc x – 3 logx c = 5

2 logc x – 31logc c

logc x2 = 5

2 logc x – 31 1logc x

2 = 5

Let logc x = u

2u – 31 1u 2 = 5

2u2 – 3 = 5u2u2 – 5u – 3 = 0

(2u + 1)(u – 3) = 0



u = – 12

or u = 3

logc x = – 12

logc x = 3

x = c–

12 =

1c x = c3

41 loga 1 xa22 = 3 loga 2 – loga (x – 2a)

loga 1 xa22 = loga 2

3 – loga (x – 2a)

loga 1 xa22 + loga (x – 2a) = loga 8

loga 31 xa22(x – 2a)4 = loga 8

xa2

(x – 2a) = 8

x(x – 2a) = 8a2

x2 – 2ax – 8a2 = 0(x + 2a)(x – 4a) = 0

x = –2a or 4ax = –2a (is not

accepted)[ x = 4a

42 Simplify 2log2 a

first.

2log2 a

= 21

loga 2

= 2 loga 2

= loga 22

= loga 4

loga (3x – 4a) + loga 3x = 2log2 a

+ loga (1 – 2a)loga (3x – 4a) + loga 3x = loga 4

+ loga (1 – 2a)loga 3x(3x – 4a) = loga 4(l – 2a)

3x(3x – 4a) = 4(1 – 2a)9x2 – 12ax + 8a – 4 = 0

x = –(–12a) ± (–12a)2 – 4(9)(8a – 4)

2(9)

= 12a ± 144a2 – 288a + 144

18

= 12a ± (12a – 12)2

18

= 12a ± (12a – 12)18

= 24a – 1218

or 1218

= 4a – 23

or 23

For 0 < a < 12

, x = 4a – 23

is not accepted

because when it is substituted into the given equation, it produces loga (–ve) which is undefined.

[ x = 23

43 (a) The graph of y = |sin x| is as shown below.

y

x

1

O2��

2� 3

�2

y = sin x y = −sin x

In the non-modulus form,y = |sin x| is

f (x) =sin x,

-sin x,

0 � x � �,

� � x � 2�.

The graph of y = sin x for 0 < x < 2p is as shown below.y

x

1

–1

O2��

2� 3

�2

Hence, the function f (x) = |sin x| – sin x in the non-modulus form is:

f (x) = {sin x – sin x, 0 < x < p

–sin x – sin x, p < x < 2p

f(x) = {0, 0 < x < p–2 sin x, p < x < 2p

(b) Hence, the graph of y = f (x) = |sin x| – sin x for 0 < x < 2p is as shown below.



y

x

2

O2��

2� 3

�2

y = f (x )

The range of f (x) is {y | 0 < y < 2, y P }.

(c) By using the horizontal-line test, there are two intersection points between the horizontal line and the graph of y = f (x) = |sin x| – sin x. Hence, f (x) is not a one-to-one function.

y

x

2

O2��

2� 3

�2

y = f (x )

Two intersection points

44 (a) LHS = cos4 q + sin2 q= 1cos2 q22 + sin2 q

= 11 – sin2 q22 + sin2 q= 1 – 2 sin2 q + sin4 q + sin2 q= sin4 q + 1 – sin2 q= sin4 q + cos2 q= RHS

[ cos4 q + sin2 q sin4 q + cos2 q [Proven]

(b) LHS = sin (q + a) + cos (q – a)= sin q cos a + cos q sin a +

cos q cos a + sin q sin a

= sin q cos a + cos q cos a + sin q sin a + cos q sin a

= cos a(sin q + cos q) + sin a(sin q + cos q)

= (sin q + cos q)(cos a + sin a)= RHS

[ sin (q + a) + cos(q – a) (sin q + cos q)(cos a + sin a) [Proven]

45 LHS = sin q tan q

tan q – sin q

= sin q 1 sin q

cos q 2sin qcos q – sin q

= sin2 q

sin q – sin q cos q

= 1 – cos2 q

sin q (1 – cos q)

= (1 + cos q)(1 – cos q)

sin q (1 – cos q)

= 1 + cos q

sin q

RHS = tan q + sin qsin q tan q

=

sin qcos q + sin q

sin q 1 sin qcos q 2

= sin q + sin q cos q

sin2 q

= sin q (1 + cos q)

sin2 q

= 1 + cos q

sin q

= LHS

[ sin q tan qtan q – sin q

tan q + sin qsin q tan q

[Proven]

46 LHS = tan (P + Q) – tan P

= sin (P + Q)cos (P + Q) –

sin Pcos P

=

cos P sin (P + Q) – sin P cos (P + Q)cos P cos (P + Q)

=

cos P (sin P cos Q + cos P sin Q) –sin P (cos P cos Q – sin P sin Q)

cos P cos (P + Q)

=

cos P sin P cos Q + cos2 P sin Q –sin P cos P cos Q + sin2 P sin Q

cos P cos (P + Q)

= cos2 P sin Q + sin2 P sin Qcos P cos (P + Q)



= sin Q (cos2 P + sin2 P)cos P cos (P + Q)

= sin Q ? (l)

cos P cos (P + Q)

= sin Q

cos P cos (P + Q)

= RHS[ tan (P + Q) – tan P

sin Q

cos P cos (P + Q) [Proven]

47 LHS = csc 2q – cot 2q

= 1sin 2q

– cos 2qsin 2q

= 1 – cos 2q

sin 2q

= 1 – (1 – 2 sin2 q)

sin 2q

= 2 sin2 q

2 sin q cos q

= sin qcos q

= tan q= RHS

[ csc 2q – cot 2q tan q [Proven]

tan 22.5° = cosec 2(22.5°) – cot 2(22.5°)

= 1sin 45°

– 1

tan 45°

= 112

– 11

= 2 – 1 [Shown]

48 Since A, B and C are angles of a triangle, then A + B + C = 180°.

(a) tan A + tan B + tan C

= sin Acos A

+ sin Bcos B

+ sin Ccos C

=

sin A cos B cos C + sin B cos A cos C + sin C cos A cos Bcos A cos B cos C

=

cos C (sin A cos B + sin B cos A) + sin C cos A cos Bcos A cos B cos C

= cos C [sin (A + B)] + sin C cos A cos B

cos A cos B cos C

=

cos C [sin (180° – C)] + sin C cos A cos B

cos A cos B cos C

= cos C sin C + sin C cos A cos B

cos A cos B cos C

= sin C (cos C + cos A cos B)

cos A cos B cos C

=

sin C {cos [180° – (A + B)] + cos A cos B}

cos A cos B cos C

= sin C {– cos (A + B) + cos A cos B}

cos A cos B cos C

=

sin C (sin A sin B – cos A cos B + cos A cos B)

cos A cos B cos C

= sin A sin B sin C

cos A cos B cos C

= tan A tan B tan C [Shown]

(b) sin 2A + sin 2B + sin 2C= sin 2A + sin 2C + sin 2B

= 2 sin 12A + 2C2 2 cos 12A – 2C

2 2 + sin 2B

= 2 sin (A + C) cos (A – C) + sin 2B= 2 sin (180° – B) cos (A – C) + sin 2B= 2 sin B cos (A – C) + sin 2B= 2 sin B cos (A – C) + 2 sin B cos B= 2 sin B [cos (A – C) + cos B]

= 2 sin B 32 cos 1 A – C + B2 2

cos 1 A – C – B2 24

= 2 sin B 32 cos 1 A + B – C2 2

cos 1 A – (B + C)2 24

= 2 sin B 32 cos 1 180° – C – C2 2

cos 1 A – (180° – A)2 24

= 2 sin B 32 cos 1 180° – 2C2 2

cos 1 2A – 180°2 24

= 4 sin B [cos (90° – C) cos (A – 90°)]= 4 sin B [sin C sin A]= 4 sin A sin B sin C [Shown]



49 (a) LHS = 2 sin 1A + p42 cos 1A +

p42

= sin 2 1A + p42

= sin 12A + p22

= sin 2A cos p2 + cos 2A sin

p2

= (sin 2A)(0) + (cos 2A)(1)= cos 2A= RHS

[ 2 sin A + p4 cos A +

p4 cos 2A

[Proven]

(b) LHS = 2 cos 1B + p42 cos 1B –

p42

= cos 1B + p4 + B –

p42 +

cos 3B + p4 – 1B –

p424

= cos 2B + cos p2

= cos 2B + 0= cos 2B= RHS

[ 2 cos B + p4 cos B –

p4 cos 2B

[Proven]

50 cos 3x = cos2 x4 cos3 x – 3 cos x = cos2 x

4 cos3 x – cos2 x – 3 cos x = 0cos x (4 cos2 x – cos x – 3) = 0

cos x (4 cos x + 3)(cos x – 1) = 0

cos x = 0, – 34

, 1

When cos x = 0,x = 90°, 270°

When cos x = – 34

,

x = 138.6°, 221.4°

When cos x = 1,x = 0°, 360°

[ x = 0°, 90°, 138.6°, 221.4°, 270°, 360°

51 sin 3q + sin2 q = 2

(3 sin q – 4 sin3 q) + sin2 q – 2 = 04 sin3 q – sin2 q – 3 sin q + 2 = 0

(sin q + 1)(4 sin2 q – 5 sin q + 2) = 0sin q + 1 = 0 or 4 sin2 q – 5 sin q + 2 = 0

When sin q + 1 = 0 sin q = –1 q = 270°

For 4 sin2 q – 5 sin q + 2 = 0, there are no real roots because b2 – 4ac = (–5)2 – 4(4)(2) = –7 (< 0)[ q = 270°

4 sin2 q – 5 sin q + 2

sin q + 1 2 4 sin3 q – sin2 q – 3 sin q + 24 sin3 q + 4 sin2 q

–5 sin2 q – 3 sin q–5 sin2 q – 5 sin q

2 sin q + 22 sin q + 2

0

52 tan x + cot x = 8 cos 2x

sin xcos x + cos x

sin x = 8 cos 2x

sin2 x + cos2 xsin x cos x

= 8 cos 2x

1sin x cos x

= 8 cos 2x

22 sin x cos x

= 8 cos 2x

2sin 2x

= 8 cos 2x

1 = 4 sin 2x cos 2x1 = 2(2 sin 2x cos 2x)1 = 2 sin 4x

sin 4x = 12

Basic = p6

4x = 16

p, 56

p, 136

p, 176

p

x = 124

p, 524

p, 1324

p, 1724

p

If 0 < x < p, then 0 < 4x < 4p.

sin 3q = 3 sin q – 4 sin3 q



53 sin3 x sec x = 2 tan x

sin2 x sin x 1 1cos x 2 = 2 tan x

sin2 x tan x = 2 tan xsin2 x tan x – 2 tan x = 0

tan x (sin2 x – 2) = 0tan x = 0 or sin2 x = 2

When tan x = 0,x = 0 or p

When sin2 x = 2,

sin x = ± 2

[sin x = ± 2 is not possible because it is out of the range of –1 < sin x < 1].Hence, x = 0 or p.

54 LHS = cos 5

2 A + cos

32 A

sin 32 A + sin A

2

=

2 cos 12 15

2 A + 32 A2 cos

12 15

2 A – 32

A22 sin 1

2 132 A + 1

2 A2 cos 1

2 132 A – 1

2 A2

= 2 cos 2A cos 1

2 A

2 sin A cos 12

A

= cos 2Asin A

= cos2 A – sin2 Asin A

= cos2 Asin A

– sin2 Asin A

= cos Asin A

cos A – sin A

= cot A cos A – sin A= RHS

[ cos 5

2 A + cos 3

2 A

sin 32

A + sin 12

A cot A cos A – sin A

[Proven]

55 LHS = cos 2A – 2 cos 4A + cos 6Acos 2A + 2 cos 4A + cos 6A

= cos 6A + cos 2A – 2 cos 4Acos 6A + cos 2A + 2 cos 4A

=

2 cos 16A + 2A2 2 cos 1 6A – 2A

2 2 – 2 cos 4A

2 cos 1 6A + 2A2 2 cos 1 6A – 2A

2 2 + 2 cos 4A

= 2 cos 4A cos 2A – 2 cos 4A2 cos 4A cos 2A + 2 cos 4A

= 2 cos 4A (cos 2A – 1)2 cos 4A (cos 2A + 1)

= cos 2A – 1cos 2A + 1

= 1 – 2 sin2 A – 1

2 cos2 A – 1 + 1

= – 2 sin2 A2 cos2 A

= – sin2 Acos2 A

= – tan2 A= RHS

[ cos 2A – 2 cos 4A + cos 6Acos 2A + 2 cos 4A + cos 6A

– tan2 A

[Proven]

56 sin x – sin 3x + sin 5x = 0

sin 5x + sin x – sin 3x = 0

2 sin 5x + x2

cos 5x – x2

sin 3x = 0

2 sin 3x cos 2x – sin 3x = 0 sin 3x (2 cos 2x – 1) = 0

sin 3x = or cos 2x = 12

When sin 3x = 0 3x = 0°, 180°, 360°, 540° x = 0°, 60°, 120°, 180°

When cos 2x = 12

2x = 60°, 300° x = 30°, 150° [ x = 0°, 30°, 60°, 120° 150°, 180°

[ x = 0, 12

p, 13

p,23

p, 56

p, p

x° = x × 180p

rad.



57

q

1 − t 2

2t

1 + t 2

(a) LHS = csc q – cot q

= 1 + t2

2t – 1 – t2

2t

= 1 + t2 – 1 + t2

2t

= 2t2

2t= t

= tan q2

= RHS[ csc q – cot q tan

q2

(b) LHS = sec q – tan q

= 1 + t2

1 – t2 – 2t1 – t2

= 1 + t2 – 2t1 – t2

= t2 – 2t + 11 – t2

= (t – 1)2

(1 + t)(1 – t)

= (1 – t)2

(1 + t)(1 – t)

= 1 – t1 + t

= tan

p4 – tan

q2

1 + tan p4 tan

q2

= tan 1p4 –

q22

= RHS

[ sec q – tan q tan p4 – q2

csc q – cot q = sec q – tan q

tan q2 = tan 1p

4 – q22

q2 =

p4 –

q2 or

q2 = p + 1p

4 – q22

q = p4 or q =

54 p

58 8 sin q – 3 cos q = r sin (q – a) = r(sin q cos a – cos q sin a)= r sin q cos a – r cos q sin a

By comparison, r cos a = 8 …jr sin a = 3 …k

Squaring and adding j and k:r2 (cos2 a + sin2 a) = 82 + 32

r2(1) = 73r = 73

kj

: r sin ar cos a =

38

tan a = 38

a = tan–1 1 38 2

= 20.56°

[ 8 sin q – 3 cos q = 73 sin (q – 20.56°)

= 73 sin (q – 20.6°) [Correct to the nearest 0.1°]

The maximum value of 8 sin q – 3 cos q is 73.The minimum value of 8 sin q – 3 cos q is – 73.

8 sin q – 3 cos q = 734

73 sin (q – 20.56°) = 734

sin (q – 20.56°) = 14

q – 20.56° = 14.48°, 165.52°q = 35.0°, 168.1°

[Correct to the nearest 0.1°]

First quadrant.

Third quadrant.



59 Let 4 sin q – 3 cos q r sin (q – a)r = 42 + (–3)2 = 5

a = tan–1 1342 = 36.87°

[ 4 sin q – 3 cos q = 5 sin (q – 36.9°)

4 sin q – 3 cos q = 35 sin (q – 36.87°) = 3

sin (q – 36.87°) = 35

q – 36.87° = 36.87°, 143.13°q = 73.7°, 180.0°

60

1

O

–1

x

y

4 2 4 88

y = cos 2x

y =21

π π π π3π 7π

cos 2x = 2 cos2x – 1

2 cos2 x = 1 + cos 2x

2 2 cos2 x = 2 + 2 cos 2x

2 2 cos2 x 1 + 2 becomes

2 + 2 cos 2x 1 + 2

cos 2x 12

when cos 2x = 12

2x = p4, 2p –

p4

x = p8,

7p8

[ Solution set is

{x | 0 x p8,

7p8 x p}

maths t (chap 1)

Documents

graph of f x

othe domain of f x

x2x55 5oy25 x

x4 x 0x4

x4444 b thel range of

domain of g f

domain of f g

range of g f