CHAPTER 1 FUNCTIONS
Focus on Exam 1
(b)
x2 1 (a) g ( x) x) = 16 – x For g For g ( x) x) to be defined, 16 – 16 – x x2 0 (4 + x x)(4 )(4 – x) 0 – x)
y y
Hence, the domain of of g g ( x) x) is { x : }. : – 4 x 4, x � R }. (b)
9
−
3
(c) The range range is { y : y
x
4
x 2
x
O
3
−
−4
=
0, y
3 First, consider only only g g ( x) x) =
12 x x – – 2. The
graph of of g g ( x) x) is as shown below.
y The graph of g (x ) is actually part of a circle with the equation y 2 = 16 − x 2 ⇒ x 2 + y 2 = 42.
y y=−
1 2
x+2 y=
4 2
1 2
x−2
2
y = 16 − x
4
−
−2
x 4
(c) The range is { y : 0 : x 2 (a) f : x
x
4
O
O
y
4, y
}. R }.
�
Thus, g Thus, g ( x) x) =
x2 – 9 9
f ( x) x) = x2 – 9 9 For f ( x) For f x) to be defined, x defined, x2 – 9 9 ( x x + 3)( x – 3)
0 0
�
1 – x + 2, x < 4, 2 1 x 2, – 2, x 2
Next, consider only h( x) x) =
graph of h( x) x) is as shown below. y=
x
3
Hence, the domain of of f f ( x) x) is { x : }. : x –3 or x 3, x � R }.
4.
12 x + 2. The
y
−3
}. R }.
�
y=−
1 2
1 2
x+2
x−2 2
−4
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O
x
Fajar Sdn. Bhd. (008974-T) 2012
2
ACE AHEAD Mathematics (T) First Fi rst Term
Thus, h( x) x) =
�
1 – x – x – 2, x < – 4, 2 1 x 2, x + 2
The domain of f of f ( x) x) is { x : : x � R }. }. The range of f of f ( x) x) is { y : y 2, y � R }. }.
– 4.
(ii) g ( x) x) =
Therefore,
1 x – x – 2 y
�
�
1 1 f ( x) x) = – x + 2 – – – x x – – 2 2 2 2
for x for x < – 4,
•
y
=4 •
�
−
�
1 1 for – 4 x < 4, 4, f f(( x) x) = – x + 2 – x x + + 2 2 2 x = –
for x for x
•
�
�
1 1 f(( x) f x) = x – x – 2 – x + 2 2 2
4,
= – 4 x < – 4,
4,
�
Hence, f ( x x ) = – x , – 4,
–4
x < 4,
x 4.
(a) The graph of of f f ( x) x) is as shown below. y y
=
4
2
2
[( x x + 1)2 + 2] = g [( 1 = 2 ( x x + 1) + 2 – 2 1 , x ≠ ≠ –1 = ( x x + 1) 2 The domain of g of g ° f is is { x : : x � R, x ≠ –1}. The range of g of g ° f is is { y : y > 0, y y
y
=
x
−
4
y = g f (x )
x
O
4
−
The domain of g of g ( x) x) is { x : : x � R, x ≠ 2}. The range of g of g ( x) x) is { y : y � R, y ≠ 0}. (b) g ° f = g [ f ( x)] x)]
4
−
1
x
x
O
1 2
=
=
y
(b) The range of of f f ( x) x) is { y : – 4 y 4, y � R }. }.
=
−
x
O
1
−
−4
1 (x + 1)2
4
5 (a) (i) f ( x) x) = x x – 2 – 2 y
x) = ( x + 1)2 + 2 4 (a) (i) f ( x)
y
=
x
2
−
y O y = (x
x 2
1)2 + 2
+
3
(−1, 2) O
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x
Fajar Sdn. Bhd. (008974-T) 2012
The domain of f of f ( x) x) is { x : : x 2, x � R} . The range of f of f ( x) x) is { y : y 0, y � R} .
R} .
�
Fully Worked Solution
(ii) g ( x) = x2 – 3 y
O
(c) For f ° g to be defined, R g D f . D f R g D f
x y
=
x 2
3
−
3
}
−
–5 0
The domain of g ( x) is { x : x � R} . The range of g ( x) is { y : y –3, y � R} .
The domain of g ( x) is { x : x � R} . The range of g ( x) is { y : y –5, y � R} . (b) f ° g does not exist because R g D f .
2
x
x
2
10
7 (a) (i) f ( x) =
1 x – 2 y
= � x – 2 �2 – 3 = x – 5 (c) For f ° g to be defined, R g D f . R g D f { { 2
x2 – 5
0
−1
25 – x2 y
x
O
2
2
�x + 5�� x – 5� 0 Hence, the required set of values of x is { x : x – 5 or x 5, x � R} . 6 (a) (i) f ( x) =
5
Hence, the required set of values of x is { x : – 10 x 10, x � R} .
(b) g ° f exists because R f D g . g ° f = g [ f ( x)] = g � x – 2 �
x2 – 3
}
}
– 5
The domain of f ( x) is { x : x � R, x ≠ 2} . The range of f ( x) is { y : y � R, y ≠ 0}. 2 (ii) g ( x) = x + 4 y
5
y
=
25
−
x 2 1
O
5
−
x
2
5
O
−4
The domain of f ( x) is { x : –5 x 5, x � R} . The range of f ( x) is { y : 0 y 5, y � R}. (ii) g ( x) = x2 – 5
y
The domain of g ( x) is { x : x � R, x ≠ – 4}. The range of g ( x) is { y : y � R, y ≠ 0} . (b) f ° g = f [ g ( x)] 2 = f x + 4 1 = 2 – 2 x + 4 x + 4 = 2 – 2( x + 4)
�
x
O y 5
−
=
x 2
5
−
�
© O xford
x
…(1)
�
�
Fajar Sdn. Bhd. (008974-T) 2012
3
4
ACE AHEAD Mathematics (T) First Term
x + 4 , x ≠ –3 = –6 – 2 x
9 (a) The graph of y = f ( x) = x2 – 3 x is as shown below.
…(2)
y
Combining (1) and (2), the domain of f ° g is { x : x � R, x ≠ – 4, x ≠ –3} .
y
y = f –1( x) f ( y) = x 2 + y – 1 = x y – 1 = x – 2 y – 1 = ( x – 2)2 y – 1 = x2 – 4 x + 4 y = x2 – 4 x + 5 –1 � f ( x ) = x 2 – 4 x + 5 The domain of f –1 is the same as the range of f , i.e. { x : x 2, x � R} . The range of f –1 is the same as the domain of f , i.e. { y : y 1, y � R} . (b) The graphs of y = f ( x) and y = f –1( x) are as shown below.
=
f (x )
=
x 2
8 (a) Let
=
3 2 1
=
The graph of y f 1(x ) is the reflection of the graph of y f (x ) in the straight line y x . =
O 1 2 3 4 5
−
=
x
x
=
y
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 – 4 x + 5 and the straight line y = x. y = x2 – 4 x + 5 …(1) y = x …(2) 2 x – 4 x + 5 = x x2 – 5 x + 5 = 0 –(–5) ± (– 5)2 – 4(1)(5) x = 2(1) 5± 5 x = 2 x = 1.38 or 3.62 x = 1.38 is not accepted � x = 3.62 � y = x = 3.62 Hence, the required point of intersection is (3.62, 3.62). © O xford
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)
(b) In order for f –1 to exist, the domain of f must be restricted to only 1 x : x 1 , x � R . 2 Let y = f –1( x) f ( y) = x 2 y – 3 y = x y2 – 3 y – x = 0 2 x) y = –(– 3) + (– 3) – 4(1)(– 2(1) 3 + 9 + 4 x y = 2 3 + 9 + 4 x –1 � f ( x ) = 2
�
f (x ) =
21 4
−
f –1 does not exist because f is not a one-to-one function.
−
y
3
(112 ,
f 1(x )
5 4
3x x
O
y y
−
�
The domain of f –1 is the same as the 1 range of f , i.e. x : x –2 , x � R . 4 10 Since ( x – 2) is a factor of P ( x) = qx3 – rx2 + x – 2, P (2) = 0 3 2 q(2) – r (2) + 2 – 2 = 0 8q – 4r = 0 2q – r = 0 …(1) P ( x) has a remainder of –12 when it is divided by ( x + 1). P (–1) = –12 3 2 q(–1) – r (–1) – 1 – 2 = 0 –q – r = 3 …(2) (1) – (2): 2q – r = 0 – – q – r = 3
�
3q q From (1), 2(–1) –
= –3 = –1 r = 0 r = –2 3 2 � P ( x) = – x + 2 x + x – 2
�
Fully Worked Solution
–x2 x – 2 � –x3 – x3
+ 1 + 2 x2 + x – 2 + 2 x2 x – 2 x – 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.
11 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 3 2 m(2) – 5(2) + k (2) + 54 = 0 8m + 2k = –34 4m + k = –17 …(1) Q(–3) = 0 3 2 m(–3) – 5(–3) + k (–3) + 54 = 0 –27m – 3k = –9 9m + k = 3 …(2) (2) – (1): 5m = 20 ⇒ m = 4 From (1), 4(4) + k = –17 ⇒ k = –33 12 Since ( x + 2) is a factor of P ( x), then P (–2) = 0 3 2 (–2) + 4(–2) – h(–2) + k = 0 2h + k = –8 k = –2h – 8 …(1) When P ( x) is divided by ( x – h), the remainder is h3. P (h) = h3 h3 + 4h2 – h2 + k = h3 3h2 + k = 0 …(2) Substituting (1) into (2), 3h2 – 2h – 8 = 0 (3h + 4)(h – 2) = 0 4 h = – or 2 3
� �
16 4 4 , k = –2 – – 8 = – 3 3 3 When h = 2, k = –2(2) – 8 = –12
When h = –
13 When a polynomial P ( x) of degree n 2 is divided by 2 x2 + 3 x – 2 = (2 x – 1)( x + 2), the remainder is an expression in the form ax + b, where a and b are constants. i.e.
P ( x) = (2 x – 1)( x + 2) Q( x) + (ax + b) When P ( x) is divided by (2 x – 1), the 3 remainder is . 2
� 12 � = (0)� 12 + 2� Q( x) + 12 a + b = 32
P
a + 2b = 3 …(1)
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 …(2) Solving (1) and (2), a = 1, b = 1 Hence, the remainder when P ( x) is divided by 2 x2 + 3 x – 2 is ax + b = x + 1 .
14 6 x4 – 25 x3 – 12 x2 + 25 x + 6 = 0 Dividing throughout by x2,
6 x2 – 25 x – 12 +
�
6 x2 +
�
�
25 6 + = 0 x x2
�
1 1 – 25 x + – 12 = 0 2 x x
�
…(1)
2
�
1 Considering z = x – , x 2
z 2 = x2 – 2 +
1 x2
1 z 2 + 2 2 = x From (1), 6( z 2 + 2) – 25 z – 12 = 0 6 z 2 + 12 – 25 z – 12 = 0 6 z 2 – 25 z = 0 z (6 z – 25) = 0 25 z = 0 or z = 6 1 When z = 0, x – = 0 x x2 – 1 = 0 x = ±1 25 1 25 When z = , x – = 6 x 6 2 6 x – 6 = 25 x x2 +
6 x2 – 25 x – 6 = 0 25 ± (– 25)2 – 4(6)(–6) x = 2(6) 25 ± 769 12 x = –0.228 or 4.39 x =
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5
6
ACE AHEAD Mathematics (T) First Term
15
x3 + x – 2 x2 – 4 � x5 – 3 x3 – 2 x2 – 4 x x5 – 4 x3 x3 – 2 x2 – 4 x x3 – 4 x 2 – 2 x – 2 x2
+ 8 + 8 + 8 + 8 0
The remainder is 0. Since the remainder is 0, ( x2 – 4) is a factor of P ( x). ( x) = ( x2 – 4)( x3 + x – 2) � P Let Q( x) = x3 + x – 2. If x = 1, Q( x) = 1 3 + 1 – 2 = 0 � ( x – 1) is a factor of Q( x). 2 � P ( x) = ( x – 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 = 1 2 – 4(1)(2) = –7 (< 0) The zeroes of P ( x) are ±2 and 1.
16 x2 – 1 = ( x – 1)( x + 1) 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). 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). When m = 8, P ( x) = x2n – (8 + 2) x2 + 8 + 1 = x2n – 10 x2 + 9 Since ( x – 3) is a factor, then P (3) = 0 2n 3 – 10(3)2 + 9 = 0 32n = 81 32n = 3 4 2n = 4 n=2 Hence, P ( x) = x 4 – 10 x2 + 9 = ( x2 – 9)( x2 – 1) = ( x + 3)( x – 3)( x + 1)( x – 1) 17 Let
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2 x + 1 ≡ Ax( x + 1)2 + B( x + 1)2 + Cx2( x + 1) + Dx2 Letting x = –1, –1 = D(–1)2 ⇒ D = –1 Letting x = 0, B = 1 Letting x = 1, 3 = 4 A + 4 B + 2C + D 3 = 4 A + 4(1) + 2C – 1 4 A + 2C = 0 …(1) Letting x = 2, 5 = 18 A + 9 B + 12C + 4 D 5 = 18 A + 9(1) + 12C + 4(–1) 18 A + 12C = 0 3 A + 2C = 0 …(2) (1) – (2): A = 0 From (2), 3(0) + 2C = 0 ⇒ C = 0
2 x + 1 2 x + 1 2 2 = 2 ( x + x) x ( x + 1)2 A B C D ≡ + 2+ + x x x + 1 ( x + 1) 2 Fajar Sdn. Bhd. (008974-T) 2012
Hence,
18
1 1 2 x + 1 = 2 – 2 2 ( x + x) x ( x + 1) 2
4 x2 – x + 3 4 x2 – x + 3 = x3 – 1 ( x – 1)( x2 + x + 1) A Bx + C + 2 x – 1 x + x + 1 2 4 x – x + 3 ≡ A( x2 + x + 1) + ( Bx + C )( x – 1) Letting x = 1, 6 = 3 A ⇒ A = 2 Letting x = 0, 3 = A + C (–1) 3 = 2 – C C = –1 Letting x = –1, 8 = A + (– B + C )(–2) 8 = 2 + (– B – 1)(–2) 8 = 2 + 2 B + 2 2 B = 4 B = 2 4 x2 – x + 3 2 2 x – 1 � + = x3 – 1 x – 1 x 2 + x + 1 ≡
19 P ( x) = x3 + x2 + px + q P �( x) = 3 x2 + 2 x + p Since ( x – 1) is a factor of P ( x), P (1) = 0. 3 2 � 1 + 1 + p(1) + q = 0 p + q = –2 …(1) Since ( x – 1) is a factor of P �( x), P �(1) = 0. 3 � 3(1 ) + 2(1) + p = 0 p = –5 From (1): –5 + q = –2 ⇒ q = 3 � P ( x) = x3 + x2 – 5 x + 3 Since ( x – 1) is a factor of P ( x) and P �( x), thus ( x – 1)2 [or x2 – 2 x + 1] is a factor of P ( x).
Fully Worked Solution
x + 3 x – 2 x + 1 � x3 + x2 – 5 x + 3 x3 – 2 x2 + x 3 x2 – 6 x + 3 3 x2 – 6 x + 3 0 � P ( x) = ( x – 1)2( x + 3) 20 – 4 x 20 – 4 x = P ( x) ( x – 1)2( x + 3) A B C ≡ + + 2 x – 1 ( x – 1) x + 3 20 – 4 x ≡ A( x – 1)( x + 3) + B( x + 3) + C ( x – 1)2 Letting x = –3, 32 = 16C ⇒ C = 2 Letting x = 1, 16 = 4 B ⇒ B = 4 Letting x = 0, 20 = –3 A + 3 B + C 20 = –3 A + 3(4) + 2 3 A = –6 A = –2 20 – 4 x –2 4 2 � = + + 2 2 ( x – 1) ( x + 3) x – 1 ( x – 1) x + 3 2
20 Since the remainders when P ( x) is divided by ( x + 1) is 0, P (–1) = 0. P (–1) = 0 3 2 (–1) + m(–1) + 15(–1) + k = 0 –1 + m – 15 + k = 0 m + k = 16 …(1) Since the remainders when P ( x) is divided by ( x + 2) is (– 4), P (–2) = – 4. P (–2) = – 4 3 2 (–2) + m(–2) + 15(–2) + k = – 4 –8 + 4m – 30 + k = – 4 4m + k = 34 …(2) (2) – (1): 3m = 18 ⇒ m = 6 From (1): 6 + k = 16 ⇒ k = 10 � P ( x) = x3 + 6 x2 + 15 x + 10 Since the remainders when P ( x) is divided by ( x + 1) is 0, ( x + 1) is a factor of P ( x).
x2 + 5 x + 10 x + 1� x3 + 6 x2 + 15 x + 10 x3 + x2 5 x2 + 15 x 5 x2 + 5 x 10 x + 10 10 x + 10 0 2 x) = ( x + 1)( x + 5 x + 10) � P (
x + 7 x + 7 = P ( x) ( x + 1)( x2 + 5 x + 10) A Bx + C ≡ + 2 x + 1 x + 5 x + 10 x + 7 ≡ A( x2 + 5 x + 10) + ( Bx + C )( x + 1) Letting x = –1, 6 = 6 A ⇒ A = 1 Letting x = 0, 7 = 10 A + C 7 = 10(1) + C C = –3 Letting x = 1, 8 = 16 A + 2 B + 2C 8 = 16(1) + 2 B + 2(–3) 2 B = –2 B = –1 x + 7 1 – x – 3 � = + 2 2 ( x + 1)( x + 5 x + 10) x + 1 x + 5 x + 10 1 x + 3 – 2 = x + 1 x + 5 x + 10 21 –12 x2 – 7 x –10 The first inequality is –12 x2 – 7 x x2 – 7 x + 12 0 ( x – 3)( x – 4) 0 −
−
+
−
+
+
x − 4 0
x − 3 0 x
3
+
4
−
+
Therefore, ( x – 3)( x – 4) x 4. The second inequality is x2 – 7 x –10 x2 – 7 x + 10 0 ( x – 2)( x – 5) 0 −
−
+
−
+
+
0
if x
3
or
x − 5 0
x − 2 0 x
2
+
5
−
+
Therefore, ( x – 2)( x – 5)
0 2
if 2
x 5.
x 5
x
3 or
x 4
x
2
3
4
5
The required set of values of x is { x : 2 x 3 or 4 x 5}. © O xford
Fajar Sdn. Bhd. (008974-T) 2012
7
8
ACE AHEAD Mathematics (T) First Term
22 –16 x3 – 4 x2 + 4 x – 16 0 When –16 x3 – 4 x2 + 4 x – 16, x3 – 4 x2 + 4 x 0 x( x2 – 4 x + 4) 0 x ( x – 2)2 0 Since ( x – 2)2 0, in order that x( x – 2)2 0, then x 0 … (1) When x3 – 4 x2 + 4 x – 16 0, we let f ( x) = x3 – 4 x2 + 4 x – 16. f (4) = 4 3 – 4(4)2 + 4(4) – 16 = 0 Thus, ( x – 4) is a factor of f ( x). x2 + 4 x – 4� x3 – 4 x2 + 4 x – 16 x3 – 4 x2 4 x – 16 4 x – 16 0 x3 – 4 x2 + 4 x – 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 …(2) Combining (1) and (2), the required set of values of x is { x : 0 x 4}.
3 x – 5 x
23
3 x – 5 – x + 3 x 3 x – 5 – x2 + 3 x x 2 – x + 6 x – 5 x 2 x – 6 x + 5 x ( x – 1)( x – 5) x
x –
3
0 0 0
24
x – x 3 < 4
|x| <4 |x – 3| |x| < 4| x – 3| x2 < 16( x – 3)2 x2 < 16 x2 – 96 x + 144 0 < 15 x2 – 96 x + 144 0 < 5 x2 – 32 x + 48 0 < ( x – 4)(5 x – 12) −
+
+
−
−
+
5x − 12
x − 4
0
0
x
+
12 5
4
−
+
Hence, the required set of values of x is 12 or x > 4 . x : x < 5 Alternative method x –4 < < 4 x – 3 For the left-end For the right-end inequality, inequality, x x – 4 < < 4 x – 3 x – 3 x x – 4 < 0 + 4 > 0 x – 3 x – 3 x + 4( x – 3) x – 4( x – 3) > 0 < 0 x – 3 x – 3 x + 4 x – 12 –3 x + 12 > 0 < 0 x – 3 x – 3 5 x – 12 3(– x + 4) > 0 < 0 x – 3 x – 3
�
0 0
�
−
−
+
−
+
+
x − 3 0
5x − 12
0
+
+
−
−
+
+
x
−
−
−
+
−
−
+
+
−
+
+
+
x
− 5
0
x
− 1
0
x
0
+
12 5
−
3
0
+ 1
−
5
−
+
−
12 or x > 3 …(1) � x < 3 or x > 4 …(2) 5 Combining (1) and (2):
+
The required set of values of x is { x : x < 0 or 1 x 5}. We write ‘<’ and not ‘’ because x ≠ 0.
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Fajar Sdn. Bhd. (008974-T) 2012
0
4
x
−
x − 3
0
x
3
+
x <
�
−x + 4
12 5
x <
3 or
x <
12 or 5
x >
4
x >
3
x
3
4
The required set of values of x is 12 or x > 4 . x : x < 5
�
�
Fully Worked Solution
1 is as 25 The graphs of y = | x + 2| and y = x 1 + shown below. y y = x + 2 y = −x − 2
2
A −2
−1
y=
1
x+1
O
x
y = x + 2 …(1) 1 y = …(2) x + 1 Substituting (1) into (2), 1 x + 2 = x + 1 2 x + 3 x + 2 = 1 x2 + 3 x + 1 = 0 –3 ± 32 – 4(1)(1) x = 2(1) –3 ± 5 x = 2
The x-coordinate of point A is –3 + 5 x = . 2 Based on the graphs, the solution set of x for which |x + 2| > 1 is x + 1 –3 + 5 . x < –1 or x > 2
�
�
This is the set of values of x where the graph of y = | x + 2| is above the graph of 1 y = . x + 1
26
y
y=x−1 y = −x − 1
A
y = x
+ 1
y = x – 1 …(1) y = x + 1 …(2) Substituting (1) into (2), x – 1 = x + 1 ( x – 1)2 = x + 1 x2 – 2 x + 1 = x + 1 x2 – 3 x = 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}. 27 P ( x) = 2 x3 + hx2 + kx + 36 Since ( x – 3) is a factor, then P (3) = 0 3 2 2(3) + h(3) + k (3) + 36 = 0 9h + 3k = –90 3h + k = –30 …(1) P ( x) = ( x + 2) f ( x) – 30 means that the remainder when P ( x) is divided by ( x + 2) is –30. P (–2) = –30 3 2 2(–2) + h(–2) + k (–2) + 36 = –30 4h – 2k = –50 2h – k = –25 …(2) (1) + (2): 5h = –55 ⇒ h = –11 From (1): 3(–11) + k = –30 k = 3 3 Therefore, P ( x) = 2 x – 11 x2 + 3 x + 36. 2 x2 – 5 x – 12 x – 3� 2 x3 – 11 x2 + 3 x + 36 2 x3 – 6 x2 – 5 x2 + 3 x –5 x2 + 15 x –12 x + 36 –12 x + 36 0 Therefore, P ( x) = ( x – 3)(2 x2 – 5 x – 12) = ( x – 3)(2 x + 3)( x – 4)
1
−1
O
x 1
3
−1
−3
3
x
4
2
To determine the x-coordinate of point A, solve
The sets of values of x such that P ( x) 3 is x : – x 3 or x 4 . 2
�
�
© O xford
0
Fajar Sdn. Bhd. (008974-T) 2012
9
10
ACE AHEAD Mathematics (T) First Term
28 P ( x) = 2 x3 + px2 + qx + 6 Since (2 x + 1) is a factor of P ( x), then
�
� 12� = 0 1 1 1 2� – � + p� – � + q� – � + 6 = 0 2 2 2 P –
3
2
1 1 1 – + p – q + 6 = 0 4 4 2 – 1 + p – 2q + 24 = 0 p – 2q = –23 …(1) 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 9 p – 3q = 33 3 p – q = 11 …(2) p – 2q = –23 …(1) –6 p – 2q = 22 …(2) × 2
30
�
y
y
=
e x −
2
y
2e
−
=
x
1
2
−
y
x
O
1
−
y
=
e x −
−
−
1
e x
= −
−
31 (a)
y
y |ln x | =
–5 p = – 45 p =9 From (1): 9 – 2q = –23 ⇒ q = 16 ( x) = 2 x3 + 9 x2 + 16 x + 6 � P x2 + 4 x + 6 2 x + 1 � 2 x3 + 9 x2 + 16 x + 6 2 x3 + x2 8 x2 + 16 x 8 x2 + 4 x 12 x + 6 12 x + 6 0 2 Let Q( x) = x + 4 x + 6 2 4 4 2 – = x + 4 x + 2 2 2 = ( x + 2) + 2 [> 0]
e x – e –x g ( x) = 2 – x e – e x e x – e –x g (– x) = g ( x) = – = – 2 2 Since g ( x) = – g ( x), g ( x) is an odd function.
O
x
1
(b)
y
y ln x
y ln ( x ) =
=
−
1 O
−
(c)
x 1
y
2
� � � � + 6 [Shown]
P ( x) = (2 x + 1)( x2 + 4 x + 6)
y
Since x2 + 4 x + 6 is positive for all real values of x, then P ( x) < 0 only if 1 2 x + 1 < 0 ⇒ x < – . 2 1 Hence, the solution set is x : x < – . 2
�
Since f ( x) = f (– x), f ( x) is an even function.
© O xford
Fajar Sdn. Bhd. (008974-T) 2012
x 1 – 2
x
Let
ln ( x )
= −
1 – 2
32
�
e x e –x 29 f ( x) = + 2 – x e + e x e x + e –x f (– x) = = = f ( x) 2 2
O
1
−
−
+ 2 x –1 = 15 1 2 – 2
� �
+ 2 x
= 15
1 – 2
x = u u + 2u2 = 15 2u2 + u – 15 = 0 (2u – 5)(u + 3) = 0 5 u = or u = –3 2
x
Fully Worked Solution
5 When u = , 2 1 – 5 x 2 = 2 2 5 –1 x = 2 1 25 = x 4 4 x = 25
��
33
35
When u = –3, 1 – 2
x
x =
� � 1 2 log x – 3� = 5 log x �
= – 3
[Not possible because 1 – 2
c
x > 0 for all real values of x.]
c
Let logc x = u
�1u� = 5
2u – 3
2u2 – 3 = 5u 2u2 – 5u – 3 = 0 (2u + 1)(u – 3) = 0 1 u = – or u = 3 2 1 logc x = – logc x = 3 2 1 – 1 x = c 2 = x = c3 c
8 x + 6(8 – x) = 5 6 8 x + x = 5 8 Let 8 x = u 6 u+ = 5 u u2 + 6 = 5u u2 – 5u + 6 = 0 (u – 2)(u – 3) = 0 u=2 or u = 3 x 8 = 2 8 x = 3 23 x = 2 1 x lg 8 = lg 3
36 x log4 32 – y log8 2 = 4 log2 32 log2 2 x – y = 4 log2 4 log2 8
log2 25 log2 2 x – y = 4 log2 22 log2 23
lg 3 lg 8 x = 0.528
3 x = 1
x =
1 3
5 x y – = 4 2 3 15 x – 2 y = 24 …(1) 1 log2 x + log2 y5 = 2 log4 6 5 log2 6 1 log2 x + (5 log2 y) = 2 log24 5 log2 x + log2 y = log2 6
34 log2 x – log x 8 + 2 log2h + h log x 4 = 0 log2 8 log2 4 log2 x – h h + + = 0 log2 x log2 x
� � log 2 log 2 y – + h + h� = 0 y y � 3 2 y – + h + h� � = 0 y y 3
2
2
2
y2 – 3 + hy + 2h = 0 y 2 + hy + 2 h – 3 = 0 [Shown]
� �
1 1 1 When h = – , y 2 – y + 2 – – 3 4 4 4 2 4 y – y – 14 (4 y + 7)( y – 2)
Given y = log2 x, then x = 2 y. 7 – 7 4 When y = – , x = 2 = 0.297. 4 When y = 2, x = 2 2 = 4.
log2 xy = log2 6 xy = 6 6 y = …(2) x Substituting (2) into (1), 6 15 x – 2 = 24 x 15 x2 – 12 = 24 x 5 x2 – 8 x – 4 = 0 ( x – 2)(5 x + 2) = 0 2 x = 2 or – 5 2 x = – is not accepted 5 � x = 2 6 When x = 2, y = = 3 2
��
= 0
= 0 = 0 7 y = – or 2 4
2 logc x – 3 log x c = 5 logc c 2 logc x – 3 = 5 logc x
© O xford
Fajar Sdn. Bhd. (008974-T) 2012
11
12
ACE AHEAD Mathematics (T) First Term
37 (a) The graph of y = f ( x) = ln ( x + 1) is as shown below. y
y f (x ) =
1
−
=
Since it is known that f f –1( x) = x, by x – 1 comparison f –1( x) = g ( x) = ln . 2
(b) The domain of f –1 is the same as the range of f , i.e. { x : x � 1, x � R} . The range of f –1 is the same as the domain of f , i.e. { y : y � R}. (c) The graphs of y = f ( x) and y = f –1( x) are as shown below.
ln (x + 1)
x
O
y
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) = x y + 1 = e x y = e x – 1 f –1( x) = e x – 1 The domain of f –1 is the same as the range of f , i.e. { x : x � R }. The range of f –1 is the same as the domain of f , i.e. { y : y –1, y � R }. (c) g ° f –1 = g [ f –1( x)] = g (e x – 1) = e x – 1 + 1
x
=
y
y f (x ) =
=
1 + 2e x
3 1
−
x
O 1
3 y f 1(x ) =
39
−
=
ln x
(
−
1
2
)
(a) The graph of y = |sin x| is as shown below. In the non-modulus form, y |sin x | is
1 x 2
=
=e The domain of g ° f –1 is the same as the domain of f –1, i.e. { x : x � R} . The range of g ° f –1 is { y : y > 0, y � R} .
y y
=
sin x
y
=
sin x
−
f (x )
sin x ,
−
1 =
e 2
p
2
x
1
3 p 2
� x � 2 p
y
38 (a) f ° g = f [ g ( x)]
� � x –2 1 ��
= f ln
ln
� x –2 1�
= 1 + 2e x – 1 = 1 + 2 2 = x
�
© O xford
�
Fajar Sdn. Bhd. (008974-T) 2012
x 2p .
2p
The graph of y = sin x for 0 is as shown below.
x
O
p
x p
y
0 x p ,
1
O
y
sin x , =
1
x
O p
2 –1
p
2p
3 p
2
Fully Worked Solution
= 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]
Hence, the function f ( x) = |sin x| – sin x in the non-modulus form is: f ( x) = f ( x ) =
{ {
sin x – sin x, –sin x – sin x, 0,
–2 sin x ,
0
� x
<
p
p � x � 2 p
0 � x < p p � x � 2 p
(b) Hence, the graph of y = f ( x) = |sin x| – sin x for 0 � x � 2 p is as shown below. y
41 LHS =
sin q tan q tan q – sin q
2
sin � cos � q
sin q y
=
=
f (x )
x
O p
p
2
3 2
p
2p
The range of f ( x) is { y : 0 � y � 2, y � R} . (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.
sin q – sin q cos 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 =
y
q
Two intersection points 2
=
tan q + sin q sin q tan q sin q + sin q cos q sin � cos � q
sin q y
=
f (x ) x
O p
2
p
3 2
p
=
sin q + sin q cos q sin2 q
=
sin q (1 + cos q ) sin2 q
=
1 + cos q sin q
2p
40 (a) LHS = cos 4 q + sin 2 q 2 2 = �cos q � + sin 2 q 2 2 = �1 – sin q � + sin 2 q = 1 – 2 sin2 q + sin 4 q + sin 2 q = sin 4 q + 1 – sin2 q = sin 4 q + cos2 q = RHS 4 2 4 2 � cos q + sin q sin q + cos 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
q
= LHS �
sin q tan q tan q – sin q
+ sin q [Proven] sin q tan q
tan
q
42 LHS = tan ( P + Q) – tan P sin ( P + Q) sin P – = cos ( P + Q) cos P
=
cos P sin ( P + Q) – sin P cos ( P + Q) cos P cos ( P + Q) © O xford
Fajar Sdn. Bhd. (008974-T) 2012
13
14
ACE AHEAD Mathematics (T) First Term
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 Q cos 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)
�
sin 3q cos 3q – 2 sin q cos q
45 LHS =
�
= RHS � tan ( P + Q) – tan P sin Q cos P cos ( P + Q)
sin 3q cos 3q – sin q cos q 3 sin q – 4 sin3 q – = sin q 4 cos3 q – 3 cos q cos q
44 LHS =
�
�
Fajar Sdn. Bhd. (008974-T) 2012
3 A sin A + sin 2 2
1 2 = 1 2 sin 2
=
[Proven]
5 3 cos A + cos A 2 2
2 cos
[Proven]
43 LHS = cosec 2q – cot 2q 1 cos 2q – = sin 2q sin 2q 1 – cos 2q = sin 2q 1 – (1 – 2 sin2 q ) = sin 2q 2 sin2 q = 2 sin q cos q sin q = cos q = tan q = RHS [Proven] � cosec 2q – cot 2q tan q tan 22.5° = cosec 2(22.5°) – cot 2(22.5°) 1 1 – = sin 45° tan 45° 1 1 – = 1 1 2 [Shown] = 2 – 1
© O xford
= 3 – 4 sin2 q – 4 cos2 q + 3 = 6 – 4 (sin2 q + cos 2 q ) = 6 – 4(1) = 2 = RHS
�52 A + 32 A� cos 12 �52 A – 32 A� �32 A + 12 A� cos 12 �32 A – 12 A�
1 2 cos 2 A cos A 2
1 2 sin A cos A 2 cos 2 A = sin A cos2 A – sin2 A = sin A 2 cos A sin2 A – = sin A sin A cos A cos A – sin A sin A = cot A cos A – sin A = RHS 5 3 cos A + cos A 2 2 cot A cos A – sin A 3 1 sin A + sin A [Proven] 2 2 =
�
cos 2 A – 2 cos 4 A + cos 6 A cos 2 A + 2 cos 4 A + cos 6 A cos 6 A + cos 2 A – 2 cos 4 A = cos 6 A + cos 2 A + 2 cos 4 A 6 A + 2 A 6 A – 2 A 2 cos cos 2 2 – 2 cos 4 A = 6 A + 2 A 6 A – 2 A 2 cos cos 2 2 + 2 cos 4 A 2 cos 4 A cos 2 A – 2 cos 4 A = 2 cos 4 A cos 2 A + 2 cos 4 A
46 LHS =
�
� �
�
�
� �
�
Fully Worked Solution
2 cos 4 A (cos 2 A – 1) 2 cos 4 A (cos 2 A + 1) cos 2 A – 1 = cos 2 A + 1 1 – 2 sin2 A – 1 = 2 cos2 A – 1 + 1 – 2 sin2 A = 2 cos2 A – sin2 A = cos2 A = – tan2 A = RHS cos 2 A – 2 cos 4 A + cos 6 A cos 2 A + 2 cos 4 A + cos 6 A
48 Since A, B and C are angles of a triangle, then A + B + C = 180°. (a) tan A + tan B + tan C
=
�
=
sin A cos B cos C + sin B cos A cos C + sin C cos A cos B = cos A cos B cos C cos C (sin A cos B + sin B cos A) + sin C cos A cos B = cos A cos B cos C
– tan2 A
cos x + cos 2 x – cos 3 x – cos 4 x sin x + sin 2 x + sin 3 x + sin 4 x cos 2 x + cos x – (cos 4 x + cos 3 x) = sin 2 x + sin x + (sin 4 x + sin 3 x) 2 x + x 2 x – x 2 cos cos 2 2 4 x + 3 x 4 x – 3 x – 2 cos cos 2 2 = 2 x + x 2 x – x 2 sin cos 2 2 4 x + 3 x 4 x – 3 x cos + 2 sin 2 2 3 x x 7 x x 2 cos cos – / 2 cos cos / 2 2 2 2 = 3 x x 7 x x / 2 sin cos / sin cos + 2 2 2 2 2 3 x 7 x cos – cos 2 2 = 3 x 7 x sin + sin 2 2 3 x + 7 x 3 x – 7 x –2 sin sin 4 4 = 2 sin 3 x + 7 x cos 3 x – 7 x 4 4 –sin (– x) = cos (– x) sin x = cos x = tan x = RHS cos x + cos 2 x – cos 3 x – cos 4 x tan x sin x + sin 2 x + sin 3 x + sin 4 x [Proven]
�
� � � � � � � � �� � �� � � � � � � � � �
=
[Proven]
47 LHS =
� � � � � � � � � � � � � �
sin A sin B sin C + + cos 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 2 A + sin 2 B + sin 2C = sin 2 A + sin 2C + sin 2 B = 2 sin
�2 A +2 2C � cos �2 A –2 2C � + sin 2 B
= 2 sin ( A + C ) cos ( A – C ) + sin 2 B = 2 sin (180° – B) cos ( A – C ) + sin 2 B = 2 sin B cos ( A – C ) + sin 2 B = 2 sin B cos ( A – C ) + 2 sin B cos B = 2 sin B [cos ( A – C ) + cos B]
�
� A – C2 + B� A – C – B cos � �� 2
= 2 sin B 2 cos
© O xford
Fajar Sdn. Bhd. (008974-T) 2012
15
16
ACE AHEAD Mathematics (T) First Term
cos x (4 cos x + 3)(cos x – 1) = 0
�
C � A + B – � 2 A – ( B + C ) cos � �� 2 180° – C – C = 2 sin B �2 cos � � 2 A – (180° – A) cos � �� 2 180° – 2C = 2 sin B �2 cos � � 2 2 A – 180° cos � �� 2 = 2 sin B 2 cos
cos x = 0, – When cos x = 0, x = 90°, 270° 3 , 4 x = 138.6°, 221.4°
When cos x = –
When cos x = 1, x = 0°, 360°
= 4 sin B [cos (90° – C ) cos ( A – 90°)] = 4 sin B [sin C sin A] [Shown] = 4 sin A sin B sin C
� x
p
sin 3q = 3 sin q – 4 sin3
p
4 sin3 q – sin2 q – 3 sin q + 2 = 0 (sin q + 1)(4 sin2 q – 5 sin q + 2) = 0 sin q + 1 = 0 or 4 sin2 q – 5 sin q + 2 = 0
p
p
+ cos 2 A sin
When sin q + 1 = 0 sin q = –1 q = 270°
p
2 2 = (sin 2 A)(0) + (cos 2 A)(1) = cos 2 A = RHS �
2 sin A +
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°
cos A + � cos 2 A 4� 4 [Proven]
p
p
�
(b) LHS = 2 cos B +
� 4�
p
cos B –
4 sin2 q – 5 sin sin
4�
p
q + 1
= cos 2 B + cos
�
50
2 cos B +
4
q q
q + 2 q + 2
0
p
2
52
� cos B – 4 � cos 2 B p
Fajar Sdn. Bhd. (008974-T) 2012
q
2 sin 2 sin
[Proven]
cos 3 x = cos 2 x 4 cos3 x – 3 cos x = cos 2 x 4 cos3 x – cos2 x – 3 cos x = 0 cos x (4 cos2 x – cos x – 3) = 0
© O xford
sin2
p
= cos 2 B + 0 = cos 2 B = RHS p
q + 4
–5 sin2 q – 3 sin –5 sin2 q – 5 sin
p
p
q + 2
� 4 sin3 q – sin2 q – 3 sin q + 2 4 sin3
� 4 + B – 4 � + cos � B + – � B – �� 4 4
= cos B +
p
q
(3 sin q – 4 sin3 q ) + sin 2 q – 2 = 0
p
= sin 2 A cos
= 0°, 90°, 138.6°, 221.4°, 270°, 360°
51 sin 3q + sin 2 q = 2
� 4� cos � A + 4 � = sin 2 � A + � 4 = sin �2 A + � 2
49 (a) LHS = 2 sin A +
3 ,1 4
sin x – sin 3 x + sin 5 x = 0 sin 5 x + sin x – sin 3 x = 0 5 x + x 5 x – x 2 sin cos – sin 3 x = 0 2 2 2 sin 3 x cos 2 x – sin 3 x = 0 sin 3 x (2 cos 2 x – 1) = 0 1 sin 3 x = 0 or cos 2 x = 2 When sin 3 x = 0 3 x = 0°, 180°, 360°, 540° x = 0°, 60°, 120°, 180°
�
� �
�
Fully Worked Solution
1 – t 2 cos x = 1 + t 2 1 cot x = tan x
1 When cos 2 x = , 2 2 x = 60°, 300° x = 30°, 150° � x = 0°, 30°, 60°, 120°, 150°, 180° 1 1 2 5 � x = 0, p , p , p , p , p 6 3 3 6
=
cot x = cos x 1 – t 2 1 – t 2 = 2t 1 + t 2 (1 – t 2)(1 + t 2) = (1 – t 2)(2t ) (1 – t 2)(1 + t 2) – (1 – t 2)(2t ) = 0 (1 – t 2)(1 + t 2 – 2t ) = 0 (1 – t 2)(t 2 – 2t + 1) = 0 (1 – t 2)(t – 1)2 = 0 (1 + t )(1 – t )(1 – t )2 = 0 (1 + t )(1 – t )3 = 0 [Shown]
x° = x × 180 rad. p
53 (a)
3 sin –1 x + cos –1 x =
p
� 2 – sin x� =
p
3 sin –1 x +
p
–1
2 sin –1 x = p – 2 sin –1 x = sin –1 x =
x =
p
2
p
1 + t = 0 or (1 – t )3 = 0 t = –1 t = 1
2 p p
4
1 2
cos –1 x + sin –1 x = sin –1
(b)
� 2 – sin x� + sin p
–1
–1
x = sin –1 p
2
When t = –1, x tan = –1 2 basic = 45° x = 180° – 45° 2 x = 135° 2 x = 270° � x = 90°, 270°
4
x = sin
= sin –1
�32 – x� 3 � 2 – x� � 32 – x�
1
3 sin = – x 2 2
x =
−
+
t
2
q 1
1 2
2 tan
+
x = 90°
2t
3 – x 2
x 2 tan x = 54 x 1 – tan2 t 1 2 x Letting t = tan , 2 x 2t tan x = t 1 1 – t 2 From the right-angled triangle above,
When t = 1, x tan =1 2 basic = 45° x = 45° 2
55
p
1=
1 – t 2 2t
2
2t
−
t
2
(a) LHS = cosec q – cot q 1 + t 2 1 – t 2 – = 2t 2t 1 + t 2 – 1 + t 2 = 2t 2 2t = 2t = t = tan
2
�
q
2 = RHS cosec q – cot
© O xford
q
tan
q
2
Fajar Sdn. Bhd. (008974-T) 2012
17
18
ACE AHEAD Mathematics (T) First Term
(b) LHS = sec q – tan q 1 + t 2 2t – = 1 – t 2 1 – t 2 1 + t 2 – 2t = 1 – t 2 t 2 – 2t + 1 = 1 – t 2 (t – 1)2 = (1 + t )(1 – t ) (1 – t )2 = (1 + t )(1 – t )
2t 2 = 1 + t = 1 +
= 1 + tan 0° < 0° <
tan – tan 4 2 p
q
1<
� 4 – 2 � p
sec q – tan q tan
tan
First quadrant. q
2 q
= =
q
2
q
4
2
2
p
=
p
q
tan
p
q
q
p
5 or q = p 4
4
p
� 4 – 2� + � – � 4 2
= tan
q
or
4 – 2 �
q = sec q –
p
–
2
< 45° q
2 q
2
< tan 45° < 1
t
[Shown]
q
Squaring and adding (1) and (2): r 2 (cos2 a + sin 2 a) = 8 2 + 3 2 r 2(1) = 73 r =
Third quadrant.
73
(2) r sin a 3 : = (1) r cos a 8
2
2t
3 8
a
=
a
= tan –1
� 38 �
= 20.56°
q 1
−
t
2
1 + sin q + cos q 1 + cos q 1 + cos q sin q = + 1 + cos q 1 + cos q 2t 1 + t 2 1 = + 1 – t 2 1+ 1 + t 2 2t = 1 + 2 1 + t + 1 – t 2 © O xford
< 2
1 + sin q + cos q < 2 1 + cos q
tan +
2
57 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 …(1) r sin a = 3 …(2)
56 1
q
q
= RHS
cosec q – cot
q
1 < 1 + tan
q
1 + tan tan 4 2
�
q < 90°
0 < tan
p
= tan
2
tan 0° < tan
1 – t = 1 + t =
q
Fajar Sdn. Bhd. (008974-T) 2012
� 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 . q =
73 4
73 sin (q – 20.56°) =
73 4
8 sin q – 3 cos
Fully Worked Solution
1 4 q – 20.56° = 14.48°, 165.52° q = 35.0°, 168.1° [Correct to the nearest 0.1°]
sin (q – 20.56°) =
1 58 Sketch the graphs of y = | x – 2| and y = . x
x =
2± 8 2 ± 2 2 = 2 2 = 1 ± 2 =
x = 1 – 2 is not accepted because x must be positive. � x
y
y =
x
2
2
y = x − 2
This is the range of values of x where the graph of y = | x – 2| is below the graph 1 of y = . x
P Q
y =
= 1 +
Hence, the solution set for the inequality | x – 2| < 1 is { x : 0 < x < 1 + 2 , x ¹ 1}. x
1
y = −x + 2
O
– (–2) ± (–2)2 – 4(1)(–1) 2(1)
x 1
1
x
2 1+
2
59 (a) For f : x
To determine the x-coordinates of the points of intersection of the g raphs of 1 y = | x – 2| and y = , solve the following x simultaneous equations. Case 1 (for point P ) 1 y = x y = – x + 2
…(1) …(2)
Substituting (1) into (2): 1 x + 2 = – x 1 = – x2 + 2 x x2 – 2 x + 1 = 0 ( x – 1)2 = 0 x = 1 Case 2 (for point Q) 1 y = x y = x – 2
Substituting (1) into (3): 1 = x – 2 x 1 = x2 – 2 x x2 – 2 x – 1 = 0
… (1) … (3)
x , the domain is x + 1
{ x : x � R, x ≠ –1}. x 2 For g : x + , the domain is x { x : x � R, x ≠ 0}.
�
x (b) g ° f = g f ( x) = g x – 1 x + 2 x + 1 = x x + 1 x + 2( x + 1) = x 3 x + 2 = x 2 = 3 + x , x ≠ 0
�
Other than x ≠ 0, the domain of g ° f also has to follow the domain of f , i.e. x � R , x ≠ –1. Hence, the domain of g ° f is { x : x � R , x ¹ 0, x ¹ –1}. 2 If x ≠ –1, then g f ( x) ≠ 3 + , i.e. (–1) 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 © O xford
Fajar Sdn. Bhd. (008974-T) 2012
19
20
ACE AHEAD Mathematics (T) First Term
Let f ( x) = x3 – x2 – 4 x + 4 f (1) = 1 3 – 12 – 4(1) + 4 = 0 Therefore, ( x – 1) is another factor of P ( x).
cannot take the value 3. Hence, the range of g ° f is { y : y � R, y ¹ 1, y ¹ 3} . 2 (c) For h : x 3+ , x the domain is { x : x � R, x ¹ 0} and the range is { y : y � R, y ¹ 3} .
x2 – 4 x – 1 � x3 – x2 – 4 x + 4 –( x3 – x2) – 4 x + 4 –(– 4 x + 4)
y h (x ) = 3 +
2 x
3
2 – 3
O
0 x
Hence, P ( x) = ( x + 3)( x – 1)( x2 – 4) ( x + 3)( x – 1)( x + 2)( x – 2) 12 y4 – 8 y3 – 7 y2 + 2 y + 1 = 0
(d) h ¹ g ° f because the domain and the range of g ° f are not the same as the domain and the range of h. 60 (a) P ( x) = x4 + ax3 – 7 x2 – 4ax + b Since ( x + 3) is a factor of P ( x), P (–3) = 0. (–3)4 + a(–3)3 – 7(–3)2 – 4a(–3) + b = 0 81 – 27a – 63 + 12a + b = 0 –15a + b = –18 …(1) When p( x) is divided by ( x – 3), the remainder is 60. Therefore, P (3) = 60 4 3 2 3 + a(3) – 7(3) – 4a(3) + b = 60 81 + 27a – 63 – 12a + b = 60 15a + b = 42 …(2)
(1) + (2), 2b = 24 b = 12 From (1), –15a + 12 = –18 a=2 4 3 2 � P ( x) = x + 2 x – 7 x – 8 x + 12 (b)
© O xford
x3 – x2 – 4 x + 4 x + 3 � x4 + 2 x3 – 7 x2 – 8 x + 12 –( x4 + 3 x3) – x3 – 7 x2 –(– x3 – 3 x2) – 4 x2 – 8 x –(– 4 x2 – 12 x) 4 x + 12 –(4 x + 12) 0 Fajar Sdn. Bhd. (008974-T) 2012
4
3
2
�� �� �� ��
1 1 1 1 12 – 8 – 7 2 1 0 + x x x x + = 12 – 8 x – 7 x2 + 2 x3 + x4 = 0 ( x + 3)( x – 1)( x + 2)( x – 2) = 0 x = –3, 1, –2 or 2 1 Letting y = 1 –3, 1, –2 or 2 = x y 1 1 1 y = – , 1, – or 3 2 2 61 Let 4 sin q – 3 cos q r sin (q – a)
42 + (–3)2 = 5 3 a = tan –1 = 36.87° 4 � 4 sin q – 3 cos q = 5 sin (q – 36.9°) r =
��
4 sin q – 3 cos q = 3 5 sin (q – 36.87°) = 3 3 sin (q – 36.87°) = 5 q – 36.87° = 36.87°, 143.13° q = 73.7°, 180.0°
� xa � = 3 log 2 – log ( x – 2a) x log � � = log 2 – log ( x – 2a) a x log � � + log ( x – 2a) = log 8 a x log �� �( x – 2a)� = log 8 a
62 loga
2
a
2
a
a
2
a
a
a
3
a
2
a
a
a
Fully Worked Solution
x ( x – 2a) = 8 a2 x( x – 2a) = 8a2 x2 – 2ax – 8a2 = 0 ( x + 2a)( x – 4a) = 0 x = –2a or 4a x = –2a is not accepted � x = 4a
3 x2 + 5 x = (1 – x)(1 + x)3 3 x + 5 x (1 – x)(1 + x)3
� � � � � � � 1 1 = � x + x � – 7� x + x � – 6 3
y 2 – y – 6
1 A– x + 1 B x + (1 C x)2 +
+
y + 1 � y 3 + 0 y 2 – 7 y – 6 –( y 3 + y 2) – y 2 – 7 y –(– y 2 – y)
D (1 + x)3 A(1 + x)3 + B(1 – x)(1 + x)2 + C (1 – x)(1 + x) + D(1 – x) +
3 x2 + 5 x (1 – x)(1 + x)3 3 x2 + 5 x
0
A(1 + x)3 + B(1 – x)(1 + x)2
+ C (1 – x)(1 + x) + D(1 – x) Letting x = 1, 8 = 8 A ⇒ A = 1 Letting x = –1, –2 = D(2) ⇒ D = –1 Letting x = 0, 0 = A + B + C + D 0 = 1 + B + C – 1 B + C = 0 …(1) Letting x = 2, 22 = 27 A + B(–1)(9) + C (–1)(3) + D(–1) 22 = 27 – 9 B – 3C + 1 –6 = –9 B – 3C 3 B + C = 2 …(2) (2) – (1), 2 B = 2 ⇒ B = 1 From (1), 1 + C = 0 ⇒ C = –1 1 3 64 Consider the expansion of x + , as x follows.
�
x +
1 x
�
3
= x3 + 3C1 x2
�� 1 x
1
�
�
3
+ C2 x
3 1 = x3 + 3 x + x + 3 x 1 1 = x3 + 3 x + x + 3 x 3 1 1 1 3 x + – 3 x + � x + 3 = x x x
�
–6 y – 6 –(–6 y – 6)
(1 – x)(1 + x3)
�
� � + x1 1 x
2
3
f ( y) = 0 ( y + 1)( y 2 – y – 6) = 0 ( y + 1)( y + 2)( y – 3) = 0 y = –1, –2 or 3 When y 1 x + x 2 x + 1 2 x + x + 1
= –1, = –1 x = – = 0
–1 ± 12 – 4(1)(1) –1 ± –3 x = = 2(1) 2(1) (No real roots) When
y = –2, 1 x + = –2 x x2 + 1 = –2 x x2 + 2 x + 1 = 0 ( x + 1)( x + 1) = 0 x = –1 When
�
� �
�
f ( y) = y 3 – 7 y – 6 f (–1) = 0 � ( y + 1) is a factor of f ( y).
3 x2 + 5 x 3 x2 + 5 x 63 = (1 – x2)(1 + x)2 (1 – x)(1 + x)(1 + x)2
2
4 1 f ( x) = x3 – 4 x – 6 – + 3 x x 1 1 = x3 + 3 – 4 x + x – 6 x 1 3 1 1 x = + x – 3 x + x – 4 x + x – 6
�
y = 3, 1 x + = 3 x x2 + 1 = 3 x x2 – 3 x + 1 = 0
© O xford
Fajar Sdn. Bhd. (008974-T) 2012
21
22
ACE AHEAD Mathematics (T) First Term
1 (–1) – k = 0 2 1 –2 + 4 – – k = 0 2
2 x = – (–3) ± (–3) – 4(1)(1) = 3 ± 5 2 2(1)
2(–1)3 + 4(–1)2 +
Hence, the real roots of f ( x) = 0 are –1 and 3± 5 . 2
3 – k = 0 2 3 k = 2
65 tan x + cot x = 8 cos 2 x
sin x cos x 8 cos 2 x cos x + sin x = 2
1 3 (b) P ( x) = 2 x 3 + 4 x 2 + x – 2 2 3 2 1 3 x + 1 2 x 3 + 4 x 2 + x – 2 2 3 2 –(2 x + 2 x ) 2 x 2 + 2 x –
2
sin x + cos x = 8 cos 2 x sin x cos x
�
1 = 8 cos 2 x sin x cos x 2 = 8 cos 2 x 2 sin x cos x
1 2 x 2 + x 2 2 –(2 x + 2 x)
2 = 8 cos 2 x sin 2 x
3 3 – x – 2 2
1 = 4 sin 2 x cos 2 x 1 = 2(2 sin 2 x cos 2 x) 1 = 2 sin 4 x sin 4 x = Basic
66 f : x
=
1 2
�
3 3 – – x – 2 2
0 Hence, P ( x) = ( x + 1) 2 x 2 + 2 x –
�
2
=
A\ B means A – B or A ∩ B�.
R \{0}
g : x 2 x – 1, x � R f ° g = fg ( x) = f (2 x – 1) 1 1 , x ¹ = 2 x – 1 2
68 y =
1 ( x + 1)(2 x + 3)(2 x – 1) 2
4 x – 1
��
4 , x > 1, x – 1 y = 4 – , x < 1. x – 1 As y
�
�, x –
1 0 x 1 Thus, x = 1 is the asymptote. As x ± �, y 0. 3 y = 3 – x As y ± �, x 0.
�
The domain of f ° g is x : x
�
R, x ¹
Fajar Sdn. Bhd. (008974-T) 2012
�
1 . 2
1 67 (a) P ( x) = 2 x 3 + 4 x 2 + x – k 2 Since ( x + 1) is a factor of P ( x), then P (–1) = 0
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3 � 2� 4 x + 4 x – 3 = ( x + 1) � � 2
If 0 < x < p , then 0 < 4 x < 4p .
p
6 1 5 13 17 4 x = p , p , p , p 6 6 6 6 1 5 13 17 x = p , p , p , p 24 24 24 24
1 , x x
�
Thus, x = 0 (the y-axis) is the asymptote.
Fully Worked Solution
As x ± �, y 3. Thus, y = 3 is the asymptote.
tan
AB = AC 2 + BC 2
y
y
=
y
3
=
4
x
y
4 =
−
x
A
y
3
=
3 −
1
−
1
O
x
=
1 – 2t 2 + t 4 + 4t 2 2
x 3
= 4 x = 0 = 0 = 0 1 x = or 3 3
1 x = is not accepted. 3 Thus, x = 3 The solution set for which
2 tan
q
1 – tan2
q
2
2
2t
4
q 1
� �
−
t
C
2
2
�
= 18.43°
q =
36.9° [correct to one decimal place]
When t = –7 tan
q
2 q
� �
2
= –7 = 98.13°
q =
70
196.3° [correct to one decimal place]
x 1 � x + 1 x + 1 1 0 x 1 – � 0 1 0 x + 1 x + 1 x – 1 1 1 � 0 x + 1 Hence, the required set of values of x is { x : x < –1 or x � 1} . −
−
+
−
+
+
−
+
2
t
q =
q
4 3 > 3 – x x – 1 is given by the part of the graph where the 4 curve y = is above the curve x – 1 3 y = 3 – , that is, { x : 0 < x < 1 or 1 < x < 3}. x
� �
+
1 + 2t + t
�
(3 x – 3)( x – 1) 3 x – 6 x + 3 – 4 x 3 x2 – 10 x + 3 (3 x – 1)( x – 3) 2
1
2t [Shown] 1 + t 2 1 – t 2 and cos q = [Shown] 1 + t 2 10 sin q – 5 cos q = 2 2t 1 – t 2 10 – 5 = 2 1 + t 2 1 + t 2 10(2t ) – 5(1 – t 2) = 2(1 + t 2) 20t – 5 + 5t 2 = 2 + 2t 2 3t 2 + 20t – 7 = 0 (3t – 1)(t + 7) = 0 1 t = or t = –7 3 1 When t = , 3 q 1 tan = 2 3 sin
The x-coordinate of point A is obtained by solving the following equations simultaneously. 4 y = …(1) x – 1 3 y = 3 – …(2) x 4 3 = 3 – x x – 1 4 3 x – 3 = x x – 1
69 tan q =
(1 – t ) + (2t )
= (1 + t 2)2 A = 1 + t 2 Based on � ABC ,
4
3
2
= =
x
B
2 2
1
−
3 −
2t 1 – t 2
q =
© O xford
x −
x + x
−
+
Fajar Sdn. Bhd. (008974-T) 2012
23
24
ACE AHEAD Mathematics (T) First Term
2 first. 71 Simplify log2 a 2 2 = log2 a 1 loga 2 = 2 loga 2 = loga 2 2 = loga 4 2 log2 a + loga (1 – 2a) loga (3 x – 4a) + loga 3 x = loga 4 + loga (1 – 2a) loga 3 x(3 x – 4a) = loga 4(l – 2a) 3 x(3 x – 4a) = 4(1 – 2a) 9 x2 – 12ax + 8a – 4 = 0 loga (3 x – 4a) + loga 3 x =
–(–12a) ± (–12a)2 – 4(9)(8a – 4) x = 2(9) 12a ± 144a2 – 288a + 144 = 18 12a ± (12a – 12)2 = 18 12a ± (12a – 12) 18 24a – 12 12 or = 18 18 4a – 2 2 or = 3 3 =
1 4a – 2 , x = is not accepted 2 3 because when it is substituted into the given equation, it produces log a (–ve) which is undefined. For 0 < a <
� x
=
2 3
72 (a) Since ( x + 2) is a factor, then P (–2) = 0 4 3 6(–2) – a(–2) – b(–2)2 + 28(–2) + 12 = 0 96 + 8a – 4b – 56 + 12 = 0 8a – 4b = –52 2a – b = –13 …(1) Since ( x – 2) is a factor, then P (2) = 0
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Fajar Sdn. Bhd. (008974-T) 2012
6(2)4 – a(2)3 – b(2)2 + 28(2) + 12 = 0 96 – 8a – 4b + 56 + 12 = 0 –8a – 4b = –164 2a + b = 41 …(2) (1) + (2):
4a = 28 a=7 From (1), 2(7) – b = –13 b = 27 P ( x) = ( x + 2)( x – 2) g ( x) 6 x2 – 7 x – 3 x2 – 4 � 6 x4 – 7 x3 – 27 x2 + 28 x + 12 (–) 6 x4 – 24 x2 –7 x3 – 3 x2 + 28 x (–) –7 x3 + 28 x –3 x2 (–) –3 x2
+ 12 + 12 + 12 0
P ( x) = ( x + 2)( x – 2)(6 x2 – 7 x – 3) = ( x + 2)( x – 2)(2 x – 3)(3 x + 1) (b) P ( x) = ( x + 2)( x – 2)(2 x – 3)(3 x + 1) = (2 x – 3)[( x + 2)( x – 2)(3 x + 1) ] = (2 x – 3)[( x2 – 4)(3 x + 1)] = (2 x – 3)(3 x3 + x2 – 12 x – 4) = (2 x – 3)(3 x3 – 41 + 37 + x2 – 12 x)
� Q( x)
Q( x) = x2 – 12 x + 37 = x2 – 12 x + (–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) = 10 2 – 12(10) + 37 = 17 The graph of y = Q( x) for x � [–2, 10] is as shown below. y (–2, 65)
(10, 17) (6, 1) x O
Fully Worked Solution
Hence, the corresponding range for x � [–2, 10] is [1, 65]. 73 sin3 x sec x = 2 tan x
sin2 x sin x
�
�
1 2 tan x cos x =
sin2 x tan x = 2 tan x sin2 x tan x – 2 tan x = 0 tan x (sin2 x – 2) = 0 tan x = 0 or sin 2 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 .
© O xford
Fajar Sdn. Bhd. (008974-T) 2012
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