Page 41 - ACE YR IGCSE A TOP APPR' TO MATHS

2
9 + 6x – x 2 (b) –9x – 3x
[
= [2] 1 1 2 1 2
x 2 = –9 x + x + ( ) ( )]
–
2
(b) gf(100) 3 2 3 3
[(
1
= 3 + 1 = –9 x + 1 ) ]
–
√100 – 4 9 9
3 ( 1 ) 2
= [2] = –9 x + + 1
2 9
x
(c) √ – 4 = 5 a = –9, b = – 1 , c = 1
x
(
√ = 9 9 1
x = 81 [2] Maximum point − 9 , 1 ) [4]
(d) Let g (x) = y (c) 3x + 1 = 6 + 4
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–1
3 x
+ 1 = x
2
y 3x + x = 6x + 4
2
g (x) = 3 3x – 5x – 4 = 0
–1
x – 1 –(–5) + √(–5) – 4(3)(–4)
2
3 = 1 x = 2(3)
x – 1 x – 2
2
x = 2.26 or x = −0.59
3x – 6 = x – 1 4 (a) f(–2) = 3(–2) – (–2) – 2 [3]
2
2
3x – x – 5 = 0 = 12 [1]
2
–(–1) + √(–1) – 4(3)(–5) (b) 3x – x – 2 = 2
2
2
x =
2(3) 3x – x – 4 = 0
2
x = 1.47 or x = –1.14 [4] (3x – 4)(x + 1) = 0
4
2 (a) fg(x) = 6 ( x – 3 ) – 2 x = 4 or x = –1 [2]
3
–1
= 30 – 2x [2] (c) Let g (x) = y
x – 3 4 – 5y = x
(b) fg(x) = x y = x – 4
30 – 2x = x –5
x – 3 g (x) = 4 – x [2]
–1
30 – 2x = x – 3x 5
2
2
x – x – 30 = 0 (shown) [2] (d) 3(4 – 5x) – x – 2 = 0
2
2
(c) y = x – x – 30 5x – 7x + 2 = 0
2
= (x – 6)(x + 5) (5x – 2)(x – 1) = 0
y x = 2 or x = 1 [2]
5
(e) (i) B [1]
(ii) F [1]
2
x 5 (a) gf(x) = (x – 2) – 25
–5 6 = x – 4x – 21 [2]
2
(b) gf(x) = 0
x – 4x – 21 = 0
2
(x – 7)(x + 3) = 0
–30 (c) x = 7 or x = –3 [2]
[3] gf(x)
(d) (i) x – x – 30 2
2
(
= x – 1 ) – 121 –3 7 x
2
4
a = 1 , b = – 121 [3]
2 4
(ii) Minimum value ( 1 2 , – 121 ) [1]
4
3 (a) fh(x) = (3x + 1) – (3x + 1) 2 –21 [3]
= 3x + 1 – (9x + 6x + 1) (d) gf(x) = x – 4x – 21
2
2
= –9x – 3x (shown) [2]
2
= (x – 2) – 25
2
a = –2, b = –25 [3]
Answers 161
Answers.indd 161 15/03/2022 11:08 AM