Page 38 - ACE YR IGCSE A TOP APPR' TO ADD MATH

(b) Solve the simultaneous equations to get: 23. (2x − 1)(bx + cx + d)
2
a = −1, b = −15 = 2bx + (2c − b)x + (2d − c)x − d
3
2
f(x) = x − x − 17x − 15 Form the equations:
3
2
(x − x − 17x − 15)÷[(x + 3)(x + 1)] 2b = 4
2
3
= (x – x − 17x − 15) ÷ (x + 4x + 3) 2c − b = −8
3
2
2
= x – 5 2d − c = a
2
3
x – x – 17x – 5
2
x + 4x + 3 x – x – 17x – 15 −d = 18
2
3
3
2
x + 4x + 3x Solve the simultaneous equations to get:
x – 5x – 20x – 15 a = −33
2
4
x – 5x – 20x – 15 b = 2
4
2
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f(x) = (x + 3)(x + 1)(x – 5) c = −3
[5] d = −18
(c) y [5]
24. f(1) = 12
1 + a + b − 4a + 12 = 12
−3a + b = −1
f(4) = 60
3
2
4
x (4) + a(4) + b(4) − 4a(4) + 12 = 60
–3 –1 5
3a + b = −13
–15 Solve the simultaneous equations to get:
a = −2, b = −7
[5]
[3]
3
(d) x + x − 4x − 4 = x − x − 17x − 15
3
2
2
2
2x + 13x + 11 = 0 6 Logarithmic and Exponential Functions
,
A(–1, 0); B 1 –11 –945 2 1. y = 4
2
8
b
[3]
b = log y
4
20. (−1) + a(−1) − 7(−1) + 6 = (−2) + a(−2) − 7(−2) + 6 x 2
3
2
2
3
3a = 0 log 4y
4
a = 0 = 2 log x − (log 4 + log y)
[3] 4 4 4
= 2a − (1 + b)
21. f(k) = 2 = 2a − 1− b
k + k(k) − 5k = 2 [5]
3
k + k − 5k − 2 = 0 2. 2 + 8(log 4) = log x
3
2
When k = 2, Let u = log x 4
x
(2) + (2) − 5(2) − 2 = 0 4 1
3
2
(k + k − 5k − 2) ÷ (k − 2) 2 + 8 1 2 = u
3
2
u
= k + 3k + 1 u − 2u − 8 = 0
2
2
k = 2 or k = –3 + √5 or k = –3 – √5 (u − 4)(u + 2) = 0
2 2 [6] u = 4 or u = −2
1
22. 3x − 11x − 6x + 8 = 0 x = 256 or x = 16
2
3
3x + 3x + x – 2 [5]
2
3
x – 4 3x – 11x – 6x + 8
2
3
3x – 12x 3. log x = log (2x + 3)
2
3
9
3
3x – 11x – 6x log (2x + 3)
3
2
3
3x – 11x – 4x log x = log 3 2
3
2
3
3x – 11x – 2x + 8 3
3
2
3x – 11x – 2x + 8 1
3
2
x = (2x + 3) 2
(x − 4)(3x + x − 2) = 0 x − 2x − 3 = 0
2
2
(x − 4)(3x − 2)(x + 1) = 0 x = 3 or x = −1 (rejected)
x = 4 or x = –1 or x = 2 [4]
3
[5]
Cambridge IGCSE
TM
178 Ace Your Additional Mathematics
Answers Add Math.indd 178 14/03/2022 12:29 PM