Page 35 - ACE YR IGCSE A TOP APPR' TO ADD MATH

QR + PR = PQ 24. 3x√15 = x√5 + √15
2
2
2
28 + 6√3 + 12 – 6√3 = 40 √5(3x√3) = √5(x + √3)
3x√3 – x = √3
It is proven that triangle PQR is a right-angled x(3√3 – 1) = √3
triangle using the Pythagorean theorem.
∠R is 90°. x = √3
[2] 3√3 – 1
1 + 3√3 9 + √3
(b) tan ∠QPR = =
√3 – 3 26
= 1 + 3√3 × √3 + 3 a = 9, b = 3, c = 26
√3 – 3 √3 + 3 [4]
= √3 + 3 + 9 + 9√3 25. 1 + 1 + 1
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3 – 9 1 + x b + x a 1 + x c + x b 1 + x a + x c
–6 – 5√3 x c x c x a x a x b x b
=
3 1 1 1
[3] = + +
x + x + x a x + x + x b x + x + x c
b
a
c
a
b
c
19. Let y = 3 x x c x a x b
x 2
x
2
x
8(3 ) + (3 ) · (3 ) = 3 + 6 x + x + x b
c
a
2
8y + 9y − y + 6 = 0 = x + x + x c
b
a
y = 0.515 or y = −1.29 = 1
x
x
3 = 0.515 or 3 = −1.29 (rejected) [3]
x = −0.604
[5]
8
20. 2u + = 17 5 Factors of Polynomials
u
2u – 17u + 8 = 0 1. When x = −1
2
u = 1 or u = 8 x − 3x − x + 3
3
2
2 3 2
2 = 2 or 2 = 8 = (−1) − 3(−1) − (−1) + 3
–1
x
x
x = –1 or x = 3 = 0
2
[4] x – 4x + 3
x + 1 x – 3x – x + 3
3
2
21. 864 = 2 × 3 x + x
5
3
2
3
√864 = √2 × 3 –4x – x
3
5
2
= 12√2√3 –4x – 4x
2
[3] 3x + 3
22. 3 × 30√3 – 9 3x + 3
2 3 √3
3
2
= 15√3 – 9√3 x − 3x − x + 3 = 0
2
3 (x + 1)(x − 4x + 3) = 0
= 12√3 (x + 1)(x − 3)(x − 1) = 0
[2] x = 1 or x = −1 or x = 3
23. 3 – 2 2 3 [5]
4 2. When x = 2
2 3
2
2
2 4 −(2) − (m + 1)(2)+ m + m = 0
3 – (2 × 2 ) m − m − 6 = 0
3
3
2
= 4 (m − 3)(m + 2) = 0
2 3 m = 3 or m = −2
= – 1 [3]
4
2 3 3. 4x + 2x − 5x − 7
2
5
4
a = 2 = 16 = 4 1 2 5 + 2 1 2 4 – 5 1 2 2 – 7
1
1
4
1
[3] 2 2 2
= –8
[2]
Answers 175
Answers Add Math.indd 175 14/03/2022 12:29 PM