Page 19 - 1202 Question Bank Mathematics Form 4

Answers Complete Answers (Paper 1)

https://bit.ly/3KypArA

CHAPTER 1 Value of c = 2, y-intercept is 2
(b) Axis of symmetry at x = 3 + (–2) Thus, the maximum point is (0, 2).
Paper 1 2 When f(x) = 0,
1 2
1. B 2. A 3. C 4. D 5. C = 2 –x + 2 = 0
6. B 7. A 8. B 9. D 10. A ( ) ( ) ( ) x = ±! 2
1
1
1 2
11. B 12. A 13. D 14. D 15. B f 2 = 2 – 2 – 6 x = 1.4 or x = –1.4
16. A 17. C 18. C 19. D 20. D = – 25 The graph:
21. C 22. A 23. D 24. A 25. B 4 f(x)
26. A 27. D Thus, minimum point is ( 1 2 , – 25 ) .
4
Paper 2 (c) From the graph on (a), the roots for 2
Section A the graph function are x = –2 or x = 3.
x
2
1. (a) Quadratic expression in one variable. 8. (a) f(x) = x – 6x + 8 –1.4 0 1.4
(b) Not a quadratic expression in one Value of a = 1 . 0, shape
variable because there is a variable Value of c = 8, y-intercept = 8
with a power which is not a whole When f(x) = 0,
2
2
number. x – 6x + 8 = 0 (e) f(x) = x – 2
(c) Not a quadratic expression in one (x – 4)(x – 2) = 0 Value of a = 1 . 0, shape
variable because there are two x = 2 or x = 4 Value of b = 0, axis of symmetry is
variables, x and y. The graph: the y-axis
(d) Quadratic expression in one variable. f(x) Value of c = –2, y-intercept is –2
(e) Not a quadratic expression in one Thus, the minimum point is (0, –2).
variable because there is a variable When f(x) = 0,
2
with a power which is not a whole 8 –x + 2 = 0
number. x = ±! 2
x = 1.4 or x = –1.4
2. (a) (b) (c) (d)
The graph:
3. (a) a = 2, b = –3, c = 5 x
(b) a = 1, b = 4, c = 0 0 2 4 f(x)
(c) a = 3, b = 0, c = –7
2
(d) a = –3, b = –4, c = 2 (b) f(x) = x – 4x + 4
(e) a = 3, b = –3, c = 0 Value of a = 1 . 0, shape
Value of c = 4, y-intercept = 4 CHAPTER 1
4. (a) Maximum point : (–1, 4) When f(x) = 0, x – 4x + 4 = 0 0 x
2
–1.4
1.4
©PAN ASIA PUBLICATIONS
Axis of symmetry, x = –1
2
(x – 2) = 0
(b) Minimum point : (0, –4) x = 2 –2
Axis of symmetry, x = 0
The graph:
5. (a) Given f(x) = 2x – 5x + c. Section B
2
Substitute the values of x = –1 and f(x) 9. Area of rectangle = 66 cm 2
f(x) = 5 into the quadratic function: (x + 2)(x + 7) = 66
2
5 = 2(–1) – 5(–1) + c x + 2x + 7x + 14 = 35
2
5 = 2 + 5 + c x + 9x + 14 – 35 = 0
2
c = –2
4 x + 9x – 21 = 0
2
(b) Given f(x) = 2x – 5x + c.
2

Substitute the values of x = 3 and 10. By using Pythagoras Theorom,
2
2
f(x) = 7 into the quadratic function: 0 2 x r = p + q 2
7 = 2(3) – 5(3) + c 11. x + 3x = –2(–3 – x)
2
2
2
7 = 18 – 15 + c (c) f(x) = x – 4 x + 3x = 6 + 2x
2
c = 7 – 3 Value of a = 1 > 0, shape x + 3x – 2x – 6 = 0
2
2
= 4 Value of b = 0, axis of symmetry is x + x – 6 = 0
the y-axis
6. Let Sufi’s age = x – 2 Value of c = –4, y-intercept is –4 (x – 2)(x + 3) = 0
x(x – 2) = 35 x = 2, x = –3
x – 2x = 35 Thus, the minimum point is (0, –4). 12. – 3 = x
2
2
x – 2x – 35 = 0 When f(x) = 0, x – 4 = 0 2x + 1 x – 2
2
(x + 2)(x – 2) = 0 –3(x – 2) = x(2x + 1)
2
7. (a) f(x) = x – x – 6 x = 2 or x = –2 –3x + 6 = 2x + x
2
Value of a = 1 > 0, shape The graph: 2x + x + 3x – 6 = 0
2
Value of c = –6, y-intercept = –6 2x + 4x – 6 = 0
2
When f(x) = 0, f(x) x + 2x – 3 = 0
2
2
x – x – 6 = 0 (x – 1)(x + 3) = 0
(x + 2)(x – 3) = 0 x = 1, x = –3
x = –2 or x = 3
The graph: x 13. (a) p = 1
2
–2 0 2 (b) f(x) = x + 5x + q
f(x)
From the graph, when x = 0, f(x) = 4.
–4 Substitute x = 0 and f(x) = 4 into
quadratic function:
x 2
2
–2 0 3 (d) f(x) = –x + 2 4 = 1(0) + 5(0) + q
Value of a = –1 . 0, shape q = 4

–6 Value of b = 0, axis of symmetry is
the y-axis
115
Answers 1202QB Maths Form 4.indd 115 21/02/2022 6:27 PM