SPM Mathematics (Model Test Paper)
Section B
[48 marks]
Answer any four questions from this section.
12 (a) Complete Table 1 in the answer space for the equation y = 2x2− 8x + 1. [2 marks]
(b) For this part of the question, use the graph paper provided. You may use a flexible curve rule.
By using a scale of 2 cm to 1 unit of the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x2− 8x + 1 bagi 0 ≤ x ≤ 6. [4 marks]
By using a scale of 2 cm to 1 unit of the x-axis and 2 cm to 5 units on the y-axis, draw the graph of y = 2x2− 8x + 1 bagi 0 ≤ x ≤ 6. [4 marks]
(c) From your graph, find
(i) the value of y when x = 3,
(ii) the value of x when y = 2, [2 marks]
(i) the value of y when x = 3,
(ii) the value of x when y = 2, [2 marks]
(d) Draw a suitable straight line on your graph to find all the values of x which satisfy the equation 2x2− 4x − 11 = 0 for 0 ≤ x ≤ 6. State the values of x. [4 marks]
Answer:
(a)
x |
0 |
0.5 |
1 |
2 |
3 |
4 |
5 |
5.5 |
6 |
y |
1 |
−2.5 |
−5 |
−7 |
−5 |
1 |
25 |
Table 1
Answer and solution:
(a)
x |
5 |
5.5 |
y |
11 |
17.5 |
(b)
(c)(i)
From the graph, when x = 3, y = −5
(c)(ii)
From the graph, when y = 2, x = 4.15
(d)
y = 2x2 − 8x + 1 ---- (1)
2x2 − 4x− 11 = 0
0 = 2x2 − 4x– 11 ---- (2)
(1) minus (2),
The suitable straight line is y = −4x + 12
x |
0 |
2 |
y |
12 |
4 |
From the graph, x = 3.6
Diagram 8.1
The transformation T is a translation
The transformation R is a reflection in the line y = 3.
State the coordinates of the image of point K under each of the following transformations.
(i) T
(ii) TR [3 marks]
(b) Diagram 8.2 shows three trapeziums, ABCD, PQRS and PTUV drawn on a Cartesian plane.
Diagram 8.2
(i) Trapezium PTUV is the image of trapezium ABCD under the combined transformation VW. Describe in full, the transformation W and the transformation V. [ 6 marks]
(ii) Given that the area of trapezium ABCD is 14 cm2, calculate the area, in cm2, of the shaded region. [3 marks]
Answer and solution:
(a)(i) (3, 2) → T → (1, 5)
(a)(ii) (3, 2) → R → (3, 4) → T → (1, 7)
(b)(i)
W : Rotation, 90o clockwise at centre (0 , −1)
V: Enlargement, scale factor of 2, centre of enlargement at point P (2 , 1)
(ii)
Area of PTUV = (scale factor)2 × Area of object
= 22 × 14
= 56 cm2
Area of shaded region = 56 – 14 = 42 cm2