**SPM Mathematics (Model Test Paper)**

*Section***B**

[48 marks]

Answer any

**four**questions from this section.**Complete Table 1 in the answer space for the equation**

12 (a)

12 (a)

*y*= 2

*x*

^{2}− 8

*x*+ 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

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*= 2*x*^{2}− 8*x*+ 1 bagi 0 ≤*x*≤ 6. [4 marks](c) From your graph, find

(i) the value of

(ii) the value of

(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 2*x*^{2}− 4*x*− 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*= 2

*x*

^{2}− 8

*x*+ 1 ---- (1)

2

*x*^{2}− 4*x*− 11 = 00 = 2

*x*^{2}− 4*x*– 11 ---- (2)(1) minus (2),

The suitable straight line is

*y*= −4*x*+ 12

x |
0 |
2 |

y |
12 |
4 |

**From the graph,**

*x*= 3.6**13**. (

*a*) Diagram 8.1 shows point

*K*( 3, 2 ) marked on a Cartesian plane.

Diagram 8.1

The transformation

*T**is a translation $\left(\begin{array}{l}-2\\ \text{}3\end{array}\right).$*The transformation

**is a reflection in the line***R**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

**. Describe in full, the transformation**

*VW***and the transformation**

*W*

*V*.*[ 6 marks]*

(ii) Given that the area of trapezium

*ABCD*is 14 cm^{2}, calculate the area, in cm^{2}, of the shaded region. [3 marks]

*Answer and solution:*

**(a)(i)**(3, 2) →

**T**→ (1, 5)

**(3, 2) →**

(a)(ii)

(a)(ii)

**R**→ (3, 4) →

**T**→ (1, 7)

(b)(i)

(b)(i)

**: Rotation, 90**

*W*^{o}clockwise at centre (0 , −1)

**: Enlargement, scale factor of 2, centre of enlargement at point**

*V**P*(2 , 1)

**(ii)**

Area of

*PTUV*= (scale factor)^{2}× Area of object= 2

^{2 }× 14= 56 cm

^{2}**Area of shaded region = 56 – 14 = 42 cm**

^{2}