 # SPM Additional Mathematics (Model Test Paper)

SPM Additional Mathematics (Model Test Paper)

Section B
[40 marks]
Answer any four questions from this section.

Question 7
(a)$\text{Prove that}{\left(\frac{\mathrm{cos}ec\text{}x-\mathrm{sec}x}{\mathrm{sec}x\text{}\mathrm{cos}ec\text{}x}\right)}^{2}=1-\mathrm{sin}2x$
[3 marks]
(b)(i) Sketch the graph of y = 1 – sin2x for 0 ≤ x ≤ 2π.

(b)(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation for 0 ≤ x ≤ 2π.
State the number of soultions.
[7 marks]

(a)
$\begin{array}{l}LHS\\ ={\left(\frac{\mathrm{cos}ec\text{}x-\mathrm{sec}x}{\mathrm{sec}x\text{}\mathrm{cos}ec\text{}x}\right)}^{2}\\ ={\left(\frac{\overline{)\mathrm{cos}ec\text{}x}}{\mathrm{sec}x\text{}\overline{)\mathrm{cos}ec\text{}x}}-\frac{\overline{)\mathrm{sec}x}}{\overline{)\mathrm{sec}x}\text{}\mathrm{cos}ec\text{}x}\right)}^{2}\\ ={\left(\frac{1}{\mathrm{sec}x}-\frac{1}{\mathrm{cos}ec\text{}x}\right)}^{2}\\ ={\left(\mathrm{cos}x-\mathrm{sin}x\right)}^{2}\\ ={\mathrm{cos}}^{2}x-2\mathrm{cos}x\mathrm{sin}x+{\mathrm{sin}}^{2}x\\ =1-\mathrm{sin}2x\text{(RHS)}\end{array}$

(b)(i) (b)(ii)
$\begin{array}{l}2-{\left(\frac{\mathrm{cos}ec\text{}x-\mathrm{sec}x}{\mathrm{sec}x\text{}\mathrm{cos}ec\text{}x}\right)}^{2}=\frac{x}{\pi }\\ 2-\left(1-\mathrm{sin}2x\right)=\frac{x}{\pi }←\overline{)\text{From 7(a)}}\\ 2-y=\frac{x}{\pi }←\overline{)\text{From 7(b)(i)}}\\ y=2-\frac{x}{\pi }\text{(suitable straight line)}\end{array}$

 x 0 π 2π y 2 1 0

From the graph, there is 3 number of solutions.

Question 8
Diagram 4 shows part of a curve x = y2 + 2. The gradient of a straight line QR is –1. Find
(a) the equation of PQ,   [2 marks]
(b) the area of shaded region, [4 marks]
(c) the volume of revolution, in terms of π, when the shaded region is rotated through 360o about the y-axis. [4 marks]

(a)
– 2 = –1 (x – 6)
= –x + 6 + 2
y = –x + 8

(b)
= –x + 8, at y-axis, x = 0
= 8
Point R = (0, 8)
Area of shaded region
= Area under the curve + Area of triangle
$\begin{array}{l}=\underset{0}{\overset{2}{\int }}x\text{}dy+\frac{1}{2}\left(8-2\right)\left(6\right)\\ =\underset{0}{\overset{2}{\int }}\left({y}^{2}+2\right)\text{}dy+\frac{1}{2}\left(6\right)\left(6\right)\\ ={\left[\frac{{y}^{3}}{3}+2y\right]}_{0}^{2}+18\\ =\left[\frac{{2}^{3}}{3}+2\left(2\right)-0\right]+18\\ =\frac{20}{3}+18\\ =24\frac{2}{3}uni{t}^{2}\end{array}$

(c)
Volume of revolution
= Volume under the curve + Volume of cone
$\begin{array}{l}=\pi \underset{0}{\overset{2}{\int }}{x}^{2}\text{}dy+\frac{1}{3}\pi {r}^{2}h\\ =\pi \underset{0}{\overset{2}{\int }}{\left({y}^{2}+2\right)}^{2}\text{}dy+\frac{1}{3}\pi {\left(6\right)}^{2}\left(6\right)\\ =\pi \underset{0}{\overset{2}{\int }}\left({y}^{4}+4{y}^{2}+4\right)\text{}dy+72\pi \\ =\pi {\left[\frac{{y}^{5}}{5}+\frac{4{y}^{3}}{3}+4y\right]}_{0}^{2}+72\pi \\ =\pi \left[\frac{{2}^{5}}{5}+\frac{4{\left(2\right)}^{3}}{3}+4\left(2\right)-0\right]+72\pi \\ =\pi \left[\frac{32}{5}+\frac{32}{3}+8\right]+72\pi \\ =\frac{376}{15}\pi +72\pi \\ =97\frac{1}{15}\pi {\text{unit}}^{3}\end{array}$