# SPM Additional Mathematics (Model Test Paper)

SPM Additional Mathematics (Model Test Paper)

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

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]

Answer and Solution:
(a)
y– 2 = –1 (x – 6)
y= –x + 6 + 2
y = –x + 8

(b)
y= –x + 8, at y-axis, x = 0
y= 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
$\begin{array}{l}=\pi \underset{0}{\overset{2}{\int }}{x}^{2}\text{}dy+\frac{1}{3}\pi {r}^{2}h\\ =\pi \underset{0}{\overset{2}{\int }}{\left({y}^{2}+2\right)}^{2}\text{}dy+\frac{1}{3}\pi {\left(6\right)}^{2}\left(6\right)\\ =\pi \underset{0}{\overset{2}{\int }}\left({y}^{4}+4{y}^{2}+4\right)\text{}dy+72\pi \\ =\pi {\left[\frac{{y}^{5}}{5}+\frac{4{y}^{3}}{3}+4y\right]}_{0}^{2}+72\pi \\ =\pi \left[\frac{{2}^{5}}{5}+\frac{4{\left(2\right)}^{3}}{3}+4\left(2\right)-0\right]+72\pi \\ =\pi \left[\frac{32}{5}+\frac{32}{3}+8\right]+72\pi \\ =\frac{376}{15}\pi +72\pi \\ =97\frac{1}{15}\pi {\text{unit}}^{3}\end{array}$