SPM Additional Mathematics 2018, Paper 2 (Question 10 & 11)

Question 10 (10 marks):
Diagram 7 shows the straight line 4y = x – 2 touches the curve x = y² + 6 at point P.

Find
(a) the coordinates of P,
(b) the area of the shaded region,
(c) the volume of revolution, in terms of π, when the region bounded by the curve and the straight line x = 8 is revolved through 180^o about the x-axis.


Solution:
(a)
$\begin{array}{l}4y=x-\mathrm{2\dots \dots \dots }(1)\\ x=y^2+\mathrm{6\dots \dots \dots }(2)\\ \\ \text{Substitute (2) into (1):}\\ 4y=\left(y^2+6\right)-2\\ y^2-4y+4=0\\ \left(y-2\right)\left(y-2\right)=0\\ y-2=0\\ y=2\\ \\ \text{Substitute }y=2\text{ into (2):}\\ x={\left(2\right)}^2+6\\ x=10\\ \text{Thus, }P=\left(10,2\right).\end{array}$


(b)
$\begin{array}{l}\text{At x-axis, }y=0\\ \therefore 4y=x-2\\ 0=x-2\\ x=2\\ \\ \text{Area of shaded region}\\ \text{= Area of triangle}-\text{Area under the curve}\\ =\frac{1}{2}\left(10-2\right)\left(2\right)-{\displaystyle {\int }_6^{10}ydx}\\ =8-{\displaystyle {\int }_6^{10}\sqrt{x-6}dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y=\sqrt{x-6}\end{array}}\\ =8-{\displaystyle {\int }_6^{10}{\left(x-6\right)}^{\frac{1}{2}}dx}\\ =8-{\left[\frac{{\left(x-6\right)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_6^{10}\\ =8-{\left[\frac{2{\left(x-6\right)}^{\frac{3}{2}}}{3}\right]}_6^{10}\\ =8-\left[\frac{2{\left(10-6\right)}^{\frac{3}{2}}}{3}-\frac{2{\left(6-6\right)}^{\frac{3}{2}}}{3}\right]\\ =8-\frac{16}{3}\\ =\frac{8}{3}{\text{ units}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume of revolution}\\ =\pi {\displaystyle {\int }_6^8{y}^{2}dx}\\ =\pi {\displaystyle {\int }_6^8\left(x-6\right)dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y^2=x-6\end{array}}\\ =\pi {\left[\frac{x^2}{2}-6x\right]}_6^8\\ =\pi \left[\left(32-48\right)-\left(18-36\right)\right]\\ =2\pi {\text{ units}}^3\end{array}$

Question 11 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment. A straight line will be obtained when a graph of $\frac{y^2}{x}\text{ against }\frac{1}{x}$ is plotted.


(a) Based on Table 1, construct a table for the values of $\frac{1}{x}\text{ and }\frac{y^2}{x}.$  
$\begin{array}{l}\left(b\right)\text{ Plot }\frac{y^2}{x}\text{ against }\frac{1}{x}\text{, using a scale of 2 cm to 0}\text{.1 unit on the }\frac{1}{x}\text{-axis}\\ \text{ and 2cm to 2 units on the }\frac{y^2}{x}\text{-axis}\text{.}\\ \text{ Hence, draw the line of best fit}\text{.}\end{array}$

(c) Using the graph in 11(b)
(i) find the value of y when x = 2.7,
(ii) express y in terms of x.


Solution:
(a)


(b)


(c)(i)
$\begin{array}{l}\text{When }x=2.7,\frac{1}{x}=0.37\\ \text{From graph,}\\ \frac{y^2}{x}=5.2\\ \frac{y^2}{2.7}=5.2\\ y=3.75\end{array}$



(c)(ii)
$\begin{array}{l}\text{Form graph, }y\text{-intercept, }c\text{ = –4}\\ \text{gradient, }m=\frac{16-\left(-4\right)}{0.8-0}=25\\ Y=mX+c\\ \frac{y^2}{x}=25\left(\frac{1}{x}\right)-4\\ y=\sqrt{25-4x}\end{array}$