SPM Additional Mathematics 2018, Paper 1 (Question 11 – 13)

Question 12 (4 marks):
Diagram 5 shows a circle with centre O.

Diagram 5

PR and QR are tangents to the circle at points P and Q respectively. It is given that the length of minor arc PQ is 4 cm and $OR=\frac{5}{\alpha }\text{ cm}\text{.}$  
Express in terms of α,
(a) the radius, r, of the circle,
(b) the area, A, of the shaded region.


Solution:
(a)
$\begin{array}{l}\text{Given }{s}_{PQ}=4\\ r\alpha =4\\ r=\frac{4}{\alpha }\text{ cm}\end{array}$

(b)

$\begin{array}{l}PR=\sqrt{{\left(\frac{5}{\alpha }\right)}^2-{\left(\frac{4}{\alpha }\right)}^2}\\ PR=\sqrt{\frac{9}{{\alpha }^2}}\\ PR=\frac{3}{\alpha }\\ \\ A=\text{ Area of shaded region}\\ A=\text{ Area of quadrilateral }OPRQ-\\ \text{Area of sector }OPQ\\ =2\left(\text{Area of }△OPR\right)-\frac{1}{2}{r}^2\theta \\ =2\left[\frac{1}{2}\times \frac{3}{\alpha }\times \frac{4}{\alpha }\right]-\left[\frac{1}{2}\times {\left(\frac{4}{\alpha }\right)}^2\times \alpha \right]\\ =\frac{12}{{\alpha }^2}-\frac{8}{\alpha }\\ =\frac{12-8\alpha }{{\alpha }^2}{\text{ cm}}^2\end{array}$

Question 13 (3 marks):
Diagram 6 shows the graph of a straight line $\frac{x^2}{y}\text{ against }\frac{1}{x}.$  

Diagram 6

Based on Diagram 6, express y in terms of x.


Solution:

$\begin{array}{l}m=\frac{4-\left(-5\right)}{6-0}=\frac{3}{2}\\ c=-5\\ Y=\frac{x^2}{y}\text{, }X=\frac{1}{x}\\ \\ Y=mX+c\\ \frac{x^2}{y}=\frac{3}{2}\left(\frac{1}{x}\right)+\left(-5\right)\\ \frac{x^2}{y}=\frac{3}{2x}-5\\ \frac{x^2}{y}=\frac{3-10x}{2x}\\ \frac{y}{x^2}=\frac{2x}{3-10x}\\ y=\frac{2x^3}{3-10x}\end{array}$