WAEC Further Mathematics Past Questions & Answers - Page 124

Given the endpoints of the diameter |EF|, the midpoint is the centre of the circle

= $(\frac{-2 + 3}{2} , \frac{-1 + 2}{2}) = (\frac{1}{2} , \frac{1}{2})$

The radius is the distance from the centre to any point on the circle. Using $(\frac{1}{2}, \frac{1}{2})$ and $(3, 2)$;

$r^{2} = (3 - \frac{1}{2})^{2} + (2 - \frac{1}{2})^{2} = \frac{25}{4} + \frac{9}{4}$

$r^{2} = \frac{34}{4}$

The equation of a circle is given as:

$(x - a)^{2} + (y - b)^{2} = r^{2}$, (a, b) as the centre of the circle.

$= (x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{34}{4}$

$x^{2} - x + \frac{1}{4} + y^{2} - y + \frac{1}{4} = \frac{17}{2}$

= $x^{2} - y^{2} - x - y - 8 = 0$ 

The gradient of the line y = 2x + 5 ;

$\frac{\mathrm d y}{\mathrm d x} = 2$

$\frac{y - 3}{x - (-4)} = \frac{y-3}{x+4} = 2$

$y - 3 = 2(x + 4) \implies y = 2x + 11$

$\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x\tan y}$

$\tan x = \frac{5}{12} ; \tan y = \frac{3}{4}$

$\tan (x + y) = \frac{\frac{5}{12} + \frac{3}{4}}{1 - (\frac{5}{12} \times \frac{3}{4}})$

= $\frac{\frac{14}{12}}{\frac{33}{48}}$

= $\frac{56}{33}$

$\cos(a + b) = \cos a\cos b - \sin a\sin b$

$\cos75° = \cos(30 + 45) = (\cos30)(\cos45) - (\sin30)(\sin45)$

= $(\frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{2}) - (\frac{1}{2} \times \frac{\sqrt{2}}{2})$

= $\frac{\sqrt{6} - \sqrt{2}}{4}$

= $\frac{\sqrt{2}(\sqrt{3} - 1)}{4}$

$\begin{pmatrix} 1 & 2 \\ 5 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix}$

= $\begin{pmatrix} 1\times 0 + 2\times 1 & 1\times 1 + 2\times3 \\ 5\times0 + 1\times1 &  5\times1 + 1\times 3 \end{pmatrix}$

= $\begin{pmatrix} 2 & 7 \\ 1 & 8 \end{pmatrix}$