WAEC Further Mathematics Past Questions & Answers - Page 101

The angle subtended by the minor arc = $\frac{\pi}{3} radians$

The angle subtended by the major arc = $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Perimeter of the major arc = $r\theta + 2r$

= $12 \times \frac{5\pi}{3} + 2(12) = 20\pi + 24$

= $4(5\pi + 6)$

$(x = \frac{nx_{1} + mx_{2}}{n + m}, y = \frac{ny_{1} + my_{2}}{n + m})$

Given P(-2, 3) and Q(4, 9),

$(\frac{2(4) + 3(-2)}{2 + 3}, \frac{2(9) + 3(3)}{2 + 3})$

= $(\frac{2}{5}, \frac{27}{5})$

= $(\frac{2}{5}, 5\frac{2}{5})$

$\int_{0}^{2} (8x - 4x^{2}) \mathrm {d} x  = (\frac{8x^{1 + 1}}{2} - \frac{4x^{2+1}}{3})|_{0}^{2}$

= $(4x^{2} - \frac{4x^{3}}{3}) |_{0}^{2}$

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

= $16 - \frac{32}{3} = \frac{16}{3}$

$s = ut + \frac{at^{2}}{2}$

This movement is against gravity, so it is negative.

$s = ut - \frac{gt^{2}}{2}$

$s = 20m, u = 25ms^{-1}$

$20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}$

$5t^{2} - 25t + 20 = 0$

$5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0$

$5t - 20 = \text{0 or t - 1 = 0}$

$t = \text{1 or 4}$

$P = begin{pmatrix} y - 2 & y - 1 \\ y - 4 & y + 2 \end{pmatrix}$

$|P| = (y - 2)(y + 2) - (y - 1)(y - 4) = (y^{2} - 4) - (y^{2} - 5y + 4) = -23$

$5y - 8 = -23 \implies 5y = -23 + 8 = -15$

$y = \frac{-15}{5} = -3$