WAEC Further Mathematics Past Questions & Answers - Page 98

$PQ = \begin{pmatrix} 7 - 3 \\ 4 - 1 \end{pmatrix}$

$= \begin{pmatrix} 4 \\ 3 \end{pmatrix}$

$\hat{n} = \frac{\overrightarrow{PQ}}{|PQ|}$

$|PQ| = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5$

$\hat{n} = \frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}$ 

P(winning) = $\frac{2}{10}$

P(both tickets winning) = $\frac{2}{10} \times \frac{1}{9} = \frac{1}{45}$

$P = \begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix}; Q = \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix}; R = \begin{pmatrix} -5 & 25 \\ -8 & 26 \end{pmatrix}$ 

PQ = $\begin{pmatrix} 3 & 4 \\ 2 & x \end{pmatrix} \begin{pmatrix} 1 & 3 \\ -2 & 4 \end{pmatrix} = \begin{pmatrix} -5 & 25 \\ 2 - 2x & 6 + 4x \end{pmatrix} = R$

$\implies 2 - 2x = -8; -2x = -8 - 2 = -10$

$6 + 4x = 26 \implies 4x = 26 - 6 = 20$

$\implies x = 5$

Arranging the scores in ascending order, we have: 2, 5, 12, 17, 21, 29, 33, 41, 43, 45.

The upper quartile = 41.

$2\sin^{2}\theta = 1 + \cos \theta \implies 2(1 - \cos^{2}\theta) = 1 + \cos \theta$

$2 - 2\cos^{2}\theta = 1 + \cos \theta$

$2 - 2\cos^{2}\theta - 1 - \cos \theta = 0$

$2\cos^{2}\theta + \cos \theta - 1 = 0$

$2\cos^{2}\theta + 2\cos\theta - \cos \theta - 1 = 0 \implies 2\cos \theta(\cos \theta + 1) - 1(\cos \theta + 1) = 0$

$(2\cos \theta - 1)(\cos \theta + 1) = 0 \implies \cos \theta = \frac{1}{2}$

$\theta = \cos^{-1} \frac{1}{2} = 60°$