WAEC Further Mathematics Past Questions & Answers - Page 141

On the y-intercept, x=0. At this point, y = 5(0) - 2y = -6; y = 3, so the y- intercept has coordinates (0,3).

The equation of the circle is given as $(x-a)^{2} + (y-b)^{2} = r^{2}$, where (a,b) is the centre and r is the radius.

The radius of the circle through the y- intercept = $\sqrt{(4-0)^{2} + (5-3)^{2}} = \sqrt{20}$

The equation of the circle is $(x-4)^{2} + (y-5)^{2} = 20$; expanding, we have

$x^{2} + y^{2} + 8x - 10y + 21 = 0$

$\cos(x+y) = \cos x\cos y - \sin x\sin y$

Given $\sin$ of an angle implies we have the value of the opposite and hypotenuse of the right-angled triangle. We find the adjacent side using Pythagoras' theorem.

$Adj^{2} = Hyp^{2} - Opp^{2}$

For triangle with angle x, $adj = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12$

For triangle with angle y, $adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15$

$\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}$

$\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}$

= $\frac{140}{221}$

$B.B^{-1} = 1$, let $B^{-1} = \begin{pmatrix}  a & b  \\  c & d  \end{pmatrix}$

$\begin{pmatrix}  2 & 5  \\  1 & 3  \end{pmatrix}$$\begin{pmatrix}  a & b  \\  c & d  \end{pmatrix}$ = $\begin{pmatrix}  1 & 0  \\  0 & 1  \end{pmatrix}$

Multiplying $B \times B^{-1}$, we have the following equations:

$2a+5c = 1......... (1)$; $a+3c = 0 ........(2)$

$2b+5d = 0 ..........(3)$; $b+3d = 1..........(4)$

Solving the equations simultaneously, we have

$a = 3; b = -5; c = -1; d = 2 \implies B^{-1} = \begin{pmatrix}  3 & -5  \\  -1 & 2  \end{pmatrix}$

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}$

$\alpha + \beta = \frac{-b}{a}$; $\alpha\beta = \frac{c}{a}$

$\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2(\alpha\beta)$

From the equation, a = 2, b = -3, c =4

$\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}$

$\alpha\beta = \frac{4}{2} = 2$

$\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}$

$\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}$

$\alpha + \beta = \frac{-b}{a}$

From the equation, a = 2, b = -3 and c = 4

$\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}$