WAEC Further Mathematics Past Questions & Answers - Page 118

The gradient of the equation = $\frac{y_{1} - y_{0}}{x_{1} - x_{0}} = \frac{\frac{5}{2} - 0}{\frac{5}{4} - 0}$

= $\frac{\frac{5}{2}}{\frac{5}{4}} = 2$

The line passes through the origin, therefore the equation of the line is y = mx

$y = 2x$

When x = 4, y = 2 x 4 = 8.

$Gradient = \frac{y_{1} - y_{0}}{x_{1} - x_{0}}$

Line passing through $(0, 0)$ & $(1\frac{1}{4}, 2\frac{1}{2})$,

$Gradient = \frac{2\frac{1}{2} - 0}{1\frac{1}{4} - 0} = \frac{\frac{5}{2}}{\frac{5}{4}}$

$y = x^{2} + 6x - 12$

$\frac{\mathrm d y}{\mathrm d x} = 2x + 6 = 0$

$2x = -6 \implies x = -3$

$y(-3) = (-3)^{2} + 6(-3) - 12 = 9 - 18 - 12 = -21$

$T_{n} = ar^{n-1}$ (for an exponential sequence)

$T_{1} = 36 = a$

$T_{2} = ar = 36r = p$

$T_{3} = ar^{2} = 36r^{2} = \frac{9}{4}$

$T_{4} = ar^{3} = 36r^{3} = q$

$36r^{2} = \frac{9}{4} \implies r^{2} = \frac{\frac{9}{4}}{36} = \frac{1}{16}$

$r = \sqrt{\frac{1}{16}} = \frac{1}{4}$

$p = 36 \times \frac{1}{4} = 9 ; q = \frac{9}{4} \times \frac{1}{4} = \frac{9}{16}$

$p + q = 9 + \frac{9}{16} = 9\frac{9}{16}$ 

$f(x) = \frac{2x - 1}{x + 2}$

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

$x = \frac{2y - 1}{y + 2} \implies x(y + 2) = 2y - 1$

$xy - 2y = -1 - 2x  \implies y = \frac{-1 - 2x}{x - 2}$

$f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2$