WAEC Further Mathematics Past Questions & Answers - Page 140

No of boys in the class = 7; Girls = 20-7 = 13

No of selection = $^{13}C_{3} \times ^{7}C_{2} = \frac{13!}{(13-3)!3!} \times \frac{7!}{(7-2)!2!}$

= $286\times21 = 6006$

$\sqrt{128}  = \sqrt{64\times2} = 8\sqrt{2}$

$\sqrt{32} = \sqrt{16\times2} = 4\sqrt{2}$

Simplifying, we have $\frac{8\sqrt{2}}{4\sqrt{2} - 2\sqrt{2}} = \frac{8\sqrt{2}}{2\sqrt{2}}$

= 4

Expanding $(x+1)(x-2) = x^{2} - 2x + x - 2 = x^{2} - x - 2$

$\int_{-1}^{0} (x^{2} - x - 2) \mathrm{d}x = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x]_{-1}^{0}$

= $[\frac{0}{3} - \frac{0}{2} - 2\times0 - (\frac{-1^{3}}{3} - \frac{-1^{2}}{2} - 2\times-1)]$

= $0 + \frac{1}{3} + \frac{1}{2} - 2 = \frac{-7}{6}$

Note: This can also be solved using integration by parts. 

$\int uv \mathrm{d}x = u\int v \mathrm{d}x - \int u'(\int v \mathrm{d}x)\mathrm{d}x$.

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

Using quotient rule, $\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}$, we have

$\frac{dy}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^{2}} = \frac{(1 - x +1 +x)}{(1-x)^{2}}$

= $\frac{2}{(1-x)^{2}}$.

$f(x) = 5x^{2} - 4x + 3$

$f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1$

When x = 1, f(x) = y = $5(1^{2}) - 4(1) + 3 = 4$

The coordinates are (1,4)