WAEC Further Mathematics Past Questions & Answers - Page 140

The nth term of a GP is given by: \(T_{n} = ar^{n-1}\).

\(T_{3} = ar^{3-1} = ar^{2} = 81\).......(1)

\(T_{7} = ar^{7-1} = ar^{6} = 16 \) ...... (2)

Dividing (2) by (1), we have \(r^{4} = \frac{16}{81} = (\frac{2}{3})^{4} \implies r = \frac{2}{3}\)

Putting \(r = \frac{2}{3}\) in equation (1), we have \(81 = a \times (\frac{2}{3}\)^{2} = a \times \frac{4}{9} \implies a = \frac{729}{4}\)

\(T_{5} = ar^{5-1} = ar^{4} = \frac{729}{4} \times (\frac{2}{3})^{4}\)

= \(\frac{729}{4} \times \frac{16}{81} = 36\)

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

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

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