WAEC Further Mathematics Past Questions & Answers - Page 100

\(\frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{A}{1 -  x} + \frac{Bx + C}{x^{2} + 1}\)

= \(\frac{A(x^{2} + 1) + (Bx + C)(1 - x)}{(1 - x)(x^{2} + 1)}\)

\(\implies x^{2} + x + 4 = A(x^{2} + 1) + (Bx + C)(1 - x)\)

\(x^{2} + x + 4 = Ax^{2} + A + Bx - Bx^{2} - Cx + C\)

\(\implies (A - B)x^{2} = x^{2}; A - B = 1 ...... (i)\)

\((B - C)x = x; B - C = 1 ..... (ii)\)

\(A + C = 4 ...... (iii)\)

Solving the above simultaneous equations by any of the known methods, we get

\(A = 3, B = 2, C = 1\)

\(\therefore  \frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{3}{1 - x} + \frac{2x + 1}{x^{2} + 1}\)

Given: \(\frac{\mathrm d A}{\mathrm d t} = 4 mm^{2}/s\)

\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)

\(A = \pi r^{2} \implies \frac{\mathrm d A}{\mathrm d r} = 2\pi r\)

\(\implies 4 = 2\pi r \times \frac{\mathrm d r}{\mathrm d t}\)

\(\frac{\mathrm d r}{\mathrm d t} = \frac{4}{2\pi r} = \frac{4 \times 7}{2 \times 22 \times 8}\)

= \(0.07954 mm/s \approxeq 0.08 mm/s\)

\(S.D = \sqrt{\frac{(x - \bar{x})^{2}}{n}}\)

\(S.D = \sqrt{\frac{160}{5}} = \sqrt{32}\)

= \(GH¢ 5.656 \approxeq GH¢ 5.66\)

Given a force F, the horizontal component = \(F\cos \theta\)

R = \(-50\cos 30 + 80\cos 45\)

= \(-50(\frac{\sqrt{3}}{2}) + 80(\frac{\sqrt{2}}{2})\)

= \( -25\sqrt{3} + 40\sqrt{2} = -43.30 + 56.67 \)

= \(13.37N \approxeq 13N\)

Since Ozo is a boy and must be in the sub-committee, we have 2 spaces for the boys and 2 for the girls.

= \(^{4}C_{2} \times ^{3}C_{2} \)

= \(\frac{4!}{(4 - 2)! 2!} \times \frac{3!}{(3 - 2)! 2!} = 6 \times 3 = 18\)