Question Library

\(\int \frac{1 + x}{x^{3}} \mathrm d x\) = \(\int (\frac{1}{x^{3}} + \frac{x}{x^{3}}) \mathrm d x\) = \(\int (x^{-3} + x^{-2}) \mathrm d x\) = \(\frac{-1}{2x^{2}} - \frac{1}{x} + k\)

\(\int_{0} ^{\frac{\pi}{2}} \sin x \mathrm d x\) = \(- \cos x |_{0} ^{\frac{\pi}{2}\) = \(- \cos (\frac{\pi}{2}) - (- \cos 0)\) = \(0 + 1\) = 1

The expression(P n Q) u R is shaded by identifying two specific areas: 1. (P n Q): the overlapping region between circles P and R 2. u R: the entire area of the circle R

\(\frac{3\sqrt{5} \times 4\sqrt{6}}{2 \sqrt{3} \times 3\sqrt{2}}\) = \(\frac{\sqrt{5} \times 2\sqrt{6}}{\sqrt{2} \times \sqrt{3}}\) = \(\frac{\sqrt{5} \times 2 \sqrt{6}}{\sqrt{6}}\) = 2\(\sqrt{5}\)

(a)(i) R is the required region. (iii) Coordinates of the point of intersection of the two graphs = (1.1, 3.8) (b) Assuming \(y - \frac{3x}{4} = 3\) If x = 0, \(y - \frac{3}{4}(0) = 3 \implies y = 3\) If y = 0, \(0 - \frac{3}{4}x = 3 \implies -3x = 12 ; x = - 4\) x 0 -4 y 3 0 Assuming \(y + 2x = 6\) If x = 0, \(y + 2(0) = 6 \implies y = 6\) If y = 0, \(0 + 2x = 6 \implies 2x = 6 ; x = 3\) x 0 3 y 6 0 (b) \((x + 2)(x + c) = x^{2} + 2x + cx + 2c\) \(x^{2} + (2 + c)x + 2c = x^{2} + bx + 18\) \( 2c = 18 ; c = 9\) \(b = 2 + c = 2 + 9 = 11\) (b, c) = (11, 9).

First, convert to base 10 112\(_6\) = (1 x 6\(^2\)) + (1 x 6\(^1\)) + (2 x 6\(^0\)) = 36 + 6 + 2 = 44\(_{10}\) \(\begin{array}{c|c} 3 & 44 \\ \hline 3 & 14 & r 2 \\ \hline 3 & 4 & r 2 \\ \hline 3 & 1 & r 1 \\ \hline 3 & 0 & r 1 \end{array}\) Ans = 1122 = D

\(\frac{2\frac{2}{3} \times 1\frac{1}{2}}{4\frac{4}{5}}\) \(\frac{8}{3} \times \frac{3}{2} \div \frac{24}{5}\) \(\frac{8}{3} \times \frac{3}{2} \times \frac{5}{24}\) \(\frac{5}{6}\)

\(\frac{21}{9} = \frac{7}{3}\) = 2.33 (to 3 sig. figs)

Amount spent on children = \(\frac{15}{100} \times 3500 = N525\) Amount spent = Amount on children + Additional amount on food Amount Spent = N525 + N1950 = N2475 Therefore amount left = N3,500 - N2,475 = N1,025

\(27^{x + 2} \div 9^{x + 1} = 3^{2x}\) \(3^{3(x + 2)} \div 3^{2(x + 1)} = 3^{2x}\) \(3^{3x + 6} \div 3^{2x + 2} = 3^{2x}\) \(\therefore (3x + 6) - (2x + 2) = 2x\) \(x + 4 = 2x \implies x = 4\)

log3x2 = -8 x2 = 3-8 x = \(3^{-8 \times \frac{1}{2}}\) x = 3-4 x = \(\frac{1}{3^4}\) x = \(\frac{1}{81}\)

\((\sqrt{6} + 2)^2 - (\sqrt{6} - 2)^2\) = \([(\sqrt{6} + 2) + (\sqrt{6} - 2)][(\sqrt{6} + 2) - (\sqrt{6} - 2)]\) = \((\sqrt{6} + 2 + \sqrt{6} - 2)(\sqrt{6} + 2 - \sqrt{6} + 2)]\) = \((2\sqrt{6})(4)\) = \(8\sqrt{6}\)

P = {2,3,5} Q = {2,3,6,9} P \(\cap\) Q = {2,3}

n(f \(\cap\) v) + n(f) + n(v) + n(f \(\cap\) v) = 46 3 + 19 + 23 + x = 46 22 + 23 + x = 46 45 + x = 46 x = 46 - 45 x = 1

w = \(\frac{v(2 + cn)}{1 - cn}\) 2v + cnv = w(1 - cn) 2v + cnv = w - cnw 2v - w = -cnv - cnw Multiply through by negative sign -2v + w = cnv + cnw -2v + w = n(cv + cw) n = \(\frac{-2v + w}{cv + cw}\) n = \(\frac{1}{c}\frac{-2v + w}{v + w}\) Re-arrange... n = \(\frac{1}{c}\frac{w - 2v}{v + w}\)