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(a) x 0 1 2 3 4 5 Total f 1 4 9 8 5 3 30 fx 0 4 18 24 20 15 81 (i) Mean(\(\bar{x}\) = \(\frac{\sum fx}{\sum f}\) = \(\frac{81}{30}\) = 2.7 (ii) \(N = \sum f = 30\) \(median = \frac{1}{2}(N + 1)\) = \(\frac{1}{2}(30 + 1)\) =15.5 = The 15th and 16th position. \(\therefore median = \frac{3 + 3}{2} = 3\) (b) Sectorial angle for the mark 2 = \(\frac{9}{30} \times 360^\circ\) = 108^\circ.

(a) \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\) \(\frac{x^{2}}{2} - x + \frac{1}{2} = \frac{x^{2} - 2x + 1}{2}\) = \(\frac{(x - 1)^{2}}{2}\) \(\therefore (\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1}) = (\frac{(x - 1)^{2}}{2})(\frac{1}{x - 1})\) = \(\frac{x - 1}{2}\) (b) (i) TQ = PR (North of Q) \(\implies \sin 29 = \frac{PR}{40}\) \(PR = 40 \sin 29 = 19.39 km\) (ii) QR = East of Q \(\frac{QR}{40} = \cos 29\) \(QR = 40 \cos 29\) = 34.98km \(\approxeq\) 35km. (c) Let the son's share = x. Each daughter's share = \(\frac{3}{4}x\) For the three daughters = \(3 \times \frac{3}{4} = \frac{9}{4}\) \(x + \frac{9}{4}x = 5720 \implies \frac{13}{4}x = 5720\) \(x = \frac{5720 \times 4}{13} = N1760\)

(a) Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\) = \(\frac{96}{360} \times \frac{22}{7} \times 5^{2}\) = \(20.95 cm^{2}\) Area of triangle formed from the sector = \(\frac{1}{2} r^{2} \sin \theta\) = \(\frac{1}{2} \times 5^{2} \times \sin 96\) = \(12.43 cm^{2}\) \(\therefore \text{The area of minor segment} = 20.95 - 12.43\) = \(8.52 cm^{2}\) (b) Area of minor sector = \(\frac{\theta}{360} \times \pi r^{2}\) = \(\frac{80}{360} \times \frac{22}{7} \times 7^{2}\) = \(34.22 cm^{2}\) Area of the square = \((14)^{2} = 196 cm^{2}\) Area of the circle = \(\pi r^{2} = \frac{22}{7} \times 7^{2} = 154 cm^{2}\) Area of the shaded portion in the square = \(196 - 154 = 42 cm^{2}\) Total area of the shaded portions = \(42 + 34.22\) = \(76.22 cm^{2} \approxeq 76.2 cm^{2}\)

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(a) (i) Let the length of the longitude be \(\theta\). Longitude difference = \(50 + \theta\) Distance along the parallel of latitude (KL) = \(\frac{50 + \theta}{360} \times 2\pi R \cos 60\) \(3520 = \frac{50 + \theta}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 60\) \(3520 = \frac{50 + \theta}{360} \times 2 \times \frac{22}{7} \times 3200\) \(50 + \theta = \frac{3520 \times 360 \times 7}{22 times 2 \times 3200}\) \(50 + \theta = 63^\circ\) \(\theta = 63^\circ - 50^\circ = 13^\circ\) (ii) Let the latitude of N be \(\alpha\). The distance along the parallel of longitude (KN) = \(\frac{60 + \alpha}{360} \times 2 \times \frac{22}{7} \times 6400\) \(10951 = \frac{60 + \alpha}{360} \times 2 \times \frac{22}{7} \times 6400\) \(60 + \alpha = \frac{10951 \times 360 \times 7}{2 \times 22 \times 6400}\) \(60 + \alpha = 98^\circ\) \(\alpha = 98^\circ - 60^\circ = 38^\circ\) The latitude of N = 38^\circS. (b) Let the man's income tax be x. 80% of x = \(\frac{1200}{0.25} = N4800\) \(0.8x = N4,800\) \(x = \frac{4800}{0.8} = N6000\)

(a) \(y = 2x^{2} - 7x - 3\) x -2 -1 0 1 2 3 4 5 \(2x^{2}\) 8 2 0 2 8 18 32 50 \(-7x\) 14 7 0 -7 -14 -21 -28 -35 \(-3\) -3 -3 -3 -3 -3 -3 -3 -3 \(y\) 19 6 -3 -8 -9 -6 1 12 (b) (c)(i) Minimum value of y = -9 (ii) \(y = 2x^{2} - 7x - 3\) Gradient = \(\frac{\mathrm d y}{\mathrm d x} = 4x - 7\) Gradient at x = 1 : \(4(1) - 7 = 4 - 7 = -3\). (d) \(2x^{2} - 7x - 5 = x + 4\) \(2x^{2} - 7x - 5 + 2 = x + 4 + 2\) \(2x^{2} - 7x - 3 = x + 6\) \(\implies y = x + 6\) x -2 -1 0 1 2 3 4 5 6 6 6 6 6 6 6 6 6 y 4 5 6 7 8 9 10 11 From the graph, x = -0.9 or 5.