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(a)(i) \(\log_{10} 45 = \log_{10} (3 \times 3 \times 5)\) = \(\log_{10} (3^{2} \times 5)\) = \(\log_{10} 3^{2} + \log_{10} 5\) = \(2 \log_{10} 3 + \log_{10} 5\) = \(2(0.477) + 0.699\) = \(0.954 + 0.699 = 1.653\) (ii) \(x^{0.8265} = 45\) Taking the log of both sides, \(\log_{10} x^{0.8265} = \log_{10} 45\) \(0.8265 \log_{10} x = \log_{10} 45\) \(\log_{10} x = \frac{1.653}{0.8265}\) \(\log_{10} x = 2\) \(x = 10^{2} = 100\) (b) \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\) No Log 2.067 \(0.0348\) = 0.3513 - 0.0348 \(\bar{2}.5416 +\) 0.538 \(\bar{1}.7308\) \(\bar{2}.2724\) = \(\bar{2}.2724\) = \(2.0789 \div 2 = 1.0395\) Antilog - 10.95 \(\therefore \sqrt{\frac{2.067}{0.0348 \times 0.538}} = 10.95\)

(a) Draw RC produced to E parallel to DE. \(\therefore < BCR = 52^\circ\) (alternate angles BA//RC) also < RCD = 48^\circ (alternate angles RC//DE) \(x^\circ = 360^\circ - (52^\circ + 48^\circ) = 360^\circ - 100^\circ = 260^\circ\) (angles at a point = 360^\circ) (b) \(y - 2x < 5 ... (i)\) \(2y + 2x \geq 4 \implies y + x \geq 2 ... (ii)\) \(y + 2x \leq 10 ... (iii)\) From (i), if \(y - 2x = 5\), when x = 0, y = 5, when x = 2, y = 9. From (ii), if \(y + x = 2\). when x = 0, y = 2, when x = 3, y = -1 From (iii), if \(y + 2x = 10\), when x = 0, y = 10, when x = 2, y = 6.

(a) Given : Circle ABC, centre O. To prove: < AOB = < ACB Construction : Join CO produced to P. Proof: With the lettering in the figure, OA = OB (radii) \(x_{1} = x_{2}\) (base angles of isosceles triangle) \(\therefore < AOP = x_{1} + x_{2}\) (exterior angle of triangle AOC) \(\therefore < AOP = 2x_{2} (x_{1} = x_{2})\) (base angles of an isosceles triangle) Similarly, \(< BOP = 2y_{2}\) \(\therefore < AOB = 2x_{2} + 2y_{2} \) = \(2(x_{2} + y_{2})\) \(\implies < AOB = 2 \times < ACB\) (proved) (b) From the figure, \(< ACB = \frac{130^\circ}{2} = 65^\circ\) (angle subtended at the centre of the circle) \(< ACO = 26^\circ\) (Base angles of isosceles triangle ACO) \(\therefore < BCO = 65^\circ - 26^\circ = 39^\circ\) \(< OBC = 39^\circ\) (base angles of isosceles triangle OBC) \(\therefore < COB = 180^\circ - (39^\circ + 39^\circ) = 102^\circ\)

(a) 5, 9, 13, ... is an A.P with the first term = 5 and common difference = 9 - 5 = 13 - 9 = 4. \(T_{n} = a + (n - 1)d\) (terms of an A.P) \(T_{25} = 5 + (25 - 1) \times 4\) = \(5 + 24 \times 4\) = \( 5 + 96 = 101\) (b) \(\frac{8}{9}, \frac{-4}{3}, 2, ...\) is a G.P with first term = \(\frac{8}{9}\) \(r = \frac{T_{2}}{T_{1}} = -\frac{4}{3} \div \frac{8}{9} = \frac{-4}{3} \times \frac{9}{8} = -\frac{3}{2}\) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{5} = (\frac{8}{9})(-\frac{3}{2})^{5 - 1}\) = \(\frac{8}{9} \times \frac{81}{16} = \frac{9}{2}\) (c) \(T_{n} = ar^{n - 1}\) (terms of a G.P) \(T_{3} = ar^{2} = 48 ... (1)\) \(T_{6} = ar^{5} = 14\frac{2}{9} ... (2)\) \((2) \div (1) : r^{3} = \frac{128}{9} \div 48 = \frac{128}{9} \times \frac{1}{48}\) \(r^{3} = \frac{8}{27}\) \(r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\) From (1), \(ar^{2} = 48 \implies a \times (\frac{2}{3})^{2} = 48\) \(4a = 48 \times 9 \implies a = \frac{48 \times 9}{4} = 108\) \(T_{2} = 108 \times \frac{2}{3} = 72\) \(T_{4} = 48 \times \frac{2}{3} = 32\) \(\therefore\) The first four terms of the sequence are \(108, 72, 48, 32\).

(a) \(\theta\) 0^\circ 30^\circ 60^\circ 90^\circ 120^\circ 150^\circ 180^\circ y -1.0 1.732 2.098 0 -2.098 -1.732 1.0 (b) (c)(i) Solution of \(3 \sin 2 \theta - \cos \theta = 0\) is at y = 0 = \(9.30^\circ, 80^\circ, 174^\circ\) (ii) Maximum value of y = 2.098 \(\approxeq\) 2.1 (to 1 decimal place)

Marks Upper class boundary Freq Cum Freq 0-9 9.5 2 2 10-19 19.5 3 5 20-29 29.5 4 9 30-39 39.5 6 15 40-49 49.5 13 28 50-59 59.5 10 38 60-69 69.5 5 43 70-79 79.5 3 46 80-89 89.5 2 48 90-99 99.5 2 50 (a) (b)(i) Upper quartile \(Q_{3}\) = 57 marks (ii) Pass mark if 60% of the class passed = 42 marks

(a) Angular difference = 33^\circ + 23^\circ = 56^\circ Distance PQ along parallel of latitude = \(\frac{56}{360} \times 2 \pi r\) where r = R cos 55^\circ \(\therefore |PQ| = \frac{56}{360} \times 2 \times \frac{22}{7} \times 6400 \cos 55\) = \(3589.46 km \approxeq 3589 km\) (to the nearest degree) (b) \(|PO|^{2} = r^{2} + r^{2} - 2r^{2} \cos 56^\circ = R^{2} + R^{2} - 2R^{2} \cos \theta \) = \(2r^{2} - 2r^{2} \cos 56^\circ = 2R^{2} - 2R^{2} \cos \theta\) But \(r = R cos 55^\circ = 6400 \times 0.5736 = 3671\) \(\therefore 2r^{2} (1 - \cos 56^\circ) = 2R^{2} (1 - \cos \theta)\) \(2 \times 3671^{2} ( 1 - 0.5592) = 2 \times 6400^{2} (1 - \cos \theta)\) \(1 - \cos \theta = \frac{2 \times 3671^{2} \times 0.4408}{2 \times 6400^{2}}\) \(1 - \cos \theta = 0.1259\) \(\cos \theta = 1 - 0.1259 = 0.8741\) \(\theta = \cos^{-1} (0.8741) = 29.06^\circ\) \(\therefore\) Distance PQ along great circle = \(\frac{29.06}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(3247.34 km \approxeq 3247km\)

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option A.

The sentence expresses surprise at the fact that he lost the presidential election. This implies that most people expected him to win. Option B best captures this idea, as it states that it was widely believed that he would win.

wonderful means inspiring delight, pleasure, or admiration; extremely good; marvellous

No expanded working has been attached for this question yet. The verified answer is Option B.

Aminu failure surprised people because he was not a dull student and wasn't expected to have failed.