Question Library

(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 = \(\frac{29.06}{360} \times 2 \times \frac{22}{7} \times 6400\) = \(3247.34 km \approxeq 3247km\)

(\(\frac{3}{4} + \frac{1}{3}) \times 4\frac{1}{3} \div 3\frac{1}{4}\) = \(\frac{13}{12} \times \frac{13}{3} \times\frac{4}{13}\) = \(\frac{13}{9}\) or 1\(\frac{4}{9}\))

2y\(^2\) + xy - 3x\(^2\) 2y\(^2\) + 3xy - 2xy - 3x\(^2\) y(2y + 3x) - x(2y + 3x) = (y - x)(2y + 3x)

Given that 2, 6, 12, 20...? the nth term = n\(^2\) + n check: n = 1, u1 = 2 n = 2, u2 = 4 + 2 = 6 n = 3, u3 = 9 + 3 = 12 ∴ n = 4, u4 = 16 + 4 = 20

Identity(e) : a \(\ast\) e = a m \(\ast\) e = m...(i) m \(\ast\) e = me + m + e Because m \(\ast\) e = m : m = me + m + e m - m = e(m + 1) e = \(\frac{0}{m + 1}\) e = 0

Using pythagora's theorem: hyp\(^2\) = opp\(^2\) + adj\(^2\) hyp\(^2\) = 12\(^2\) + 5\(^2\) → 144 + 25 hyp\(^2\) = 169 hyp = \sqrt169 hyp = 13

Using sinØ = \(\frac{6\sqrt3}{12}\) → \(\frac{\sqrt3}{2}\) sinØ = \(\frac{\sqrt3}{2}\) or 60^\circ The bearing of Z from X = [270 - 60]^\circ → 210^\circ

H will represent \(\frac{108}{630}\) x \(\frac{360^o}{1}\) ≈ 62^\circ

U = {1,2,3,4}; P = {2,3}; Q = {2,4}; P∩Q = {2} (P∩Q)' = {1,3,4}