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x = (all prime factors of 44) and y = (all prime factors of 60) ∴ x = (2, 11), y = (2, 3, 5) X ∪ Y = (2, 3, 5, 11), X ∩ Y = (2)

\(9p^{2} - q^{2} + 6qr - 9r^{2}\) = \(9p^{2} - (q^{2} - 6qr + 9r^{2})\) = \(9p^{2} - (q^{2} - 3qr - 3qr + 9r^{2})\) = \(9p^{2} - (q(q - 3r) - 3r(q - 3r))\) = \(9p^{2} - (q - 3r)^{2}\) = \((3p + (q - 3r))(3p - (q - 3r))\) = \((3p + q - 3r)(3p - q + 3r)\)

Given S = ut + \(\frac{1}{2} at^2\) S = ut + \(\frac{1}{2} at^2\) ∴ 2S = 2ut + at2 = at2 + 2ut - 2s = 0 t = \(\frac{-2u \pm 4u^2 + 2as}{2a}\) = -2u \(\pi\) \(\frac{\sqrt{u^2 4u^2 + 2as}}{2a}\) = \(\frac{1}{a}\) (-u + \(\sqrt{U^2 - 2as}\))

Let the amounts invested at 4% and 5% respectively be x and y. \(\therefore x + y = 280 ... (i)\) Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\) Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\) \(\therefore 0.04x + 0.05y = 12.80\) \(\implies 4x + 5y = 1280 ... (ii)\) From (i), \(x = 280 - y\). Put into (ii), \(4(280 - y) + 5y = 1280\) \(1120 - 4y + 5y = 1280\) \(1120 + y = 1280 \implies y = 1280 - 1120 = N160\) \(\therefore\) N160 was invested at the rate of 5% per annum.

x + 1 is a factor of x3 + 3x2 + kx + 4 Let f(x) = x3 + 3x2 + kx + 4 ∴ f(-1) = (-1)3 + 3(-1)2 + k(-1) + 4 = 0 -1 + 3 - k + 4 = 0 ∴ k = 6

\(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\) \(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\) A(x + 2) + B(x - 1) = 3 when x = 1, 3A = 3 \(\to\) a = 1 when x = -2, -3B = 3 \(\to\) B = -1 = \(\frac{1}{x - 1} - \frac{1}{x + 2}\)

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

\(\frac{dm}{dt}\) = 4 - 0.6t \(\int\)dm = \(\int\)(4 - 0.6t)dt m = \(4t - 0.3t^2 + c\), when t = 0, m = 2g ∴ c = 2 m = \(4t - 0.3t^2 + 2\), when t = 5 minutes m = \(4(5) - 0.3(5)^2 + 2 = 20 - 7.5 + 2\) = 14.5

f(x) = x\(^3\) - 12x + 11 \(\frac{df(x)}{dx)}\) = 3x\(^2\) - 12 = 0 ∴ 3x\(^2\) - 12 = 0 \(\to\) x\(^2\) = 4 x = \(\pm\)2, f(+2) = 8 - 24 + 11 = -15 = f(-2) = (-8) + 24 + 11 = 35 - 8 = 27 ∴ maximum value = 27

\(\frac{dv}{dt}\) = \(\pi\)(20 - t2)cm2S-1 \(\int\)dv = \(\pi\)(20 - t2)dt V = \(\pi\) \(\int\)(20 - t2)dt V = \(\pi\)(20 \(\frac{t}{3}\) - t3) + c when c = 0, V = (20t - \(\frac{t^3}{3}\)) after t = 2 seconds V = \(\pi\)(40 - \(\frac{8}{3}\) = \(\pi\)\(\frac{120 - 8}{3}\) = \(\frac{112}{3}\) = 37.33\(\pi\)

Treat the couple as one block, giving us 4 units to arrange (the couple + 3 others). Arrange the 4 units: 4! = 4 x 3 x 2 x 1 = 24. Arrange the couple within their block: 2! = 2 x 1 = 2. Total arrangements: 4! \times 2! = 24 \times 2 = 24 \times 2 = 48. Thus, the total number of arrangements is 48.

Total Mathematics courses = 4 Total Physics courses = 3 Courses to choose = 5 Courses in Physics to choose = 2 Courses in Mathematics to choose = 3 Mathematics = \(^{4} C_{3} = 4\) Physics = \(^{3} C_{2} = 3\) \(\therefore\) He can choose in 4 x 3 = 12 ways.

PSR = 80^\circ \(\therefore\) SRP = 20^\circ \(\implies\) RPQ = 20^\circ x = 140^\circ

(a) Class Interval (x) Frequency (f) Cum. Freq 1 - 10 6 6 11 - 20 6 12 21 - 30 12 24 31 - 40 11 35 41 - 50 10 45 51 - 60 5 50 (b)(i) Median = N30.50 (ii) 80th percentile = N45 (c) The modal weekly profit = Lower class boundary of modal class + \(\frac{(f - f_{1}) \times c}{(f - f_{1}) + (f - f_{2})}\) where \(f_{1}\) = frequency above the modal class ; \(f_{2}\) = frequency below modal class; c = class width; f = frequency of modal class. \(\therefore \text{Modal profit} = 20.5 + \frac{(12 - 6) \times 10}{(12 - 6) + (12 - 11)}\) = \(20.5 + \frac{60}{7}\) = \(N29.07\)

\(\frac{27}{5^1}\)(3)-3 x \(\frac{(1)^{-1}}{5}\) = \(\frac{27}{5}\) x \(\frac{1}{3^3}\) x \(\frac{1}{\frac{1}{5}}\) = \(\frac{27}{5}\) x \(\frac{1}{27}\) x \(\frac{1}{5}\) = \(\frac{1}{25}\)