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\(y = x \sin x\) \(\frac{\mathrm d y}{\mathrm d x} = x \cos x + \sin x\) \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = x (- \sin x) + \cos x + \cos x\) = \(2 \cos x - x \sin x\)

\(\frac{dm}{dt}\) = 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\(\pi\)

\(\int_{\frac{\pi}{12}} ^{\frac{\pi}{4}} 2 \cos 2x \mathrm {d} x\) = \([\frac{2 \sin 2x}{2}]|_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\) = \(\sin 2x |_{\frac{\pi}{12}} ^{\frac{\pi}{4}}\) = \(\sin 2(\frac{\pi}{4}) - \sin 2(\frac{\pi}{12})\) = \(\sin \frac{\pi}{2} - \sin \frac{\pi}{6}\) = \(1 - \frac{1}{2} = \frac{1}{2}\)

Mean \(\bar{x}\) = \(\frac{\sum fx}{\sum f}\) = \(\frac{5.2}{1}\) = \(\frac{8 + 4y + 36 + 40}{4 + y + 6 + 5}\) = \(\frac{5.2}{1}\) = \(\frac{84 + 4y}{15 + y}\) = 5.2(15 + y) = 84 + 4y = 84 + 4y = 78 + 5.2y = 5.2y - 4y = 84 - 78 1.2y = 6 y = \(\frac{6}{1.2}\) = \(\frac{60}{12}\) = 5

From the table, the mode = 1. The median = 2.

(x) = \(\frac{5 + 6 + 7}{3}\) = \(\frac{18}{3}\) = 6 \(\begin{array}{c|c} scores(X) & \text{d = (x - x) deviation} & (deviation)^2\\\hline 5 & 5 - 6 & 1\\ 6 & 6 - 6 & 0 \\ 7 & 7 - 6 & 1\\ \hline & & 2\end{array}\) S.D \(\sqrt{\frac{\sum d^2}{n}}\) where d = deviation = (x - x) = \(\sqrt{\frac{2}{3}}\)

\(\begin{array}{c|c} & & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline a & 1 & 1, 1 & 1, 2 & 1, 3 & 1, 4 & 1, 5 & 1, 6\\\hline & 2 & 1,1 & 2, 2 & 2, 3 & 2, 4 & 2, 5 & 2, 6 \\\hline B & 3 & 3, 1 & 3, 2 & 3, 3 & 3, 4 & 3, 5 & 3, 6\\\hline & 4 & 4 , 1 & 4, 2 & 4, 3 & 4, 4 & 4, 5 & 4, 6\\\hline Die & 5 & 5, 1 & 5, 2 & 5, 3 & 5, 4 & 5, 5 & 5, 6 \\\hline & 6 & 6, 1 & 6, 2 & 6, 3 & 6, 4 & 6, 5 & 6, 6\end{array}\) Probability of obtaining total score of 8 = \(\frac{5}{36}\)

Prob of P = \(\frac{3}{4}\) Prob of Q = \(\frac{1}{6}\) Prob of both P or Q = \(\frac{3}{4} + \frac{1}{6} = \frac{11}{12}\)

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.

Given the graph and the curve y = ax2 + bx + c the roots are x - 2 and 2 while its equation (x + 2)(x - 2) = y y = x2 - 4 i.e. y = x2 + 0x - 4 a = 1, b = 0 and c = -4

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

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