Explanation

(a)To find sin q =\(\frac{2\sqrt{2}}{3}\) and cos p = \(\frac{\sqrt{3}}{3}\). Then, substituting

sin(p - q) = sin p cos q - cos p sin q = (\(\frac{1}{2} \times -\frac{1}{3}\)) - (\(\frac{2\sqrt{2}}{3} \times \frac{\sqrt{3}}{2}\)

Which simplified to -\(\frac{1}{6} - \frac{2\sqrt{6}}{6}\) and can be written as \(\frac{-1 -2\sqrt{6}}{6}\)

(b), candidates were expected to obtain the following table

Then, using the table and applying the formula;

\(\int^4_1\frac{2}{\sqrt{x + 3}}\)dx = 0.5{\(\frac{1}{2}\) + 0.9428 + 0.8944 + 0.8528 + 0.8165 + 0.7845 + \(\frac{0.7559}{2}\)}

= 0.5 (5.16895)

= 2.584475

= 2.58 correct to two decimal places.

Correct Answer: B. \(\frac{-y}{2y + x}\)

Explanation

\(y^2 + xy = 5\)

By implicit differentiation

\(=2y\frac{dy}{dx}+y+x\frac{dy}{dx}=0\)

\(=2y\frac{dy}{dx}+x\frac{dy}{dx}=-y\)

Factor out \(\frac{dy}{dx}\)

\(=\frac{dy}{dx}(2y+x)=-y\)

\(∴\frac{dy}{dx}=\frac{-y}{2y + x}\)

Correct Answer: A. \(\begin{pmatrix} 1 & 1 \\ -1.5 & -2 \end{pmatrix}\)

Explanation

Let A = \(\begin{pmatrix} 4 & 2 \\ -3 & -2 \end{pmatrix}\);

|A| = -8 - (-6) = -8 + 6

|A| = -2

A\(^{-1}\) = \(\frac{1}{-2}\) = \(\begin{pmatrix} -2 & 2- \\ 3 & 4 \end{pmatrix}\)

= \(\begin{pmatrix} 1 & 1 \\ -1.5 & -2 \end{pmatrix}\)

Explanation

\(\overrightarrow{MN} = \overrightarrow{RQ}\)

\(\begin{pmatrix} 4 - 10 \\ - 6 - 2 \end{pmatrix} = \begin{pmatrix} x - 8 \\ y - 16 \end{pmatrix}\)

\(\implies 4 - 10 = x - 8\)

\(-6 = x - 8 \implies x = -6 + 8 = 2\)

\(\implies - 6 - 2 = y - 16\)

\(-8 = y - 16 \implies y = -8 + 16 = 8\)

The coordinate of R is (2, 8).

Explanation

Given :

\(v = 50kmh^{-1} ; s = 20 km ; a = 6kmh^{-2} ; u = ?\)

\(v^{2} = u^{2} + 2as\)

\(50^{2} = u^{2} + 2(6)(20)\)

\(2500 = u^{2} + 240 \implies u^{2} = 2500 - 240 = 2260\)

\(u = \sqrt{2260} = 47.54 kmh^{-1}\)

To find time, we use

\(v = u + at \)

\(50 = 47.54 + 6t\)

\(t = \frac{50 - 47.54}{6} = 0.41h\)

\(\approxeq 0.4hour\)

Explanation

p*e = e*p = p

p + e - \(\frac{pe}{2}\) = p

e - \(\frac{pe}{2}\) = 0

e(\(1 - \frac{p}{2}\)) = 0

e = 0

p*p\(^{-1}\) = e

p - p\(^{-1}\) - \(\frac{pp^{-1}}{2}\) = 0

p\(^{-1}\) - \(\frac{pp}{2}\) = -p

p\(^{-1}\) (1 - \(\frac{p}{2}\)) = -p

p\(^{-1}\) = \(\frac{-p}{1 - \frac{p}{2}}\)

Inverse of -1 \(\to\) p = -1

p\(^{-1}\) = \(\frac{-1}{1 - (\frac{-1}{2})} = \frac{-1}{\frac{3}{2}}\)

= \(\frac{-2}{3}\)

(b) m*7 = m + 7 - \(\frac{7m}{2}\), m*5 = m + 5 - \(\frac{5m}{2}\)

m*7 = m*5, m + 7 - \(\frac{7m}{2}\) = m + 5 - \(\frac{5m}{2}\)

7 - 5 = -\(\frac{5m}{2} + \frac{7m}{2}\)

2 = \(\frac{2m}{2}\)

m = 2

Correct Answer: A. \(\frac{-3}{4}\)

Explanation

\(9^{2x + 1} = 81^{3x + 2}\)

\(9^{2x + 1} = (9^{2})^{3x + 2}\)

\(9^{2x + 1} = 9^{6x + 4}\)

Equating powers,

\(2x + 1 = 6x + 4 \implies -3 = 4x\)

\(\therefore x = \frac{-3}{4}\)

Correct Answer: A. \(\sqrt{2}\) - 1

Explanation

\(\frac{1}{\sqrt{2} + 1}\) x \(\frac{\sqrt{2} - 1}{\sqrt{2} - 1}\)

= \(\frac{\sqrt{2} - 1}{2 - 1}\)

= \(\frac{\sqrt{2} - 1}{1} = \sqrt{2} - 1\)

Correct Answer: C. \(\frac{5}{36}\)

Explanation

probability of selecting three balls with alternating colours = probability of selecting a blue ball (without replacement) x probability of selecting a red ball (without replacement) x probability of selecting a blue ball (without replacement)

In this case, we will be using the first instance (both would give the same answer since the number of blue and red balls are equal)

first: probability of selecting a red ball (without replacement) = \(\frac{5}{10}\) = \(\frac{1}{2}\)

second: the probability of selecting a blue ball (without replacement) = \(\frac{5}{9}\)

third: probability of selecting a red ball (without replacement) = \(\frac{4}{8}\) = \(\frac{1}{2}\)

Therefore, the probability of selecting three balls with alternating colours = \(\frac{1}{2}\) x \(\frac{5}{9}\) x \(\frac{1}{2}\)

= \(\frac{5}{36}\)

Correct Answer: D. \(16 ms^{-2}\)

Explanation

\(v(t) = 3t^{2} - 2t + 1\)

\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6t - 2\)

\(a(3) = 6(3) - 2 = 18 - 2 = 16 ms^{-2}\)