Explanation

(a) \((p + qx)^{7} = p^{7} + 7p^{6} qx + 21p^{5} (qx)^{2} + 35p^{4} (qx)^{3} + 35p^{3} (qx)^{4} + 21p^{2} (qx)^{5} + 7p(qx)^{6} + (qx)^{7}\)

The coefficient of \(x^{2}\) = \(21p^{5} q^{2}\)

The cofficient of \(x^{3}\) = \(35p^{4} q^{3}\)

\(\therefore 21p^{5} q^{2} = 35p^{4} q^{3}\)

\(\implies 21p = 35q\)

\(q = \frac{21}{35} p = \frac{3}{5} p\).

(b) N180 + N260 + N340 + ...

This sequence is a linear sequence with first term = a = 180, d = 80.

\(S_{16} = ?\)

\(S_{n} = \frac{n}{2} (2a + (n - 1) d)\)

\(S_{16} = \frac{16}{2} (2(180) + (16 - 1) 80\)

= \(8(360 + 1200)\)

= \(8(1560)\)

= \(N12,480\)

He would have contributed N12,480 by the 16th week.

Correct Answer: B. 54

Explanation

\(T_{n} = a + (n - 1)d\)

\(380 = 9 + (n - 1)7\)

\(380 = 9 + 7n - 7 \implies 380 = 2 + 7n\)

\(7n = 380 - 2 = 378 \therefore n = \frac{378}{7} = 54\)

Explanation

\(h = 20t - \frac{3}{2}t^{2} - \frac{2}{3}t^{3}\)

Velocity, \(\frac{\mathrm d h}{\mathrm d t} = 20 - 3t - 2t^{2}\)

At point of rest, v = 0

\(2t^{2} + 3t - 20 = 0\)

\(\implies 2t^{2} - 5t + 8t - 20 = 0\)

\(t(2t - 5) + 4(2t - 5) = 0 \implies t = \frac{5}{2}; -4\)

The value of time cannot be negative hence \(t = \frac{5}{2}secs\)

(b) Maximum height, \(h_{m}\) is at time \(t = \frac{5}{2}s\).

\(h_{m} = 20(\frac{5}{2}) - \frac{3}{2}(\frac{5}{2})^{2} - \frac{2}{3}(\frac{5}{2})^{3}\)

= \(50 - \frac{75}{8} - \frac{250}{16} = \frac{400}{16} = 25 m\)

Correct Answer: C. \(\begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\)

Explanation

\(PQ = \begin{pmatrix} 7 - 3 \\ 4 - 1 \end{pmatrix}\)

\(= \begin{pmatrix} 4 \\ 3 \end{pmatrix}\)

\(\hat{n} = \frac{\overrightarrow{PQ}}{|PQ|} \)

\(|PQ| = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\)

\(\hat{n} = \frac{1}{5}\begin{pmatrix} 4 \\ 3 \end{pmatrix} = \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}\)

Correct Answer: C. \(1, -\frac{1}{3}\)

Explanation

At turning point, \(\frac{\mathrm d y}{\mathrm d x} = 0\).

Given \(x^{3} - x^{2} - x + 6 \)

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 2x - 1 = 0 \)

\(3x^{2} - 3x + x - 1 = 0 \implies (3x + 1)(x - 1) = 0\)

\(x = \frac{-1}{3}, 1\)

Correct Answer: A. \(\frac{4}{7}\)

Explanation

P(one blue, other red) = P(1st red then blue) or P(1st blue then red)

= \((\frac{4}{7} \times \frac{3}{6}) + (\frac{3}{7} \times \frac{4}{6})\)

= \(\frac{2}{7} + \frac{2}{7} = \frac{4}{7}\)

Correct Answer: C. 2 cm

Explanation

The area of sector in radians = \(\frac{r^{2} \theta}{2}\)

\(3 cm^{2} = \frac{1.5 \times r^{2}}{2}\)

\(r^{2} = \frac{3 \times 2}{1.5} = 4\)

\(r = 2 cm\)

Correct Answer: C. 12\(ms^{-2}\)

Explanation

\(\frac{\mathrm d s(t)}{\mathrm d t} = v(t)\) and \(\frac{\mathrm d v(t)}{\mathrm d t} = a(t)\)

\(\therefore v(t) = \frac{\mathrm d (12t^{2} - 2t^{3})}{\mathrm d t} = 24t - 6t^{2}\)

\(\frac{\mathrm d (24t - 6t^{2})}{\mathrm d t} = 24 - 12t = a(t)\)

\(a(1) = 24 - 12(1) = 24 - 12 = 12ms^{-2}\)

Correct Answer: D. x = 0 or x = 3

Explanation

\(3^{2x-2} - 28(3^{x-2}) + 3 = 0\)

\(\frac{3^{2x}}{3^2} - \frac{28.3^x}{3^2} + 3 = 0\)

\(\frac{3^{2x}}{9} - \frac{28.3^x}{9} + 3 = 0\)

let p = 3\(^x\)

\(\frac{p^{2}}{9} - \frac{{28p}}{9} + 3 = 0\)

multiply through by 9

p\(^2\) - 28p + 27 = 0

p\(^2\) - p - 27p + 27 = 0

p (p - 1) - 27(p - 1) = 0

(p-1)(p-27) = 0

p = 1 or 27

when p = 1

p = 3\(^x\)

3\(^x\) = 1

3\(^x\) = 3\(^0\)

x = 0

when p = 27

3x = 27

3x = 33

x = 3

Correct Answer: D. { }

Explanation

\(P = {n^{2} + 1: n = 0,2,3} \therefore P = {1, 5,10}\)

\(Q = {n + 1: n = 2,3,5} \therefore Q = {3, 4, 6}\)

\(P \cap Q = { }\)