Explanation

(a) 5 men, 3 women

Number of ways of forming the committee = \(^{8}C_{3}\)

= \(\frac{8!}{(8 - 3)! 3!}\)

= 56 ways.

(b) p(no woman) = \(\frac{^{5}C_{3}}{^{8}C_{3}}\)

= \(\frac{10}{56}\)

\(\therefore\) p( at least one woman) = 1 - p(no woman)

= \(1 - \frac{10}{56}\)

= \(\frac{46}{56}\)

= \(\frac{23}{28}\).

Explanation

(a) \(x = 3i - j ; y = 2i + kj\)

\(x \cdot y = |x||y| \cos \theta\)

\((3i - j) \cdot (2i + kj) = (\sqrt{3^{2} + (-1)^{2}})(\sqrt{2^{2} + k^{2}}) \cos \theta\)

\(6 - k = (\sqrt{10})(\sqrt{4 + k^{2}})(\frac{\sqrt{5}}{5})\)

\(5(6 - k) = \sqrt{10(4 + k^{2})(5)}\)

\(30 - 5k = \sqrt{200 + 50k^{2}}\)

\((30 - 5k)^{2} = 200 + 50k^{2}\)

\(900 - 300k + 25k^{2} = 200 + 50k^{2}\)

\(900 - 200 - 300k + 25k^{2} - 50k^{2} = 0\)

\(25k^{2} + 300k - 700 = 0 \implies k^{2} + 12k - 28 = 0\)

\(k^{2} - 2k + 14k - 28 = 0 \implies k(k - 2) + 14(k - 2) = 0\)

\(\text{k = 2 or -14}\).

(b)

ABCD is a parallelogram if AD = BC.

\(\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD}\)

= \(\begin{pmatrix} -5 \\ -1 \end{pmatrix} + \begin{pmatrix} 4 \\ -7 \end{pmatrix}\)

= \(\begin{pmatrix} -1 \\ -8 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}\)

= \(\begin{pmatrix} 5 \\ 1 \end{pmatrix} + \begin{pmatrix} -6 \\ -9 \end{pmatrix}\)

= \(\begin{pmatrix} -1 \\ -8 \end{pmatrix}\).

\(\overrightarrow{AD} = \overrightarrow{BC}\). ABCD is a parallelogram.

Explanation

(a) \(y = 4x^{3}\)

\(\frac{\mathrm d y}{\mathrm d x} = 12x^{2}\)

\(12x^{2} = 108 \implies x^{2} = 9\)

\(\therefore x = \pm 3\)

When x = -3, \(y = 4(-3^{3}) = -108\)

When x = 3, \(y = 4(3^{3}) = 108\)

\(\therefore P = (-3, -108) ; Q = (3, 108)\)

(b)(i) \(\sin (A - B) = \sin A \cos B - \sin B \cos A\)

\(\sin 15 = \sin (45 - 30) = \sin 45 \cos 30 - \sin 30 \cos 45\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) - (\frac{1}{2})(\frac{\sqrt{2}}{2})\)

= \(\frac{\sqrt{6} - \sqrt{2}}{4}\)

\(\cos (A - B) = \cos A \cos B + \sin A \sin B\)

\(\cos 15 = \cos (45 - 30) \)

= \(\cos 45 \cos 30 + \sin 45 \sin 30\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)

= \(\frac{\sqrt{6} + \sqrt{2}}{4}\)

(b)(ii) \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

\(\tan 15 = \frac{\sqrt{6} - \sqrt{2}}{4} \div \frac{\sqrt{6} + \sqrt{2}}{4}\)

= \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)

Ratinalizing, we have

\((\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}})(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}) = \frac{6 - 2\sqrt{3} - 2\sqrt{3} + 2}{6 - 2}\)

= \(\frac{8 - 4\sqrt{3}}{4}\)

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

Correct Answer: B. 2950

Correct Answer: B. 1

Correct Answer: A. 60

Correct Answer: A. 3453

Explanation

4321\(_7\) + 1234\(_7\) = 5555\(_7\)

Missing number = 12341\(_7\) - 5555\(_7\)

= 3453\(_7\)

Correct Answer: A. (1, 5)

Explanation

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

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 1 = 5\)

\(6x = 5 + 1 = 6 \implies x = 1\)

\(y = 3(1^{2}) - 1 + 3 = 3 - 1 + 3 = 5\)

\(P = (1, 5)\)

Explanation

(a) \(\tan 50° = \frac{|TF|}{20} \implies |TF| = 20 \tan 50\)

= \(20 \times 1.1918\)

= \(23.836 m\)

\(\tan 30 = \frac{|PF|}{20} \implies |PF| = 20 \tan 30\)

= \(20 \times 0.5774\)

= \(11.548 m\)

\(\therefore |PT| = 23.836 m - 11.548m\)

= \(12.288 m \approxeq 12.3 m\)

(b) \(|TH| = \frac{20}{\cos 50}\)

= \(\frac{20}{0.6428}\)

= \(31.11 m\)

(c) \(|HF| = \frac{20 \tan 50°}{\tan 40°}\)

= \(\frac{20 \times 1.1918}{0.8391}\)

= \(28.41 m\)

Correct Answer: B. 23.0 years