Correct Answer: B. \(\frac{2}{5}\)

Correct Answer: D. \(PQ\)

Explanation

A \(1 \times 2\) matrix can be multiplied by \(2 \times 2\) matrix to give a \(1 \times 2\) matrix.

Correct Answer: D. 2?3

Explanation

d = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

x1 = x, x2 = -x, y1 = 4, y2 = 3, d = 7

7 = \(\sqrt{(-x -x)^2 + (3 - 4)^2}\)

7 = \(\sqrt{(-2x)^2 + (-1)^2}\)

7 = ( \(\sqrt{4x^2 + 1}\)

square both sides

7\(^2\) = 4x\(^2\) + 1

collect like terms

4x\(^2\) = 49 - 1

4x\(^2\) = 48

x\(^2\) = \(\frac{48}{4}\)

x\(^2\) = 12

x = √12

x = 2√3

Explanation

(a) p(man wins) = p = 0.8 ; p(man loses) = q = 0.2.

4 different races : \((p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2} q^{2} + 4pq^{3} + q^{4}\)

(i) p(man wins all) = \(p^{4} = (0.8)^{4} = 0.4096\)

(ii) p(man wins no race) = \(q^{4} = (0.2)^{4} = 0.0016\)

(iii) p(man wins at most 3) = 1 - p(man wins all)

= 1 - 0.4096 = 0.5904.

(b)(i) 5 girls, 10 boys ; Total = 15.

Ways of selecting 3 boys = \(^{10}C_{3} \times ^{5}C_{2}\)

= \(120 times 10\)

= 1200

Without restriction, selection = \(^{15}C_{5}\)

= \(\frac{15!}{10! 5!}\)

= 3003

p(selecting 3 boys) = \(\frac{1200}{3003} = 0.3996\)

(ii) Ways of selecting no girls = \(^{10}C_{5} \times ^{5}C_{0}\)

= 252

p(at least one girl) = 1 - p(no girl)

= \(1 - \frac{252}{3003}\)

= \(2751}{3003} = 0.9161\)

Explanation

(a)(i)

(ii)

(b)(i) Those who scored less and equal to 32 marks = 75

Those who scored greater and equal to 74 = 508

Therefore those that scored between 32 and 74 = 508 - 75 = 433.

(ii) 18% of 550 = 99

18% failed, the number that passed = 550 - 99 = 451.

(iii) 8% passed with distinction, so 92% did not get distinction.

92% of 550 = 506.

Lowest mark for distinction = 74.

Correct Answer: D. 120

Explanation

\(\frac{^{n}P_{5}}{^{n}C_{5}} = \frac{n!}{(n-5)!} ÷ \frac{n!}{(n-5)!5!}\)

= \(\frac{n!}{(n-5)!} \times \frac{(n-5)!5!}{n!} = 5! = 120\)

Correct Answer: C. -i + 12j

Explanation

→PQ = →PR + →RQ

→PR = →PQ - →RQ

→PR = -2i + 5j - (-i - 7j)

→PR = -2i + 5j + i + 7j

→PR = -i + 12j.

Correct Answer: C. \(\frac{3}{2}\)

Explanation

\(\alpha + \beta = \frac{-b}{a}\)

From the equation, a = 2, b = -3 and c = 4

\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)

Explanation

\(P = (4N, 030°) ; R = (10N, 300°)\)

Let \(F = (a N, \theta)\) be the force that will keep the particle in equilibrium.

Then P + R + F = 0.

\(\begin{pmatrix} 4 \cos 30 \\ 4 \sin 30 \end{pmatrix} + \begin{pmatrix} 10 \cos 300 \\ 10 \sin 300 \end{pmatrix} + \begin{a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\).

\(\begin{pmatrix} 4(\frac{\sqrt{3}}{2}) \\ 4(\frac{1}{2}) \end{pmatrix} + \begin{pmatrix} 10(\frac{1}{2}) \\ 10(-\frac{\sqrt{3}}{2}) \end{pmatrix} + \begin{pmatrix} a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(\begin{pmatrix} 2\sqrt{3} + 5 \\ 2 - 5\sqrt{3} \end{pmatrix} + \begin{pmatrix} a \cos \theta \\ a \sin \theta \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \begin{pmatrix}\)

\(\implies 2\sqrt{3} + 5 + a \cos \theta = 0 \)

\(2\sqrt{3} + 5 = -a \cos \theta\)

\(2 - 5\sqrt{3} = -a \sin \theta\)

\(\tan \theta = \frac{2 - 5\sqrt{3}}{2\sqrt{3} + 5} = \frac{-6.66}{8.486}\)

\(\tan \theta = -0.7848 \implies \theta = -38.13°\)

= \(141.87°\)

\(-a \cos \theta = 8.486\)

\(-a \cos (141.87) = -a (-0.7866) = 8.486\)

\(0.7866a = 8.486 \implies a = \frac{8.486}{0.7866} = 10.79N\)

\(\therefore F = (10.79N, 141.87°)\)

Correct Answer: D. \(\frac{9}{2} m\)

Explanation

Given, \(a(t) = (3t - 2) ms^{-2}\), the first integration of a(t), with respect to t, gives v(t) (the velocity). The second integration of a(t) or first integration of v(t) gives s(t).

\(v(t) = \int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)

\(s(t) = \int (\frac{3}{2}t^{2} - 2t) \mathrm {d} t\)

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

When t = 3s,

\(s(3) = \frac{3^{3}}{2} - 3^{2} = \frac{27}{2} - 9 \)

= \(\frac{9}{2}\)