Explanation

\(F_{1} = (18N, 330°) ; F_{2} = (10N, 090°) ; F_{3} = (25N, 180°)\)

\(R = F_{1} + F_{2} + F_{3}\)

= \(\begin{pmatrix} 18 \cos 330 \\ 18 \sin 330 \end{pmatrix} + \begin{pmatrix} 10 \cos 90 \\ 10 \sin 90 \end{pmatrix} + \begin{pmatrix} 25 \cos 180 \\ 25 \sin 180 \end{pmatrix}\)

= \(\begin{pmatrix} 18 \cos 30 \\ -18 \sin 30 \end{pmatrix} + \begin{pmatrix} 10 \times 0 \\ 10 \times 1 \end{pmatrix} + \begin{pmatrix} -25 \\ 0 \end{pmatrix}\)

= \(\begin{pmatrix} 18 \times 0.866 \\ -18 \times 0.5 \end{pmatrix} + \begin{pmatrix} 0 \\ 10 \end{pmatrix} + \begin{pmatrix} -25 \\ 0 \end{pmatrix}\)

= \(\begin{pmatrix} 15.588 - 25 \\ -9 + 10 \end{pmatrix}\)

= \(\begin{pmatrix} -9.41 \\ 1 \end{pmatrix}\)

\(R = \sqrt{(-9.41)^{2} + 1^{2}}\)

= \(\sqrt{88.5481 + 1}\)

= \(\sqrt{89.5481}\)

\(R = 9.463N\)

Direction: Let \(\theta\) be the angle resultant makes with the x - axis.

\(\tan \theta = \frac{1}{-9.41} = -0.1062\)

\(\theta = \tan^{-1} (-0.1063) = -6.07°\)

The resultant has magnitude 9.463N and makes an angle of 6.07° with the negative x- axis.

Correct Answer: B. 2

Explanation

\(a * b = a^{2} + b + ab\)

\(5 * x = 5^{2} + x + 5x = 37\)

\(25 + 6x = 37 \implies 6x = 12\)

\(x = 2\)

Explanation

No. of red = 10, No of green = 8

Total = 18

P(RG) = P(R) x P(G)

= \(\frac{10}{18} \times \frac{8}{17} = \frac{40}{152}\)

P(GR) = P(G) x P(R)

= \(\frac{8}{18} \times \frac{10}{17} = \frac{40}{153}\)

P(one red and one green)

= \(\frac{40}{153} + \frac{40}{153} = \frac{80}{153}\)

(b) P = 20% = \(\frac{20}{100}\) = 0.2

q = 1 - P = 1 - 0.2 = 0.8

n = 12

P(2 \(\leq x \leq 5\)) = P(5) - P(2)

= \(^{12}C_5(0.5)^5 (0.8)^7 - ^{12}C_2(0.2)^2(0.8)^{10}\)

= 792 x (0.2)\(^5\) (0.8)\(^7\) - 65(0.2)\(^2\)(0.8)\(^10\)

= 0.04892 - 0.2791

= 0.2302

Correct Answer: A. (1, 2)

Explanation

\(4(2^{x^2}) = 8^{x} \equiv (2^{2})(2^{x^2}) = (2^{3})^{x}\)

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

Comparing bases, we have

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

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

\(x(x - 2) - 1(x - 2) = 0\)

\((x - 1) = 0\) or \((x - 2) = 0\)

\(x = \text{1 or 2}\)

Correct Answer: C. 240

Explanation

\(^{6}C_{4}(1)^{6-4}(-2x)^{4}\) = \(15\times1\times16x^{4} = 240x^{4}\)

The coefficient of \(x^{4}\)= 240

Correct Answer: C. 18

Explanation

Since Ozo is a boy and must be in the sub-committee, we have 2 spaces for the boys and 2 for the girls.

= \(^{4}C_{2} \times ^{3}C_{2} \)

= \(\frac{4!}{(4 - 2)! 2!} \times \frac{3!}{(3 - 2)! 2!} = 6 \times 3 = 18\)

Correct Answer: D. \(\sqrt{29}\)

Explanation

\(p = \begin{pmatrix} 2 \\ -2 \end{pmatrix} , q = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)

\(\frac{1}{2}p = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\)

\(q - \frac{1}{2}p = \begin{pmatrix} 3 \\ 4 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}\)

\(|q - \frac{1}{2}p| = \sqrt{2^{2} + 5^{2}} = \sqrt{29}\)

Correct Answer: C. \(3\)

Explanation

\(\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)

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

= \(\frac{6}{3 - 1} \)

= 3

Correct Answer: B. 5

Explanation

\((m_{1} v_{1} + m_{2} v_{2}) = (m_{1} + m_{2})v\) (Inelastic momentum)

\(8x + (5 \times 2) = (8 + 5) \times 3.85\)

\(8x + 10 = 13 \times 3.85 = 50.05\)

\(8x = 50.05 - 10 = 40.05\)

\(x \approxeq 5 m/s\)

Correct Answer: D. 36

Explanation

The nth term of a GP is given by: \(T_{n} = ar^{n-1}\).

\(T_{3} = ar^{3-1} = ar^{2} = 81\).......(1)

\(T_{7} = ar^{7-1} = ar^{6} = 16 \) ...... (2)

Dividing (2) by (1), we have \(r^{4} = \frac{16}{81} = (\frac{2}{3})^{4} \implies r = \frac{2}{3}\)

Putting \(r = \frac{2}{3}\) in equation (1), we have \(81 = a \times (\frac{2}{3}\)^{2} = a \times \frac{4}{9} \implies a = \frac{729}{4}\)

\(T_{5} = ar^{5-1} = ar^{4} = \frac{729}{4} \times (\frac{2}{3})^{4}\)

= \(\frac{729}{4} \times \frac{16}{81} = 36\)