Correct Answer: D. \(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\)

Explanation

\(\frac{x^{3} + 5x + 1}{x^{3}} \equiv 1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}\)

\(\equiv \int (1 + \frac{5}{x^{2}} + \frac{1}{x^{3}}) \mathrm {d} x = \int (1 + 5x^{-2} + x^{-3}) \mathrm {d} x\)

= \((x - 5x^{-1} - \frac{1}{2}x^{-2} + c)\)

= \(x - \frac{5}{x} - \frac{1}{2x^{2}} + c\).

Correct Answer: A. 13N

Explanation

Given a force F, the horizontal component = \(F\cos \theta\)

R = \(-50\cos 30 + 80\cos 45\)

= \(-50(\frac{\sqrt{3}}{2}) + 80(\frac{\sqrt{2}}{2})\)

= \( -25\sqrt{3} + 40\sqrt{2} = -43.30 + 56.67 \)

= \(13.37N \approxeq 13N\)

Explanation

(a) \(p(head) = p = \frac{1}{2}; p(tail) = q = \frac{1}{2}\)

\((p + q)^{8} = p^{8} + 8p^{7} q + 28p^{6} q^{2} + 56p^{5} q^{3} + 70p^{4} q^{4} + 56p^{3} q^{5} + 28p^{2} q^{6} + 8pq^{7} + q^{8}\)

(i) p(exactly 8 heads) = \(p^{8} = (\frac{1}{2})^{8} = \frac{1}{256}\)

(ii) p(5 heads) = \(56p^{5} q^{3} = 56(\frac{1}{2})^{5} (\frac{1}{2})^{3} = \frac{7}{32}\)

p(6 heads) = \(28p^{6} q^{2} = 28(\frac{1}{2})^{6} (\frac{1}{2})^{2} = \frac{7}{64}\)

p(7 heads) = \(8p^{7} q = 8(\frac{1}{2})^{7} (\frac{1}{2}) = \frac{1}{32}\)

p(at least 5 heads) = \(\frac{7}{32} + \frac{7}{64} + \frac{1}{32} + \frac{1}{256} = \frac{93}{256}\)

(iii) p(at most 1 head) = p(0 head) + p(1 head)

p(0 head) = \(q^{8} = (\frac{1}{2})^{8} = \frac{1}{256}\)

p(1 head) = \(8p q^{7} = 8(\frac{1}{2})(\frac{1}{2})^{7} = \frac{8}{256}\)

p(at most 1 head) = \(\frac{1}{256} + \frac{8}{256} = \frac{9}{256}\)

(b) The word SHEEP has 5 letters, two of which are identical.

(i) Hence 4 letters from the word SHEEP can be arranged in \(\frac{^{5} P_{4}}{2!}\) ways.

= \(\frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 60\).

(ii) The two Es are identical. If one of them are taken, there are 4 letters to be arranged in \(^{4} P_{4}\) ways.

= \(\frac{4!}{(4 - 4)!} = 4! = 24\).

Correct Answer: B. 75

Explanation

No. of ways = 3 x 5 x 5 = 75

Correct Answer: A. \(\frac{3}{7}\)

Explanation

\(P(A) = \frac{7}{12}\)

\(P(A \cap B) = \frac{1}{4} = P(A) \times P(B)\) (Independent events)

\(\frac{1}{4} ÷ \frac{7}{12} = \frac{1}{4} \times \frac{12}{7} \)

= \(\frac{3}{7}\)

Correct Answer: C. 1

Explanation

\(\frac{1}{8^{2y - 3y}} = \frac{1}{8^{-y}} = 8^{y}\)

\(8^{y} = 2^{y + 2} \implies (2^{3})^{y} = 2^{y + 2}\)

\(3y = y + 2 \implies 2y = 2\)

\(y = 1\)

Correct Answer: A. 61

Explanation

x - 1 = 2

x = 3

f(2) = (3)\(^3\) + 3(3)\(^2\) + 4(3) - 5

f(2) = 27 + 27 + 12 - 5

= 61

Explanation

(a) \(p = \begin{pmatrix} 3 \\ -1 \end{pmatrix} = 3i - j ; q = \begin{pmatrix} 3 \\ 4 \end{pmatrix} = 3i + 4j\)

Let the angle be \(\theta\),

\(p . q = |p||q| \cos \theta\)

\((3i - j).(3i + 4j) = 9 - 4 = 5\)

\(|3i - j| = \sqrt{3^{2} + (-1)^{2}} = \sqrt{10}\)

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

\(5 = (\sqrt{10})(5) \cos \theta \implies \cos \theta = \frac{\sqrt{10}}{10} = 0.3162\)

\(\theta = \cos^{-1} 0.3162 = 71.567° \approxeq 71.6°\)

(b)

(i) Resolving the forces along the x and y axes, we obtain

Resultant (R) = (10i + 20j) N.

Magnitude = \(\sqrt{10^{2} + 20^{2}} = \sqrt{500}\)

= \(10\sqrt{5} \approxeq 22.4 N\)

(ii) Direction : \(\tan \theta = \frac{20}{10} = 2\)

\(\theta = \tan^{-1} 2 = 63.4°\).

Correct Answer: A. 13%

Explanation

Let the original volume be V with radius r.

\(V = \frac{4}{3}\pi r^{3}\)

45% increased volume = 145%V.

Let the %age increase in radius = m%r

\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)

\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)

\(1.45V = V(\frac{m}{100})^{3}\)

\(\implies 1.45 \times 10^{6} = m^{3}\)

\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)

\(\therefore \text{%age increase =} 113.2 - 100 = 13.2%\)

\(\approxeq 13%\)

Correct Answer: B. 2.8

Explanation

\(\mu = \frac{24+25+28+29+30+32}{6} = \frac{168}{8} = 28\)

\(S.D = \sqrt{\frac{\sum{(x - \mu)^{2}}}{n}} = \sqrt{\frac{46}{6}}\)

= \(\sqrt{7.67} \approxeq 2.8\)