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: D. \(\frac{56}{33}\)

Explanation

\(\tan (x + y) = \frac{\tan x + \tan y}{1 - \tan x\tan y}\)

\(\tan x = \frac{5}{12} ; \tan y = \frac{3}{4}\)

\(\tan (x + y) = \frac{\frac{5}{12} + \frac{3}{4}}{1 - (\frac{5}{12} \times \frac{3}{4}})\)

= \(\frac{\frac{14}{12}}{\frac{33}{48}}\)

= \(\frac{56}{33}\)

Correct Answer: D. \(\frac{11}{24}\)

Explanation

n = 3, \(\frac{1}{(n-1)!} - \frac{1}{(n+1)!} = \frac{1}{(3-1)!} - \frac{1}{(3+1)!}\)

= \(\frac{1}{2} - \frac{1}{24} = \frac{12 -1}{24}\)

= \(\frac{11}{24}\)

Correct Answer: B. \(3\)

Explanation

\(\begin{vmatrix} m-2 & m+1 \\ m+4 & m-2 \end{vmatrix} = -27\)

\((m^{2} - 4m + 4) - (m^{2} + 5m + 4) = -27\)

\(-9m = -27 \implies m = 3\)

Explanation

\(p(Ago) = \frac{4}{5} ; p(Sulley) = \frac{3}{4} ; p(Musa) = \frac{2}{3}\)

(a) p(none admitted) = \(p(Ago') \times p(Sulley') \times p(Musa')\)

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

= \(\frac{1}{60}\)

(b) p(Ago and Sulley admitted only) = \(p(Ago) \times p(Sulley) \times p(Musa')\)

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

= \(\frac{1}{5}\)

Correct Answer: D. \(\frac{160}{729}\)

Explanation

\([\frac{1}{3}(2 + x)]^{6} = (\frac{2}{3} + \frac{x}{3})^{6}\)

The coefficient of \(x^{3}\) is

\(^{6}C_{3}(\frac{2}{3})^{3}(\frac{1}{3})^{3}x^{3} = (\frac{6!}{3!3!})(\frac{8}{27})(\frac{1}{27})x^{3}\)

= \(\frac{160}{729}\)

Explanation

(i) \(m * n = m + n + 2\)

Let the identity element be e.

Then \(m * e = m + e + 2 = m\)

\(e = -2\)

(ii) Let N be the inverse of n.

Then \(n * N = e\)

\(n * N = n + N + 2 = -2\)

\(n + N = -4 \implies N = -4 - n\)

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

Explanation

\(T_{n} = ar^{n-1}\) (for a geometric progression)

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

\(T_{6} = ar^{6-1} = ar^{5} = \frac{64}{81}\)

Dividing \(T_{6}\) by \(T_{3}\),

\(\frac{ar^{5}}{ar^{2}} = \frac{\frac{64}{81}}{\frac{8}{3}} \implies r^{3} = \frac{8}{27}\)

\(\therefore r = \sqrt[3]{\frac{8}{27}} = \frac{2}{3}\)

Correct Answer: B. \(\frac{1}{\sqrt{29}}(-2i + 5j)\)

Explanation

The unit vector, \(\hat{n}\) is given by \(\hat{n} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|}\)

= \(\frac{(-2i + 5j)}{\sqrt{(-2)^{2} + 5^{2}}} = \frac{(-2i + 5j)}{\sqrt{29}}\)

= \(\frac{1}{\sqrt{29}}(-2i + 5j)\)

Correct Answer: B. -4

Explanation

Given x = (-1, 5) for the equation \(x^{2} + kx - 5 = 0\)

\(x = -1 \implies x + 1 = 0\); \(x = 5 \implies x - 5 = 0\)

\((x + 1)(x - 5) = 0\), expanding,

\(x^{2} - 5x + x - 5 = 0 \therefore x^{2} - 4x - 5 = 0\)

\(\therefore\) k = -4.