Correct Answer: A. \(2(x^{2} - 1)\)

Explanation

\(f(x) = 2x^{2} - 3; g(x) = x + 1\)

\(g o f(x) = g (2x^{2} - 3)\)

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

Explanation

\(xy^{2} + x^{2} y = 4xy\)

Differentiating with respect to x,

\(2xy \frac{\mathrm d y}{\mathrm d x} + y^{2} + x^{2} \frac{\mathrm d y}{\mathrm d x} + 2xy = 4x \frac{\mathrm d y}{\mathrm d x} + 4y\)

\((2xy + x^{2} - 4x) \frac{\mathrm d y}{\mathrm d x} = 4y - y^{2} - 2xy\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{4y^{2} - y^{2} - 2xy}{2xy + x^{2} - 4x}\)

At (1, 3), Gradient = \(\frac{4(3) - (3)^{2} - 2(1)(3)}{2(1)(3) + 1^{2} - 4(1)}\)

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

= -1

Correct Answer: B. 41

Explanation

Arranging the scores in ascending order, we have: 2, 5, 12, 17, 21, 29, 33, 41, 43, 45.

The upper quartile = 41.

Correct Answer: D. \(2 - \sqrt{3}\)

Explanation

\(\tan 15 = \tan (60 - 45)\)

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

\(\tan (60 - 45) = \frac{\tan 60 - \tan 45}{1 + \tan 60 \tan 45}\)

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

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

Rationalizing by multiplying denominator and numerator by \(1 - \sqrt{3}\),

\(\tan 15 = 2 - \sqrt{3}\)

Correct Answer: D. 6

Explanation

\(y = x^{2} - 6x + 11\)

\( y = a(x - h)^{2} + k\)

\(a(x - h)^{2} + k = a(x^{2} - 2hx + h^{2}) + k\)

\(ax^{2} - 2ahx + ah^{2} + k = x^{2} - 6x + 11\)

Comparing, we have

\(a = 1\)

\(2ah = 6 \implies 2h = 6; h = 3\)

\(ah^{2} + k = 11 \implies (1 \times 3^{2}) + k = 11\)

\(9 + k = 11 \implies k = 2\)

\(\therefore a + h + k = 1 + 3 + 2 = 6\)

Correct Answer: D. \(48\pi cm^{2}s^{-1}\)

Explanation

Surface area of sphere, \( A = 4\pi r^{2}\)

\(\frac{\mathrm d A}{\mathrm d r} = 8\pi r\)

The rate of change of radius with time \(\frac{\mathrm d r}{\mathrm d t} = 3cm s^{-1}\)

\(\frac{\mathrm d A}{\mathrm d t} = (\frac{\mathrm d A}{\mathrm d r})(\frac{\mathrm d r}{\mathrm d t})\)

= \(8\pi \times 2cm \times 3cm s^{-1} = 48\pi cm^{2}s^{-1}\)

Correct Answer: B. 6.5

Explanation

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

Using the chain rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}\)

Let \(u = x^{2} + 3\) so that \(y = u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 2u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu\)

But \(u = x^{2} + 3\),

\(\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)\)

At \(x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)\)

= \(2 \times \frac{13}{4} = \frac{13}{2} = 6.5\)

Correct Answer: B. 20

Explanation

Let the sum of the men's ages be f, so that

\(\frac{f}{n} = 50 .... (1)\)

Also, \(\frac{f - (55 + 63)}{n - 2} = 50 - 1 = 49 .... (2)\)

From (1), \(f = 50n\)

From (2), \(f - 118 = 49(n - 2) = 49n - 98\)

\(f = 49n - 98 + 118 = 49n + 20\)

\(\therefore f = 50n = 49n + 20\)

\(50n - 49n = n = 20\)

Correct Answer: B. \(\frac{3}{8}\)

Explanation

Let the probability of getting a head = p = \(\frac{1}{2}\) and that of tail = q = \(\frac{1}{2}\)

\((p + q)^{3} = p^{3} + 3p^{2}q + 3pq^{2} + q^{3}\)

In the equation above, \(p^{3}\) and \(q^{3}\) are the probabilities of 3 heads and 3 tails respectively

while, \(p^{2}q\) and \(pq^{2}\) are the probabilities of 2 heads and one tail and 2 tails and one head respectively.

Probability of exactly 2 heads = \(3p^{2}q = 3(\frac{1}{2})^{2}(\frac{1}{2})\)

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

Explanation

Simplify \(^{n + 1}C_{4} - ^{n - 1}C_{4}\)

= \(\frac{(n + 1)!}{4! (n - 3)!} - \frac{(n - 1)!}{4! (n - 5)!}\)

= \(\frac{(n + 1)(n)(n - 1)(n - 2)(n - 3)!}{4! (n - 3)!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)(n - 5)!}{4! (n - 5)!}\)

= \(\frac{(n + 1)(n)(n - 1)(n - 2)}{4!} - \frac{(n - 1)(n - 2)(n - 3)(n - 4)}{4!}\)

= \(\frac{(n - 1)(n - 2) [n(n + 1) - (n - 3)(n - 4)]}{4!}\)

= \(\frac{(n - 1)(n - 2) [n^{2} + n - n^{2} + 7n - 12]}{24}\)

= \(\frac{(n - 1)(n - 2)[8n - 12]}{24}\)

= \(\frac{(n - 1)(n - 2)(2n - 3)}{6}\)