Correct Answer: B. 3

Explanation

\(\frac{^{n}P_{r}}{^{n}C_{r}} = \frac{\frac{n!}{(n - r)!}}{\frac{n!}{(n - r)! r!}} = \frac{90}{15} = 6\)

\(\frac{n!}{(n - r)!} \times \frac{(n - r)! r!}{n!} = r! = 6\)

\(r = 3\)

Explanation

(a) \(u = x - 2 \implies x = u + 2\)

\(\frac{x^{3} + 5}{(x - 2)^{4}}\) becomes \(\frac{(u + 2)^{3} + 5}{u^{4}}\)

= \(\frac{u^{3} + 6u^{2} + 12u + 8 + 5}{u^{4}}\)

= \(\frac{u^{3} + 6u^{2} + 12u + 13}{u^{4}}\)

= \(\frac{1}{u}\) + \(\frac{6}{u^{2}}\) + \(\frac{12}{u^{3}}\) + \(\frac{13}{u^{4}}\)

(b) \(\frac{x^{3} + 5}{(x - 2)^{4}}\) = \(\frac{1}{x - 2}\) + \(\frac{6}{(x - 2)^{2}}\) + \(\frac{12}{(x - 2)^{3}}\) + \(\frac{13}{(x - 2)^{4}}\)

Explanation

(a) \(r = \alpha i + \beta j ; s = 2i - j\)

\(|r| = \sqrt{\alpha^{2} + \beta^{2}} ; |s| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}\)

\(|r| = 3|s| = 3\sqrt{5}\)

\(\implies \alpha^{2} + \beta^{2} = (3\sqrt{5})^{2} = 45 .... (1)\)

\(m = 3i + 2j ; n = i + j\)

\((m - n) = (3i + 2j) - (i + j) = 2i + j\)

\(r\) is parallel to \((m - n)\), hence, they both have the same gradient.

\(\frac{\beta}{\alpha} = \frac{1}{2} \implies \alpha = 2\beta\)

Put \(\alpha = 2\beta\) in (1),

\((2\beta)^{2} + \beta^{2} = 45 \implies 5\beta^{2} = 45\)

\(\beta^{2} = 9 \implies \beta = 3\)

\(\alpha = 2\beta = 2(3) = 6\)

\(r = 6i + 3j\).

(b) \(r = 6i + 3j ; s = 2i - j\)

\((r - s) = (6i + 3j) - (2i - j) = 4i + 4j\)

\(|r - s| = \sqrt{4^{2} + 4^{2}} = \sqrt{32} = 4\sqrt{2}\)

\(Direction : \tan \theta = \frac{4}{4} = 1\)

\(\theta = 45°\)

(r - s) makes an angle of 45° with the x- axis.

Correct Answer: C. \(1\)

Explanation

\(y = \frac{x - 1}{2x + 1}, x \neq -\frac{1}{2}\)

Using the quotient rule, we have

\(\frac{\mathrm d y}{\mathrm d x} = \frac{(2x + 1). 1 - (x - 1). 2}{(2x - 1)^{2}}\)

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

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

At (1, 0), \(\frac{\mathrm d y}{\mathrm d x} = \frac{3}{(2(1) + 1)^{2}} = \frac{3}{9} = \frac{1}{3}\)

Equation : \(\frac{y - 0}{x - 1} = \frac{1}{3}\)

\(y = \frac{1}{3}(x - 1)\)

Correct Answer: A. 80.76°

Explanation

(b) If we let r be the retardation, then \(\frac{s}{t}\) : r = 3 : 4

when simplified, r = \(\frac{10}{3}\)

Velocity after 4 seconds = 20 + \(\frac{5}{2}\) x 4 = 30ms\(^{-1}\)

So that \(\frac{5}{2}\) : \(\frac{30}{t}\) = \(\frac{3}{4}\)

t = 9 seconds

(c) The total distance of the journey = {[\(\frac{1}{2}\) (20 + 30)4] + (30 x 8) + \(\frac{1}{2}\)(30 x 9)}

= 100 + 240 + 135 = 475 metres

Correct Answer: C. \(\frac{3}{5}\)

Explanation

Cos (x + y)

= cos x cos y - sin x sin y

cos (x + y) = \(\frac{4}{5} \times \frac{24}{25} - \frac{3}{5} \times \frac{7}{25}\)

= \(\frac{96}{125} - \frac{21}{125} = \frac{96 - 21}{125}\)

= \(\frac{75}{125}\)

= \(\frac{3}{5}\)

Explanation

(a)

Let T be the tension in the rope and W the weight of the bucket of water.

T - W = net force = ma

\(T - mg = ma\)

\(150 - 80 = 8a\)

\(a = \frac{70}{8} = 8.75 m/s^{2}\).

(b) m = 12kg ; u = 15 m/s.

v = 25 m/s ; s = 50m ; a = ?

\(v^{2} = u^{2} + 2as\)

\(625 = 225 + 100a \implies 100a = 400\)

\(a = 4m/s^{2}\)

\(t = ?\)

\(s = ut + \frac{1}{2} at^{2}\)

\(50 = 15t + \frac{1}{2}(4t^{2})\)

\(50 = 15t + 2t^{2} \implies 2t^{2} + 15t - 50 = 0\)

\(2t^{2} + 20t - 5t - 50 = 0\)

\(2t(t + 10) - 5(t + 10) = 0 \implies (t + 10)(2t - 5) = 0\)

\(t = \frac{5}{2}s \text{or t = -10s}\)

Since time cannot be negative, t = 2.5 seconds.

\(F = m\frac{(v - u)}{t}\)

= \(12(\frac{25 - 15}{2.5})\)

= \(48N\)

\(2as = v^{2} - u^{2}\)

\(2as = 35^{2} - 15^{2}\)

\(2(4)(s) = 1000\)

\(s = 125m\)

Explanation

(a) \(\frac{1}{1 - \cos \theta} + \frac{1}{1 + \cos \theta}\)

= \(\frac{(1 + \cos \theta) + (1 - \cos \theta)}{1 - \cos^{2} \theta}\)

= \(\frac{2}{1 - \cos^{2} \theta}\)

= \(\frac{2}{\sin^{2} \theta}\)

(b) \(y = x^{2} (x - 3)\)

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

= \(x^{2} + 2x^{2} - 6x = 3x^{2} - 6x\)

At stationary point, \(\frac{\mathrm d y}{\mathrm d x} = 0\)

\(3x^{2} - 6x = 0\)

\(3x(x - 2) = 0 \implies x = \text{0 or 2}\)

If x = 0, \(y = 0^{2}(0 - 3) = 0\)

If x = 2, \(y = 2^{2}(2 - 3) = -4\)

The points are (0, 0) and (2, -4).

Gradient = \(\frac{-4 - 0}{2 - 0} = \frac{-4}{2} = -2\)

Equation : \(\frac{y - (-4)}{x - 2} = -2\)

\(\frac{y + 4}{x - 2} = -2 \implies y + 4 = -2(x - 2)\)

\(y + 4 = 4 - 2x \implies y + 2x = 0\)

\(y = -2x\)

Distance : \(\sqrt{(2 - 0)^{2} + (-4 - 0)^{2}}\)

= \(\sqrt{4 + 16} = \sqrt{20}\)

= \(2\sqrt{5} units\)