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

Explanation

\(\frac{d}{dx}(\frac{x }{x + 1}\)) = \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\)

u = x, \(\frac{du}{dx}\) = 1, v = x + 1, \(\frac{dv}{dx}\) = 1

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

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

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

Correct Answer: C. \(\frac{3}{4}\)

Explanation

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

\(\frac{\mathrm d y}{\mathrm d x} = 4x - 3 = 0\) (At turning point)

\(4x = 3 \implies x = \frac{3}{4}\)

Correct Answer: B. 1.0

Explanation

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

\(\int_{1}^{2} (x - x^{-2}) \mathrm {d} x = (\frac{x^{2}}{2} + \frac{1}{x})|_{1}^{2}\)

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

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

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

= \(1.0\)

Explanation

\(< PRQ = 90°\)

Gradient of PR \(\times\) Gradient of QR = -1.

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

\(\frac{2}{(k + 1)(k + 4)} = -1\)

\(-2 = k^{2} + 5k + 4 \implies k^{2} + 5k + 4 + 2 = 0\)

\(k^{2} + 5k + 6 = 0\)

\(k^{2} + 3k + 2k + 6 = 0\)

\(k(k + 3) + 2(k + 3) = 0 \implies (k + 3)(k + 2) = 0\)

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

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) = \(p(Ago) \times p(Sulley) \times p(Musa')\)

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

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

Explanation

\(Mean, (\bar {x}) = A + \frac{\sum fd}{\sum f}\)

= \(24.5 + \frac{-300}{120}\)

= \(24.5 - 2.5\)

= 22.0

Correct Answer: A. 59°

Explanation

Cos \(\theta\) = \(\frac{4(12) + (-5)(3)}{\sqrt{12^2 + (-5)^2}\sqrt{4^2 + 3^2}}\)

Cos\(\theta\) = \(\frac{48 - 15}{(\sqrt{189})(\sqrt{25})}\)

\(\theta = cos^{-1}\) (\(\frac{33}{61.28}\))

\(\theta\) = cos\(^{-1}\)0.5367

\(\theta\) = 59\(^o\)

Correct Answer: C. \(\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\)

Explanation

\(u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}\)

\(a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}; t = 3 secs\)

\(v = u + at \implies v = \begin{pmatrix} -5 \\ 3 \end{pmatrix} + \begin{pmatrix} 3 \\ -4 \end{pmatrix} \times 3\)

= \(\begin{pmatrix} -5 \\ 3 \end{pmatrix} + \begin{pmatrix} 9 \\ -12 \end{pmatrix} = \begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}\)\)

Correct Answer: B. -8.0

Explanation

f(x) = 4x\(^3\) + px\(^2\) + 7x - 23

If f(x) is divided by (2x -5), the remainder is f(\(\frac{5}{2}\))

f\(\frac{5}{2}\) = 4\(\frac{5}{2}\)\(^3\) + p\(\frac{5}{2}\)\(^2\) + 7\(\frac{5}{2}\) - 23

Hence;

= \(\frac{125}{8}\) + \(\frac{25}{4}\) p + \(\frac{35}{2}\) - 23

28 = 250 + 250p + 70 - 92

25p = -200; p = -8.0

Correct Answer: D. \(\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}\)

Explanation

\(P = \begin{pmatrix} -2 & 1 \\ 3 & 4 \end{pmatrix}; Q = \begin{pmatrix} 5 & -3 \\ 2 & -1 \end{pmatrix}\)

= \(PQ = \begin{pmatrix} -10+2 & 6-1 \\ 15+8 & -9-4 \end{pmatrix}\)

= \(\begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix}\)

\(QP = \begin{pmatrix} -10-9 & 5-12 \\ -4-3 & 2-4 \end{pmatrix}\)

= \(\begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}\)

\(PQ - QP = \begin{pmatrix} -8 & 5 \\ 23 & -13 \end{pmatrix} - \begin{pmatrix} -19 & -7 \\ -7 & -2 \end{pmatrix}\)

= \(\begin{pmatrix} 11 & 12 \\ 30 & -11 \end{pmatrix}\)