Explanation

Area, \(A = 3x^{2}\)

Perimeter, \(P = 8x\)

\(\therefore A = \frac{3P^{2}}{64}\)

Let \(P_{1}\) and \(A_{1}\) be the new perimeter and area respectively.

Perimeter increases by 2%

\(\therefore P_{1} = 1.02P\)

\(A_{1} = \frac{3(1.02P)^{2}}{64}\)

\(\frac{A_{1}}{A} = \frac{3(1.02P)^{2}}{64} \times \frac{64}{3P^{2}}\)

= 1.0404

\(A_{1} = 1.0404A = A + 0.0404A\)

Increase in A = 0.0404A.

% increase = \(0.0404 \times 100 = 4.04%\)

Explanation

(a) \(p = (4i - 3j) ; q = (-i + 5j)\)

\(2p + 3q = \begin{pmatrix} 8i - 3i \\ -6j + 15j \end{pmatrix}\)

= \(\begin{pmatrix} 5i \\ 9j \end{pmatrix}\)

Let \(r = k \begin{pmatrix} 5i \\ 9j \end{pmatrix} = \begin{pmatrix} 5ki \\ 9kj \end{pmatrix}\)

\(|r| = \sqrt{(5k)^{2} + (9k)^{2}} = 15\)

\(\sqrt{25k^{2} + 81k^{2}} = 15 \implies 25k^{2} + 81k^{2} = 225\)

\(106k^{2} = 225 \implies k^{2} = \frac{225}{106} = 2.123\)

\(k = \sqrt{2.123} = 1.457\)

\(\therefore r = \begin{pmatrix} 5i \times 1.457 \\ 9j \times 1.457 \end{pmatrix}\)

= \(\begin{pmatrix} 7.285i \\ 13.113j \end{pmatrix}\)

(b)(i)

Resolve the forces along the x and y- axis.

Let \(F_{1} = \begin{pmatrix} 8 \cos 120° \\ 8 \sin 120° \end{pmatrix}\)

\(F_{2} = \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix}\)

\(F_{3} = \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)

\(R = F_{1} + F_{2} + F_{3}\)

= \(\begin{pmatrix} 8 \cos 120° \\ 8 \sin 120° \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)

= \(\begin{pmatrix} -8 \cos 60° \\ 8 \sin 60 \end{pmatrix} + \begin{pmatrix} 6 \cos 90° \\ 6 \sin 90° \end{pmatrix} + \begin{pmatrix} 4 \cos 30° \\ 4 \sin 30° \end{pmatrix}\)

= \(\begin{pmatrix} -8 \times 0.5 \\ 8 \times 0.866 \end{pmatrix} + \begin{pmatrix} 6 \times 0 \\ 6 \times 1 \end{pmatrix} + \begin{pmatrix} 4 \times 0.866 \\ 4 \times 0.5 \end{pmatrix}\)

= \(\begin{pmatrix} -4 + 0 + 3.464 \\ 6.928 + 6 + 2 \end{pmatrix}\)

= \(\begin{pmatrix} -0.536 \\ 14.928 \end{pmatrix}\)

\(|R| = \sqrt{(-0.536)^{2} + (14.928)^{2}} = \sqrt{0.2873 + 222.8452}\)

= \(\sqrt{223.1325} = 14.94N\)

(ii)Direction \(\theta\)

\(\tan \theta = \frac{14.928}{-0.536} = -27.851\)

\(\theta = -87.9° = 92.1°\)

The direction is N 02.1° W or a bearing of 267.9°.

Explanation

Rank coefficient = I = \(\frac{6\sum d^2}{n(n^2 - 1)}\)

= 1 - \(\frac{6(141)}{10(10^2 - 1)}\)

= 1 - \(\frac{6(141)}{10 \times 99}\)

= 1 - \(\frac{846}{990}\)

= 1 - 0.85

= 0.15

(ii)

Rank coefficient = 1 - \(\frac{6(78)}{10(10^2 - 1)}\)

= 1 - \(\frac{468}{990}\)

= 1 - 0.472

= 0.53

(iii)

Rank coefficient = 1 - \(\frac{6(78)}{10(10^2 - 1)}\)

= 0.53

(b) Examiner II and III

Correct Answer: B. \(\frac{1}{2}\)

Explanation

\(T_{n} = ar^{n - 1}\)

\(T_{4} = ar^{4 - 1} = ar^{3} = 192\)

\(T_{9} = ar^{9 - 1} = ar^{8} = 6\)

Dividing \(T_{9}\) by \(T_{4}\),

\(r^{8 - 3} = \frac{6}{192}\)

\(r^{5} = \frac{1}{32} = (\frac{1}{2})^{5}\)

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

Explanation

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

Let an increment in x = \(\Delta x\) and an increment in y = \(\Delta y\).

Then, \(y + \Delta y = (x + \Delta x)^{3} + 2(x + \Delta x)\)

\(y + \Delta y = x^{3} + 3x^{2} \Delta x + 3x (\Delta x)^{2} + (\Delta x)^{3} + 2x + 2 \Delta x ... (2)\)

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

\(\frac{\Delta y}{\Delta x} = 3x^{2} + 3x \Delta x + (\Delta x)^{2} + 2\)

\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_ {\Delta x \to 0} \frac{\Delta y}{\Delta x}\)

\(\frac{\mathrm d y}{\mathrm d x} = \lim \limits_{\Delta x \to 0} (3x^{2} + 3x \Delta x + (\Delta x)^{2} + 2 = 3x^{2} + 2\)

Explanation

\(\frac{dV}{dt}=3\frac{1}{2}cm^3s^{-1}=3.5cm^3s^{-1}\)

L = 6cm

\(\frac{dL}{dt}=?\)

\(V = L^3\)

\(\frac{dV}{dL}=3L^2\)

\(\frac{dL}{dt}=(\frac{dV}{dL})^{-1}\times \frac{dV}{dt}\)

\(\frac{dL}{dt}=(3L^2)^{-1}\times 3.5\)

At L= 6cm

\(\frac{dL}{dt}=(3(6)^2)^{-1}\times 3.5\)

\(\frac{dL}{dt} =(108)^{-1}\times 3.5\)

\(\frac{dL}{dt}=\frac{1}{108}\times 3.5\)

\(\therefore \frac{dL}{dt}0.032cms^{-1}\)

Correct Answer: A. 90720

Explanation

\((x)^{n}(x^{-1})^{8 - n} = x^{0}\)

\(x^{n} . x^{-8 + n} = x^{0} \implies n - 8 + n = 0\)

\(2n - 8 = 0 \implies n = 4\)

Constant term = \(^{8}C_{4}(2^{4})(3^{4}) = 70 \times 16 \times 81\)

= \(90720\)

Correct Answer: A. 2

Explanation

Mean (\(\bar{x}\)) = \(\frac{1+2+3+4+5+5+6+7+8+9}{10} = \frac{50}{10} = 5\)

\(MD = \frac{\sum |x - \bar{x}|}{n} = \frac{20}{10} = 2\)

Correct Answer: D. \(-15(7i + \sqrt{3}j)\)

Explanation

\(F = F\cos\theta + F\sin\theta\)

\(\implies 90N = 90\cos 120° + 90\sin 120°\)

\(120N = 120 \cos 240° + 120 \sin 240°\)

\(R = F_{1} + F_{2} \)

= \((90 \cos 120 + 120 \cos 240)i + (90\sin 120 + 120 \sin 240)j\)

= \(90(-0.5) + 120(-0.5))i + (90(\frac{\sqrt{3}}{2}) + (120(-\frac{\sqrt{3}}{2}))j\)

= \(-105i - 15\sqrt{3}j = -15(7i + \sqrt{3}j)\)

Correct Answer: C. 1 or 11

Explanation

\(d = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\)

\(13 = \sqrt{(x - 6)^{2} + (7 - 19)^{2}}\)

\(13^{2} = x^{2} - 12x + 36 + 144\)

\(169 = x^{2} - 12x + 180\)

\(x^{2} - 12x + 180 - 169 = 0 \implies x^{2} - 12x + 11 = 0\)

\((x - 1)(x - 11) = 0 \implies x = \text{1 or 11}\)