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

Explanation

\((1 + kx)^{8} = ^{8}C_{0}(1^{8})(kx)^{0} + ^{8}C_{1}(1^{7})(kx)^{1} + ...\)

The 5th term = \(^{8}C_{5 - 1}(1^{4})(kx)^{4}\)

= \(^{8}C_{4} (kx)^{4}\)

\(\implies 70k^{4} = \frac{35}{8}\)

\(k^{4} = \frac{\frac{35}{8}}{70}\)

\(k^{4} = \frac{1}{16}\)

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

Correct Answer: D. \(x^{2} + y^{2} + 4x - 6y - 12 = 0\)

Explanation

Equation of a circle with centre coordinates (a, b) : \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Area of circle = \(\pi r^{2} = 25\pi cm^{2} \implies r^{2} = 25 \)

\(\therefore r = 5cm\)

(a, b) = (-2, 3)

Equation: \((x - (-2))^{2} + (y - 3)^{2} = 5^{2}\)

\(x^{2} + 4x + 4 + y^{2} - 6y + 9 = 25 \implies x^{2} + y^{2} + 4x - 6y + 13 - 25 = 0\)

= \(x^{2} + y^{2} + 4x - 6y - 12 = 0\)

Explanation

(a) \(\frac{x^{2}}{4} + y^{2} = 1\)

Differentiating w.r.t x,

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

The gradient of the curve is

\(\frac{\mathrm d y}{\mathrm d x} = -\frac{1}{4(\frac{\sqrt{3}}{2})}\)

= \(-\frac{1}{2\sqrt{3}}\)

Equation of the tangent at \((1, \frac{\sqrt{3}}{2})\) is

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

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

\(1 - x = 2\sqrt{3} y - 3 \implies 2\sqrt{3} y + x - 4 = 0\).

(b) \(\frac{3x + 2}{x^{2} + x - 2}\)

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

\(x(x - 1) + 2(x - 1) \implies x^{2} + x - 2 \equiv (x - 1)(x + 2)\)

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

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

Comparing with the equation given, we have

\(3x + 2 = A(x + 2) + B(x - 1)\)

= \(Ax + 2A + Bx - B\)

\(\implies A + B = 3 ... (1)\)

\(2A - B = 2 ... (2)\)

(1) + (2) : \(3A = 5 \implies A = \frac{5}{3}\)

\(\frac{5}{3} + B = 3 \implies B = 3 - \frac{5}{3} = \frac{4}{3}\)

\(\therefore \frac{3x + 2}{x^{2} + x - 2} = \frac{5}{3(x - 1)} + \frac{4}{3(x + 2)}\).

Correct Answer: D. 12

Explanation

\(Gradient = \frac{y_{1} - y_{2}}{x_{1} - x_{2}} \)

\(\frac{1}{2} = \frac{5 - 9}{4 - x}\)

\(\frac{1}{2} = \frac{-4}{4 - x} \implies -8 = 4 - x\)

\(x = 4 + 8 = 12\)

Correct Answer: D. \(-\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

Explanation

\(\cos 165 = -\cos (180 - 165) = -\cos 15\)

\(\cos 15 = \cos (45 - 30)\)

\(\cos (x - y) = \cos x \cos y + \sin x \sin y\)

\(\cos (45 - 30) = \cos 45 \cos 30 + \sin 45 \sin 30\)

= \((\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2})\)

= \(\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

\(\therefore \cos 165 = -\frac{1}{4}(\sqrt{6} + \sqrt{2})\)

Correct Answer: B. -3i + 2j

Explanation

\(\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}\)

\(\overrightarrow{BA} = - \overrightarrow{AB} = -(5i + 3j)\)

= \(-5i - 3j\)

\(\overrightarrow{BC} = (-5i - 3j) + (2i + 5j)\)

= \(-3i + 2j\)

Correct Answer: B. 2

Explanation

\(\frac{1}{5^{-y}} = 25(5^{4-2y})\)

\(\implies 5^{y} = (5^{2})(5^{4-2y})\)

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

Comparing bases, we have

\(y = 6 - 2y\)

\(3y = 6 \implies y = 2\)

Explanation

(a) \(m \begin{pmatrix} 2 \\ 1 \end{pmatrix} + n \begin{pmatrix} -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ -4 \end{pmatrix}\)

\(2m - n = 5 ...(1)\)

\(m + 2n = -4 ... (2)\)

\((1) \times 2 : 4m - 2n = 10 ... (3)\)

\((2) + (3) : 5m = 6 \implies m = \frac{6}{5}\)

\(2(\frac{6}{5}) - n = 5 \implies n = \frac{12}{5} - 5 = \frac{-13}{5}\)

\(\therefore (m + n) = (\frac{6}{5} + \frac{-13}{5}) = -\frac{7}{5}\).

(b) A(-1, 3) = (-i + 3j)

B(2, -1) = (2i - j)

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\)

\(\overrightarrow{AB} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

= \(\overrightarrow{AB} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)

\(|\overrightarrow{AB}| = \sqrt{3^{2} + (-4)^{2}} = 5\)

Unit vector parallel to AB = \(\frac{1}{5} \begin{pmatrix} 3 \\ -4 \end{pmatrix}\)

\(\overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OC}\)

\(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA}\)

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

= \(\begin{pmatrix} 6 \\ 0 \end{pmatrix}\)

\(|\overrightarrow{AC}| = 6\)

Unit vector parallel to AC = \(\frac{1}{6} \begin{pmatrix} 6 \\ 0 \end{pmatrix}\)

(ii) Let the angle BAC = \(\theta\)

\(\overrightarrow{AB} \cdot \overrightarrow{AC} = |\overrightarrow{AB}||\overrightarrow{AC}| \cos \theta\)

\((3i - 4j) \cdot (6i + 0j) = |3i + 4j||6i + 0j| \cos \theta\)

\(18 = 5 \times 6 \times \cos \theta \implies \cos \theta = \frac{18}{30} = 0.6\)

\(\theta = \cos^{-1} (0.6) = 53.1°\)

Correct Answer: A. 620

Explanation

d = -3 - (-7) = 4

S\(_{20}\) = \(\frac{20}{2}\){2(-7) + (20 - 1) 4}

= 10(- 14 + 76)

= 620

Correct Answer: C. 3.6

Explanation

Mean = \(\frac{15}{5} = 3\)

SD = \(\sqrt{\frac{\sum (x - \bar {x})^{2}}{n}}\)

= \(\sqrt{\frac{66}{5}} \)

= \(\sqrt{13.2}\)

= 3.6