Correct Answer: D. 2x+3y=4

Explanation

\(2x+3y = 6 \implies 3y = 6-2x\)

\(y = \frac{6}{3} - \frac{2x}{3}\)

Parallel lines have the same gradient

\(\therefore\) Gradient of the line = \(\frac{-2}{3}\)

Line passes through (-1,2)

Equation: \(\frac{y-2}{x-(-1)} = \frac{y-2}{x+1} = \frac{-2}{3}\)

\(3y-6 = -2x-2 \implies 3y+2x = -2+6 =4\)

Correct Answer: D. 4

Explanation

\(\sqrt{128} = \sqrt{64\times2} = 8\sqrt{2}\)

\(\sqrt{32} = \sqrt{16\times2} = 4\sqrt{2}\)

Simplifying, we have \(\frac{8\sqrt{2}}{4\sqrt{2} - 2\sqrt{2}} = \frac{8\sqrt{2}}{2\sqrt{2}}\)

= 4

Correct Answer: C. \(\frac{3}{2}\)

Explanation

\(\alpha + \beta = \frac{-b}{a}\)

From the equation, a = 2, b = -3 and c = 4

\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)

Correct Answer: C. (1,4)

Explanation

\(f(x) = 5x^{2} - 4x + 3\)

\(f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1\)

When x = 1, f(x) = y = \(5(1^{2}) - 4(1) + 3 = 4\)

The coordinates are (1,4)

Explanation

(a)

(b) Lower quartile \(Q_{1}\) = 45.6cm

Upper quartile \(Q_{3}\) = 52.0cm

Semi- interquartile range = \(\frac{Q_{3} - Q_{1}}{2}\)

= \(\frac{52 - 45.6}{2}\)

= \(\frac{6.4}{2}\)

= \(3.2cm\)

Correct Answer: B. \(\frac{-7}{8}\)

Explanation

\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha\beta}\)

\(\alpha + \beta = \frac{-b}{a}\); \(\alpha\beta = \frac{c}{a}\)

\(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2(\alpha\beta)\)

From the equation, a = 2, b = -3, c =4

\(\alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2}\)

\(\alpha\beta = \frac{4}{2} = 2\)

\(\alpha^{2} + \beta^{2} = (\frac{3}{2})^{2} - 2(2)) = \frac{9}{4} - 4 = \frac{-7}{4}\)

\(\implies \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\frac{-7}{4}}{2} = \frac{-7}{8}\)

Explanation

\(\int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x\)

Let \(x + 1 = u, x - 1 = u - 2\)

When x = 3, u = 3 + 1 = 4

x = 1, u = 1 + 1 = 2

\(\therefore \int_{1}^{3} (\frac{x - 1}{(x + 1)^{2}}) \mathrm {d} x \equiv \int_{2}^{4} (\frac{u - 2}{u^{2}}) \mathrm {d} u\)

= \(\int_{2}^{4} (\frac{u}{u^{2}} - \frac{2}{u^{2}}) \mathrm {d} u\)

= \(\int_{2}^{4} (\frac{1}{u} - 2u^{-2}) \mathrm {d} u\)

= \([\ln u - \frac{2u^{-1}}{-1}]|_{2}^{4}\)

= \([\ln u + 2u^{-1}]|_{2}^{4}\)

= \([\ln u + \frac{2}{u}]|_{2}^{4}\)

= \((\ln 4 + \frac{2}{4}) - (\ln 2 + \frac{2}{2})\)

= \(\ln 4 - \ln 2 - \frac{1}{2}\)

= \(\ln (\frac{4}{2}) - \frac{1}{2}\)

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

Correct Answer: B. \(\frac{140}{221}\)

Explanation

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

Given \(\sin\) of an angle implies we have the value of the opposite and hypotenuse of the right-angled triangle. We find the adjacent side using Pythagoras' theorem.

\(Adj^{2} = Hyp^{2} - Opp^{2}\)

For triangle with angle x, \(adj = \sqrt{13^{2} - 5^{2}} = \sqrt{144} = 12\)

For triangle with angle y, \(adj = \sqrt{17^{2} - 8^{2}} = \sqrt{225} = 15\)

\(\therefore \cos x = \frac{12}{13}; \cos y = \frac{15}{17}\)

\(\cos(x+y) = (\frac{12}{13}\times\frac{15}{17}) - (\frac{5}{13}\times\frac{8}{17}) = \frac{180}{221} - \frac{40}{221}\)

= \(\frac{140}{221}\)

Correct Answer: C. 14\(ms^{-2}\)

Explanation

\(accl = \frac{\mathrm d V}{\mathrm d t}\)

\(V = 3t^{2} + 2t - 1 \therefore a = \frac{\mathrm d V}{\mathrm d t} = 6t + 2\)

\(a \text{(after 2 seconds)} = (6\times 2) + 2 = 12+2 = 14ms^{-2}\)

Correct Answer: C. 240

Explanation

\(^{6}C_{4}(1)^{6-4}(-2x)^{4}\) = \(15\times1\times16x^{4} = 240x^{4}\)

The coefficient of \(x^{4}\)= 240