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

Explanation

\(y = \frac{x}{x + 1}\)

Using quotient rule because the function is of the form \(\frac{u(x)}{v(x)}\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{v\frac{\mathrm d u}{\mathrm d x} - u\frac{\mathrm d v}{\mathrm d x}}{v^{2}}\)

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

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

Correct Answer: C. -4

Explanation

\(f : x \to x^{2} - x - 6\); \(g : x \to x - 1\)

\(g(3) = 3 - 1 = 2\)

\(f(g(3)) = f(2) = 2^{2} - 2 - 6 = 4 - 2 - 6 = -4\)

Correct Answer: A. 28

Explanation

\(f(4) = 3(4) - 2 = 12 - 2 = 10\)

\(f(-3) = 5(-3) - 3 = -15 - 3 = -18\)

\(f(4) - f(-3) = 10 - (-18) = 28\)

Explanation

\(2x^{2} - 7x + 4 = 0\)

\(\alpha + \beta = \frac{7}{2}\)

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

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

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

= \(\frac{(\frac{7}{2})^{2} - 2(2)}{2}\)

= \(\frac{(\frac{49}{4} - \frac{16}{4})}{2}\)

= \(\frac{33}{8}\)

\(\frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1\)

\(\therefore Equation : x^{2} - \frac{33}{8}x + 1 = 0\)

Correct Answer: D. 90°

Explanation

a = i + 5j and b = 5i - j

cos\(\theta\) = \(\frac{a.b}{|a|.|b|}\)

= \(\frac{(1 \times 5) + (5x - 1)}{(\sqrt{1^2 + 5^2}) (5^2 + (-1))^2}\)

= \(\frac{5 - 5}{\sqrt{26}\times \sqrt{26}}\) = 0

x = cos\(^{-1}\)(0), x = 90\(^o\)

Explanation

(a) The formula for the first principles method of calculating derivatives is

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

\(f(x) = (2x + 3)^{2} = 4x^{2} + 12x + 9\)

\(f(x + \Delta x) = 4(x + \Delta x)^{2} + 12(x + \Delta x) + 9\)

= \(4x^{2} + 8x \Delta x + 4 (\Delta x)^{2} + 12x + 12\Delta x + 9\)

\(\lim \limits_{\Delta x \to 0} \frac{(4x^{2} + 8x \Delta x + 4(\Delta x)^{2} + 12x + 12\Delta x + 9) - (4x^{2} + 12x + 9)}{\Delta x}\)

= \(\lim \limits_{\Delta x \to 0} \frac{4(\Delta x)^{2} + (8x + 12)\Delta x}{\Delta x}\)

= \(\lim \limits_{\Delta x \to 0} 4 \Delta x + (8x + 12)\)

\(\frac{\mathrm d y}{\mathrm d x} = 8x + 12\)

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

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

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

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

= \([\frac{x^{2}}{2} - x - \frac{2}{x}]_{1} ^{2} \)

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

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

= \(-1 - (-\frac{5}{2}) = \frac{3}{2}\)

Correct Answer: B. \(x^{2} + y^{2} + 6x + 16y - 23 = 0\)

Explanation

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

where (a, b) and r are the coordinates of the centre and radius respectively.

Given : \((a, b) = (-3, -8); r = 4\sqrt{6}\)

\((x - (-3))^{2} + (y - (-8))^{2} = (4\sqrt{6})^{2}\)

\(x^{2} + 6x + 9 + y^{2} + 16y + 64 = 96\)

\(x^{2} + y^{2} + 6x + 16y + 9 + 64 - 96 = 0\)

\(\implies x^{2} + y^{2} + 6x + 16y - 23 = 0\)

Correct Answer: D. \(\sqrt{26}\)

Explanation

\(r = 2i - j , s = 3i + 5j , t = 6i - 2j\)

\(2r + s - t = 2(2i - j) + (3i + 5j) - (6i - 2j) = i + 5j\)

\(|2r + s - t| = \sqrt{1^{2} + 5^{2}} = \sqrt{26}\)

Correct Answer: A. \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\)

Explanation

\(F = \begin{pmatrix} F_{x} \\ F_{y} \end{pmatrix} = \begin{pmatrix} F\cos \theta \\ F\sin \theta \end{pmatrix}\)

\((14N, 240°) = \begin{pmatrix} 14\cos 240 \\ 14\sin 240 \end{pmatrix}\)

= \(\begin{pmatrix} 14 \times -0.5 \\ 14 \times \frac{-\sqrt{3}}{2} \end{pmatrix}\)

= \(\begin{pmatrix} -7 \\ -7\sqrt{3} \end{pmatrix}\)

Explanation

The weight 50N acts at G, the centre of the beam. Given that the system is in equilibrium, therefore taking moments about the pivot, we have 25 x T = 50 x 10 x 45 and when simplified and solved for T

T = \(\frac{1500}{25}\) = 56N