Correct Answer: C. 4.5

Explanation

\(XY = \sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}}\) is the distance between a point \(X(x_{1}, y_{1})\) and \(Y(x_{2}, y_{2})\).

\(XY = \sqrt{(3 - 5)^{2} + (5 - 1)^{2}} = \sqrt{20}\)

= \(2\sqrt{5} = 4. 467 \approxeq 4.5\)

Correct Answer: B. 12.6 years

Explanation

\(Mean = \frac{\text{sum of ages}}{\text{no of pupils}}\)

Let the sum of the ages of the 15 pupils be x and the sum of the 16 pupils be y.

\(\frac{x}{15} = 14.2 \implies x = 14.2 \times 15 = 213\)

\(\frac{y}{16} = 14.1 \implies y = 14.1 \times 16 = 225.6\)

\(\text{Age of new pupil} = 225.6 - 213 = 12.6\)

Correct Answer: B. \(4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})\)

Explanation

\(y = 2(2x + \sqrt{x})^{2}\)

Let \(u = 2x + \sqrt{x}\)

\(y = 2u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 4u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2 + \frac{1}{2\sqrt{x}}\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

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

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

Correct Answer: A. 2

Explanation

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

\(T_{3} = ar^{3 - 1} = ar^{2} = 10 .... (1)\)

\(T_{6} = ar^{6 - 1} = ar^{5} = 80 .....(2)\)

Divide (2) by (1)

\(r^{5 - 2} = r^{3} = 8 \)

\(r = \sqrt[3]{8} = 2\)

Explanation

(b) Median = \(L_{1} + \frac{(\frac{N}{2} - \sum f_{1}) \times c}{f_{med}}\)

= \(54.5 + \frac{(\frac{70}{2} - 31) \times 5}{23}\)

= \(54.5 + 0.9 = 55.4\)

Correct Answer: A. 70.6°

Explanation

\(\tan \theta = \frac{m_{1} - m_{2}}{1 - m_{1}m_{2}}\)

\(m_{1} = \text{slope of 1st line } 4y = 3x + 5 \implies y = \frac{3}{4}x + \frac{5}{4}\)

\(m_{1} = \frac{3}{4}\)

\(m_{2} = \text{slope of 2nd line} 3y = 1 - 2x \implies y = \frac{1}{3} - \frac{2}{3}x\)

\(m_{2} = -\frac{2}{3}\)

\(\tan \theta = \frac{\frac{3}{4} - (-\frac{2}{3})}{1 - ((\frac{3}{4})(-\frac{2}{3}))} = \frac{\frac{17}{12}}{\frac{1}{2}}\)

\(\tan \theta = \frac{17}{6}\)

\(\theta \approxeq 70.6°\)

Explanation

(a) \((2x - 1) - p(x^{2} + 2) = 0\)

\(2x - 1 - px^{2} - 2p = 0\)

\(px^{2} - 2x + (2p + 1) = 0\)

For real roots, \(b^{2} - 4ac \geq 0\)

\(2^{2} - 4(p)(2p + 1) \geq 0\)

\(4 - 8p^{2} - 4p \geq 0 \implies 2p^{2} + p - 1 \leq 0\)

(b) \(2p^{2} - p + 2p - 1 \leq 0 \implies p(2p - 1) + 1(2p - 1) \leq 0\)

\((p + 1)(2p - 1) \leq 0 \)

\(2p - 1 \leq 0 \implies p \leq \frac{1}{2}\)

\(p + 1 \leq 0 \implies p \leq -1\)

Check : For \(p \leq 1 , \text{let p = -2}\)

\(2(-2)^{2} + (-2) - 1 = 8 - 2 - 1 = 5 \)

For \(p \leq \frac{1}{2}, \text{let p = 0}\)

\(2(0^{2}) + 0 - 1 = -1 \leq 0\)

\(\therefore -1 \leq p \leq \frac{1}{2}\).

Correct Answer: B. 4

Explanation

\(\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = 3(x - 2) - 2x = -2\)

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

\(x = -2 + 6 = 4\)

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

Explanation

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

\(h(a) = a^{3} - \frac{1}{a^{3}}\)

\(h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}\)

\(h(a) - h(\frac{1}{a}) = (a^{3} - \frac{1}{a^{3}}) - (\frac{1}{a^{3}} - a^{3}) = 2a^{3} - \frac{2}{a^{3}}\)

Correct Answer: C. \(\frac{1120}{243}\)

Explanation

\(^{10}C_{7 - 1} (2^{10 - 6}) (\frac{-1}{3})^{6}\)

\(\frac{10!}{(10 - 6)! 6!} \times 16 \times \frac{1}{243} \)

= \(210 \times 16 \times \frac{1}{729} \)

= \(\frac{1120}{243}\)