Correct Answer: A. \(\frac{8}{3}\)

Explanation

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

= \(\left. \frac{x^{3}}{3} + x^{2} + x \right |_{-1}^{1}\)

= \((\frac{1^{3}}{3} + 1^{2} + 1) - (\frac{(-1)^{3}}{3} + (-1)^{2} + (-1)) = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}\)

Explanation

\(T_{n} = a + (n - 1) d\) (for an arithmetic progression).

Given: \(T_{2} + T_{5} = 42 \)

\(T_{6} - T_{3} = 12 \)

\(T_{2} = a + d ; T_{5} = a + 4d\)

\(\implies a + d + a + 4d = 42 ; 2a + 5d = 42 ..... (1)\)

(i) \(T_{6} = a + 5d ; T_{3} = a + 2d\)

\(\implies a + 5d - a - 2d = 12 ; 3d = 12\)

\(3d = 12 \implies d = 4\)

(ii) \(2a + 5d = 42 \implies 2a + 5(4) = 42\)

\(2a + 20 = 42 \implies 2a = 22\)

\(a = 11\)

(iii) \(T_{20} = a + 19d\)

= \(11 + 19(4) \)

= \(87\)

Correct Answer: B. \(\pm 2\)

Explanation

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

\(T_{4} = ar^{3} = 2 .... (1)\)

\(T_{6} = ar^{5} = 8 ......(2)\)

Divide (2) by (1),

\(\frac{ar^{5}}{ar^{3}} = \frac{8}{2} \)

\(r^{2} = 4\)

\(r = \pm 2\)

Correct Answer: C. -3

Explanation

Using the remainder theorem, the remainder when a polynomial \(ax^{2} + bx + c\) is divided by \((x - a)\) is equal to \(f(a)\).

\(2x^{3} + 3x^{2} + qx - 1\) divided by \((x + 2)\), the remainder = \(f(-2)\)

\(\implies f(-2) = f(1)\)

\(f(-2) = 2(-2^{3}) + 3(-2^{2}) + q(-2) - 1 = -16 + 12 - 2q - 1 = -5 - 2q\)

\(f(1) = 2(1^{3}) + 3(1^{2}) + q(1) - 1 = 2 + 3 + q - 1 = 4 + q\)

\(4 + q = -5 -2q \implies 4 + 5 = -2q - q = -3q\)

\(q = -3\)

Correct Answer: C. \(2\sqrt{10}\)

Explanation

\(BC = BA + AC\)

Given, \(AB\), then \(BA = - AB\)

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

\(\therefore BC = \begin{pmatrix} -4 \\ -3 \end{pmatrix} + \begin{pmatrix} 2 \\ -3 \end{pmatrix}\)

= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\)

\(|BC| = \sqrt{(-2)^{2} + (-6)^{2}} = \sqrt{40} \)

= \(2\sqrt{10}\)

Correct Answer: D. 0°

Explanation

\(\sin \theta = \tan \theta \implies \frac{\sin \theta}{1} = \frac{\sin \theta}{\cos \theta}\)

Equating, we have

\(\cos \theta = 1 \implies \theta = \cos^{-1} 1\)

= \(0°\)

Correct Answer: A. -49

Explanation

\(g : x \to 2x^{3}\)

\(g(-2) = 2(-2^{3}) = 2(-8) = -16\)

\(f : x \to 3x - 1\)

\(f(-16) = 3(-16) -1 = -48 - 1 = - 49\)

Correct Answer: B. 4.00

Explanation

\(\log_{2}(12x - 10) = 1 + \log_{2}(4x + 3)\)

Recall, \( 1 = \log_{2}2\), so

\(\log_{2}(12x - 10) = \log_{2}2 + \log_{2}(4x + 3)\)

= \(\log_{2}(12x - 10) = \log_{2}(2(4x + 3))\)

\(\implies (12x - 10) = 2(4x + 3) \therefore 12x - 10 = 8x + 6\)

\(12x - 8x = 4x = 6 + 10 = 16 \implies x = 4\)

Correct Answer: C. \(2\frac{1}{4}\)

Explanation

\(\frac{5}{\sqrt{2}} = \frac{5 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{5\sqrt{2}}{2}\)

\(\frac{\sqrt{8}}{8} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4}\)

\(\frac{5}{\sqrt{2}} - \frac{\sqrt{8}}{8} = (\frac{5}{2} - \frac{1}{4})\sqrt{2}\)

= \(\frac{9}{4}\sqrt{2} \)

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

Correct Answer: D. 9 \(\frac{9}{16}\)

Explanation

GP : 36, P, \(\frac{q}{4}\), q, ... p + q = ?

∴ \(\frac{1}{4}\) = q ÷ \(\frac{9}{4}\) ;

\(\frac{9}{4}\) = 4q