Correct Answer: A. \(\frac{5}{4}\)

Explanation

\(8^{x} ÷ (\frac{1}{4})^{y} = 1\)

\((2^{3})^{x} ÷ (2^{-2})^{y} = 2^{0}\)

\(2^{3x - (-2y)} = 2^{0}\)

\(\implies 3x + 2y = 0 .... (1)\)

\(\log_{2}(x - 2y) = 1\)

\( x - 2y = 2^{1} = 2 ..... (2)\)

Solving equations 1 and 2,

\(x = \frac{1}{2}, y = \frac{-3}{4}\)

\((x - y) = \frac{1}{2} - \frac{-3}{4} = \frac{5}{4}\)

Correct Answer: C. \(x^{2} + 2x - 3 = 0\)

Explanation

Given the roots of an equation such that you can find the sum and product of the roots, the equation can be given as:

\(x^{2} - (\alpha + \beta)x + (\alpha \beta) = 0 \)

\(\alpha + \beta = -3 + 1 = -2\)

\(\alpha \beta = -3 \times 1 = -3\)

Equation: \(x^{2} - (-2)x + (-3) = 0 \implies x^{2} + 2x - 3 = 0\)

Correct Answer: A. \(x \leq -3\) or \(x \geq \frac{1}{2}\)

Correct Answer: D. \(1 + \sqrt{2}\)

Explanation

\(\frac{1 + \sqrt{8}}{3 - \sqrt{2}}\)

Rationalizing by multiplying through with \(3 + \sqrt{2}\),

\((\frac{1 + \sqrt{8}}{3 - \sqrt{2}})(\frac{3 + \sqrt{2}}{3 + \sqrt{2}}) = \frac{3 + \sqrt{2} + 3\sqrt{8} + 4}{9 - 2}\)

= \(\frac{3 + \sqrt{2} + 3\sqrt{4 \times 2} + 4}{7} \)

= \(\frac{7 + 7\sqrt{2}}{7} = 1 + \sqrt{2}\)

Explanation

(a) Passed : Failed = 4 : 1

\(p(Passed) = \frac{4}{5} ; p(Failed) = \frac{1}{5}\)

(i) p(at least 3 passed) = 1 - [p(0) + p(1) + p(2)]

p(0 passed) = \(^{7}C_{0} (0.8)^{0} (0.2)^{7} = 0.0000128\)

p(1 passed) = \(^{7}C_{1} (0.8)^{1} (0.2)^{6} = 0.000358\)

p(2 passed) = \(^{7}C_{2} (0.8)^{2} (0.2)^{5} = 0.0043\)

p(at least 3 passed) = \(1 - [0.0000128 + 0.000358 + 0.0043] = 0.9953\)

(ii) p(between 3 and 6 students failed) = p(4 or 5 students failed)

= \((^{7}C_{4} (0.8)^{3} (0.2)^{4}) + (^{7}C_{5} (0.8)^{2} (0.2)^{5})\)

= \(0.02867 + 0.0043008\)

= \(0.03297\)

(b) p(a six) = p = \(\frac{1}{6}\), p(not a six) = q = \(\frac{5}{6}\)

Die thrown five times;

Binomial distribution function is \((p + q)^{5} = p^{5} + 5p^{4}q + 10p^{3} q^{2} + 10p^{2} q^{3} + 5pq^{4} + q^{5}\)

p(a six three times) = \(10p^{3} q^{2}\)

= \(10(\frac{1}{6})^{3} (\frac{5}{6})^{2}\)

= \(\frac{250}{7776}\)

= 0.03215

Correct Answer: A. \(-\tan \theta\)

Explanation

\(\frac{\cos 2\theta - 1}{\sin 2\theta}\)

\(\cos (x + y) = \cos x \cos y - \sin x \sin y \implies \cos 2\theta = \cos^{2} \theta - \sin^{2} \theta\)

\(\cos^{2} \theta = 1 - \sin^{2} \theta \implies \cos 2\theta = 1 - 2\sin^{2} \theta\)

\(\sin 2\theta = 2\sin \theta \cos \theta\)

\(\therefore \frac{\cos 2\theta - 1}{\sin 2\theta} = \frac{1 - 2\sin^{2}\theta - 1}{2\sin \theta \cos \theta}\)

= \(\frac{-2 \sin^{2} \theta}{2\sin \theta \cos \theta} = \frac{- \sin \theta}{\cos \theta}\)

= \(-\tan \theta\)

Explanation

\(T_{n} = ar^{n - 1}\) (Terms of a geometric progression)

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

\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a GP)

\(\frac{a}{1 - r} = \frac{25}{2} \implies 2a = 25(1 - r)\)

\(a = \frac{25 - 25r}{2}\)

\(3 = (\frac{25 - 25r}{2})r\)

\(6 = 25r - 25r^{2}\)

\(25r^{2} - 25r + 6 = 0\)

\(25r^{2} - 15r - 10r + 6 = 0\)

\(5r(5r - 3) - 2(5r - 3) = 0 \implies (5r - 3)(5r - 2) = 0\)

\(r = \frac{2}{5} \text{or r = } \frac{3}{5}\).

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: B. 5 units

Explanation

Distance between two points (a, b) and (c, d) = \(\sqrt{(d - b)^{2} + (c - a)^{2}}

Distance between (2, 5) and (5, 9) = \(\sqrt{(9-5)^{2} + (5-2)^{2}}\)

= \(\sqrt{16 + 9} = \sqrt{25} = 5 units\)

Correct Answer: B. \(\frac{3}{1 - x} + \frac{2x + 1}{x^{2} + 1}\)

Explanation

\(\frac{x^{2} + x + 4}{(1 - x)(x^{2} + 1)} = \frac{A}{1 - x} + \frac{Bx + C}{x^{2} + 1}\)

= \(\frac{A(x^{2} + 1) + (Bx + C)(1 - x)}{(1 - x)(x^{2} + 1)}\)

\(\implies x^{2} + x + 4 = A(x^{2} + 1) + (Bx + C)(1 - x)\)

\(x^{2} + x + 4 = Ax^{2} + A + Bx - Bx^{2} - Cx + C\)

\(\implies (A - B)x^{2} = x^{2}; A - B = 1 ...... (i)\)

\((B - C)x = x; B - C = 1 ..... (ii)\)

\(A + C = 4 ...... (iii)\)

Solving the above simultaneous equations by any of the known methods, we get

\(A = 3, B = 2, C = 1\)

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