Correct Answer: A. 4

Explanation

\(f(x) = \frac{1}{2 - x}, x \neq 2\)

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

\(x = \frac{1}{2 - y}\) (Let x = f(y))

\(2x - xy = 1 \implies y = \frac{2x - 1}{x}\)

\(\therefore f^{-1}(x) = \frac{2x - 1}{x}\)

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

= \(-2 \times -2 = 4\)

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

Explanation

27^x = 9^y

3^3x = 3^2y

3x = 2y

Correct Answer: B. (3, 5)

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

Explanation

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

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

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

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

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

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

= \(-\tan \theta\)

Correct Answer: C. 60.255

Explanation

\((1.98)^{6} = (1 + 0.98)^{6} = 1 + 6(0.98) + 15(0.98)^{2} + 20(0.98)^{3} + 15(0.98)^{4} + 6(0.98)^{5} + (0.98)^{6}\)

\(\approxeq 1 + 5.88 + 14.406 + 18.823 + 13.836 + 5.424 + 0.886 \)

= \(60.255\)

Explanation

\(2^{(2y+1)} - 5(2^y) + 2\) = 0

Let p = 2\(^y\)

\(2^{2y} (2^1) - 5(2^y)\) + 2 = 0

2p\(^2\) - 5p + 2 = 0

2p\(^2\) - p - 4p + 2 = 0

p (2p - 1) - 2(2p - 1) = 0

(p - 2)(2p - 1) = 0

p = 2 or \(\frac{1}{2}\)

p = 2\(^y\)

when p = 2

2\(^y\) = 2

y = 1

when p = \(\frac{1}{2}\)

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

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

y = -1

y = -1 or 1

Explanation

(a)(i) \((1 + x)^{4} = 1 + 4(x) + 6(x)^{2} + 4(x)^{3} + x^{4}\)

= \(1 + 4x + 6x^{2} + 4x^{3} + x^{4}\)

(ii) \((\frac{5}{4})^{4} = (1 + \frac{1}{4})^{4}\)

= \(1 + 4(\frac{1}{4}) + 6(\frac{1}{4})^{2} + 4(\frac{1}{4})^{3} + (\frac{1}{4})^{4}\)

= \(1 + 1 + 0.375 + 0.0625 + 0.00390625\)

= \(2 + 0.44140625\)

\(\approxeq 2.441\)

(b) \(T_{1} = a = 7\) (first term of an A.P)

\(T_{2} = a + d = 7 + d .... (1)\)

\(T_{5} = a + 4d = 7 + 4d .... (2)\)

\(T_{1} = 7 = a\) (first term of G.P)

\(T_{2} = ar = 7r\)

\(T_{3} = ar^{2} = 7r^{2}\)

\(\implies 7r = 7 + d\)

\(7r^{2} = 7 + 4d\)

\((1) \times 4 : 28r = 28 + 4d ... (3)\)

\((3) - (2) : 28r - 7r^{2} = 28 - 7 = 21\)

\(7r^{2} - 28r + 21 = 0\)

\(7r^{2} - 21r - 7r + 21 = 0\)

\(7r(r - 3) - 7(r - 3) = 0\)

\((7r - 7)(r - 3) = 0\)

\(\text{r = 1 or 3}\)

From (1), if r = 1,

\(7 = 7 + d \implies d = 0\) (This cannot be the case)

r = 3,

\(7(3) = 7 + d \implies d = 21 - 7 = 14\)

The common difference = 14.

Correct Answer: B. 1/169

Explanation

1st card prob = 4/52 2nd card prob = 4/52

prob.both are aces = (4/52) x (4/52) = 1/169

Correct Answer: (d) None of the above

Correct Answer: C. 20