Question

The least value of 'a' for which the expression

$\frac{4}{\sin x} + \frac{1}{1 - \sin x} = a.2\left\{ {sgn}\left\lbrack \lim_{n \rightarrow \infty}\left( \frac{1}{2} \right)^{\frac{1}{n}} \right\rbrack \right\}$ has at least

one solution on the interval $\left( 0,\frac{\pi}{2} \right)$ is

• A. 9
• B. 5
• C. 8
• D. 4.5

Answer 1

Answer: (D)

4.5

Answer 2

$f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$;$f'(x) = \frac{- 4\cos x}{\left( \sin x \right)^{2}} + \frac{\cos x}{\left( 1 - \sin x \right)^{2}}$

= $\frac{- 4\cos x\left( 1 - \sin x \right)^{2} + \cos x\left( \sin x \right)^{2}}{\left( \sin x \right)^{2}\left( 1 - \sin x \right)^{2}}$

$f'(x) = \frac{- 4\cos x\left( 1 + \sin^{2}x - 2\sin x \right) + \cos x.\sin^{2}x}{\left( \sin x \right)^{2}\left( 1 - \sin x \right)^{2}}$

For maxima or minima

$f'(x) = 0$

⇒ $- 4\cos x - 4\cos x\sin^{2}x + 8\sin x\cos x + \cos x.\sin^{2}x = 0$⇒ $- 3\cos x.\sin^{2}x + 8\sin x\cos x - 4\cos x = 0$

⇒ $\cos x\left( - 3\sin^{2}x + 8\sin x - 4 \right) = 0$

⇒ cos x ≠ 0; $- 3\sin^{2}x + 8\sin x - 4 = 0$

⇒ $- 3\sin^{2}x + 6\sin x + 2\sin x - 4 = 0$

⇒ $- 3\sin x\left( \sin x - 2 \right) + 2\left( \sin x - 2 \right) = 0$

⇒ $\left( \sin x - 2 \right)\left( - 3\sin x + 2 \right)$=0

Now, $\sin x \neq 2;\sin x = \frac{2}{3}$

$\frac{4}{\sin x} + \frac{1}{1 - \sin x} = 2a$ $\left\lbrack {sgn}(1) = 1 \right\rbrack$

⇒ $\frac{4}{2/3} + \frac{1}{1 - 2/3} = 2a$

⇒ $\frac{12}{2} + 3 = 2a$

⇒ $\frac{9}{2} = a$

⇒ $a = 4.5$

so 'd' is correct