Question

The least value of $a$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$ has atleast one solution in the interval $(0,\frac{\pi }{2})$ is

• A. $5$
• B. $7$
• C. $8$
• D. $9$

Answer 1

The correct option is D $9$

$a=\frac{4}{\sin x}+\frac{1}{1-\sin x}\frac{da}{dx}=(\frac{-4}{{\sin }^{2}x}+\frac{1}{(1-\sin x{)}^2})\cos x$
$\Rightarrow \frac{da}{dx}=\frac{(3\sin x-2)(2-\sin x)}{{\sin }^{2}x(1-\sin x{)}^2}\cos x$
Critical points will occur, when $\sin x=\frac{2}{3},\sin x=0,\sin x=1,\cos x=0$
But given that $x\in (0,\frac{\pi }{2})$
$\therefore \sin x=\frac{2}{3},\text{suppose}x=x_0\in (0,\frac{\pi }{2})$ is the only possible critical point.
As, $\sin x$ is increasing in the given interval.
So,$f^{\prime }\left(x\right)$ changes sign from negative to positive as $x$ crosses $x_0$ from left to right OR $\sin x$ crosses $\frac{2}{3}$ from left to right. So $f\left(x\right)$ has local minima at $\sin x=\frac{2}{3}$
Now, value of $a$ at $\sin x=\frac{2}{3}$
$a=\frac{4}{2/3}+\frac{1}{1-2/3}$
$\Rightarrow a=6+3=9$