Question

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

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (D)

$9$

Given equation
$\frac{4}{\sin x}+\frac{1}{1-\sin x}=a$ ....(1)
$f\left(x\right)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$
$f^{\prime }\left(x\right)=\cos x(\frac{1}{{(1-\sin x)}^2}-\frac{4}{{\sin }^{2}x})$
$=\cos x\frac{(2-\sin x)(3\sin x-2)}{{\sin }^{2}x{(1-\sin x)}^2}$
For maxima or minima, $f^{\prime }\left(x\right)=0$
$\Rightarrow \sin x=2$(not possible) , $\sin x=\frac{2}{3}$ and $\cos x=0\Rightarrow x=\frac{\pi }{2}$
$\Rightarrow x={\sin }^{-1}\frac{2}{3}$
$f\left({\sin }^{-1}\frac{2}{3}\right)=6+3=9$
$f\left(0^{+}\right)\to \infty$
$f\left(\frac{{\pi }^{-}}{2}\right)\to \infty$
$\Rightarrow a=9$ (by (1))