Question

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

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is D $9$

Least value of a is minimum value of $(\frac{4}{sinx}+\frac{1}{1-sinx})$
Now,
$\frac{d}{dx}(\frac{4}{sinx}+\frac{1}{1-sinx})=\frac{cotxcosecx(4{cosec}^{2}x-8cosecx+3)}{{(cosecx-1)}^2}$
Therefor critical points between$(0,\frac{\pi }{2})$ are $2{tan}^{-1}\left(\frac{1}{2}(3-\sqrt{5})\right),2{tan}^{-1}\left(\frac{1}{2}(3+\sqrt{5})\right)$ and 0
So minimum value of a is 9 at $x=2{tan}^{-1}\left(\frac{1}{2}(3-\sqrt{5})\right)$ and $x=2{tan}^{-1}\left(\frac{1}{2}(3+\sqrt{5})\right)$