Question

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

• A. $9$
• B. $3$
• C. $0$
• D. $\pi /2$

Answer 1

Answer: (A)

$9$

Given:$\left(\frac{4}{\sin x}+\frac{1}{1-\sin x}\right)$

Differentiate the given equation with respect to x.

$\frac{d}{dx}\left(\frac{4}{\sin x}+\frac{1}{1-\sin x}\right)=\frac{da}{dx}$

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

Equate the first derivative of the function to 0

$\frac{da}{dx}=0$

$\left[\frac{-4}{{\sin }^{2}x}+\frac{1}{{\left(1-\sin x\right)}^2}\right]\cos x=0$

$\frac{-4}{{\sin }^{2}x}+\frac{1}{{\left(1-\sin x\right)}^2}=0$

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

${\left(2\left(1-\sin x\right)\right)}^2={\sin }^{2}x$

$2\left(1-\sin x\right)=\sin x$

$2-2\sin x=\sin x$

$\sin x=\frac{2}{3}$

Double differentiate the given function with respect to x

$\frac{d^{2}a}{d{x}^2}=\left(\frac{-4}{{\sin }^{2}x}+\frac{1}{{\left(1-\sin x\right)}^2}\right)\left(-\sin x\right)+\cos x\left(8{\sin }^{2}x+2\left(1-\sin x\right)\cos x\right)$

Substitute the critical value in double derivative and find out whether it is less than or greater than 0.

$\frac{d^{2}a}{d{x}^2}⩾0$

Since double derivative is greater than 0.

Substitute the value in the given function to find the value of minima.

$a=\frac{4}{\sin x}+\frac{1}{1-\sin x}$

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

$=6+3=9$