Question

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

• A. $3$
• B. $2$
• C. $8$
• D. $1$

Answer 1

The correct option is A $3$

Let f(x) = $\frac{4}{sinx}+\frac{1}{1-sinx}$

$Li{m}_{x->0}f\left(x\right)=\infty$

$Li{m}_{x->\pi /2}f\left(x\right)=\infty$

Therefore f'(x) = 0 for a x = [0,$\pi /2$]

In accordance to Rolle's theorem

$f^{\prime }\left(x\right)=\frac{-4cosx}{si{n}^{2}x}-\frac{-1}{(1-sinx{)}^2}\times -cosx$ = 0

$sinx=\frac{2}{3}$

$f"\left(x\right)<0$ $at$ $x=si{n}^{-1}\left(2/3\right)$

Therefore function attains minima at this point.

$f\left(si{n}^{-1}\right(2/3\left)\right)=9$

Hence least value is 9

answer $a=\sqrt{9}=3$