Question

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

• A. $9$
• B. $4$
• C. $1$
• D. $8$

Answer 1

The correct option is A $9$

$f\left(x\right)=\frac{4}{\sin x}+\frac{1}{1-\sin x}-a$

$f^{\prime }\left(x\right)=\frac{-1}{{\sin }^{2}x}\times \cos x+\frac{1\times \left(\cos x\right)}{{(1-\sin x)}^2}=0$ (For maxima/minima)

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

$2\sin x=1$

$\sin x=\frac{1}{2}⟶1$

Putting $1$ in $f\left(x\right),f\left(x\right)=8+\frac{1}{1/2}-a$

$f\left(x\right)=10-a$

Least value of $a-9$

$f\left(x\right)=\frac{\left(1\right)}{{\left(2\right)}^{m+m}}$

$a=9$ Answer