Question

The minimum value of $\alpha$ for which the equation $\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha$ has at least one solution in $\left(0,\frac{\pi }{2}\right)$ is _____.

Answer 1

Answer: 9

Let $f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}$
$\Rightarrow f^{\prime }(x)=0$
$\Rightarrow \sin x=2/3$
$\therefore f(x{)}_{\min }=\frac{4}{2/3}+\frac{1}{1-2/3}=9$
$f(x)\max \to \infty$
$f(x)$ is continuous function
$\therefore {\alpha }_{\min }=9$