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