The minimum value of


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



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

$\therefore \alpha_{\min }=9$


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