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 \((0, \frac{\pi}{2})\) is ____________

Hint:

For $g(x) = \frac{a^2}{f(x)} + \frac{b^2}{1-f(x)}$, the minimum value is $(a+b)^2$. Here $a=2, b=1$, so $(2+1)^2 = 9$.

Answer 1

Correct Answer: 9

Step 1: Let \(f(t) = \frac{4}{t} + \frac{1}{1-t}\) where \(t = \sin x \in (0, 1)\).
Step 2: \(f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2} = 0 \Rightarrow \frac{1}{(1-t)^2} = \frac{4}{t^2}\).
Step 3: \(\frac{1}{1-t} = \frac{2}{t} \Rightarrow t = 2 - 2t \Rightarrow 3t = 2 \Rightarrow t = 2/3\).
Step 4: Minimum value \(\alpha = f(2/3) = \frac{4}{2/3} + \frac{1}{1/3} = 6 + 3 = 9\).