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

To solve for the minimum value of \(\alpha\) where the equation \(\frac{4}{\sin x} + \frac{1}{1-\sin x} = \alpha\) has at least one solution within \((0, \frac{\pi}{2})\), we need to analyze the behavior of the function \(f(\sin x) = \frac{4}{\sin x} + \frac{1}{1-\sin x}\). Step 1: Let \(y = \sin x\), where \(0 < y < 1\). The equation becomes \(f(y) = \frac{4}{y} + \frac{1}{1-y} = \alpha\). Step 2: To find the minimum \(\alpha\), consider the function \(f(y)\) over the interval \(0 < y < 1\). To find critical points, differentiate \(f(y)\) with respect to \(y\): \[f'(y) = -\frac{4}{y^2} - \frac{1}{(1-y)^2}\] Step 3: Set \(f'(y) = 0\) to find critical points: \[-\frac{4}{y^2} = \frac{1}{(1-y)^2}\] Cross-multiply and simplify: \[4(1-y)^2 = y^2\] \[4(1 - 2y + y^2) = y^2\] \[4 - 8y + 4y^2 = y^2\] \[3y^2 - 8y + 4 = 0\] Step 4: Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for \(3y^2 - 8y + 4 = 0\): \[y = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6}\] \[y = 2\] or \[y = \frac{2}{3}\] Since \(\sin x\) must be between 0 and 1, we take \(y = \frac{2}{3}\). Step 5: Evaluate \(f(y)\) at \(y = \frac{2}{3}\): \[f\left(\frac{2}{3}\right) = \frac{4}{\frac{2}{3}} + \frac{1}{1-\frac{2}{3}} = 6 + 3 = 9\] Conclusion: The minimum value of \(\alpha\) is 9, which is within the expected range [9,9]. Therefore, the answer is 9.