Question:
The least value of \(a\), for which the function
\[
\frac{4}{\sin x} + \frac{1}{1-\sin x} = a
\]
has at least one solution in the interval \(\left(0,\frac{\pi}{2}\right)\), is:
The least value of \(a\), for which the function
\[
\frac{4}{\sin x} + \frac{1}{1-\sin x} = a
\]
has at least one solution in the interval \(\left(0,\frac{\pi}{2}\right)\), is:
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For least value problems, convert to a single variable and minimize using derivatives.
Updated On: Apr 16, 2026
- 9
- 4
- 5
- 1
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The Correct Option is A
Solution and Explanation
Concept:
Let \(t = \sin x\). In \(\left(0, \frac{\pi}{2}\right)\), \(t \in (0,1)\).
\[
a = \frac{4}{t} + \frac{1}{1-t}
\]
Step 1: Find minimum value of \(a\) for \(t \in (0,1)\). \[ f(t) = \frac{4}{t} + \frac{1}{1-t} \] \[ f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2} = 0 \] \[ \frac{1}{(1-t)^2} = \frac{4}{t^2} \Rightarrow \frac{1}{1-t} = \frac{2}{t} \Rightarrow t = 2 - 2t \Rightarrow 3t = 2 \Rightarrow t = \frac{2}{3} \]
Step 2: Minimum value. \[ f\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9 \]
Step 3: Conclusion. For the equation to have at least one solution, \(a\) must be at least the minimum value of the function: \[ a_{\text{least}} = 9 \]
Step 1: Find minimum value of \(a\) for \(t \in (0,1)\). \[ f(t) = \frac{4}{t} + \frac{1}{1-t} \] \[ f'(t) = -\frac{4}{t^2} + \frac{1}{(1-t)^2} = 0 \] \[ \frac{1}{(1-t)^2} = \frac{4}{t^2} \Rightarrow \frac{1}{1-t} = \frac{2}{t} \Rightarrow t = 2 - 2t \Rightarrow 3t = 2 \Rightarrow t = \frac{2}{3} \]
Step 2: Minimum value. \[ f\left(\frac{2}{3}\right) = \frac{4}{2/3} + \frac{1}{1-2/3} = 6 + 3 = 9 \]
Step 3: Conclusion. For the equation to have at least one solution, \(a\) must be at least the minimum value of the function: \[ a_{\text{least}} = 9 \]
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