JEE Mains · Maths · STD 12 - 6. Application of derivatives
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 ..........
- A \(5\)
- B \(9\)
- C \(6\)
- D \(3\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}\) \(\Rightarrow f^{\prime}(x)=0 \Rightarrow \sin x=2 / 3\) \(\therefore f(x)_{\min }=\frac{4}{2 / 3}+\frac{1}{1-2 / 3}=9\) \(f( x ) \max \rightarrow \infty\) \(f(x)\) is continuous function \(\therefore \alpha_{\min }=9\)
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