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JEE Mains · Maths · STD 12 - 6. Application of derivatives

\(\alpha\) का न्यूनतम मान, जिसके लिए समीकरण \(\frac{4}{\sin x}+\frac{1}{1-\sin x}=\alpha\) का अन्तराल \(\left(0, \frac{\pi}{2}\right)\) में कम से कम एक हल है, है ......... |

  1. A \(5\)
  2. B \(9\)
  3. C \(6\)
  4. D \(3\)
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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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