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AP EAMCET · Maths · Application of Derivatives

If the extreme value of the function \(f(x)=\frac{4}{\sin x}+\frac{1}{1-\sin x}\) in \(\left[0, \frac{\pi}{2}\right]\) is \(m\) and it exists at \(\mathrm{x}=\mathrm{k}\), then \(\cos \mathrm{k}=\)

  1. A \(\frac{\sqrt{\mathrm{m}}}{4}\)
  2. B \(\frac{\sqrt{\mathrm{m}+1}}{\sqrt{2}}\)
  3. C \(\frac{\sqrt{5}}{\sqrt{\mathrm{~m}}}\)
  4. D \(\frac{1}{\mathrm{~m}}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\sqrt{5}}{\sqrt{\mathrm{~m}}}\)

Step-by-step Solution

Detailed explanation

\(\text{Let } y = \sin x\text{. Then } f(x) = g(y) = \frac{4}{y} + \frac{1}{1-y}\text{.}\) \(g'(y) = -\frac{4}{y^2} + \frac{1}{(1-y)^2}\) \(g'(y) = 0 \implies \frac{4}{y^2} = \frac{1}{(1-y)^2} \implies \frac{2}{y} = \frac{1}{1-y}\text{ (since } y \in (0,1)\text{)}\)…