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JEE Mains · Maths · STD 12 - 7.2 definite integral

माना \(\left[0, \frac{\pi}{2}\right]\) पर परिभाषित एक अवकलनीय फलन \(\mathrm{f}\) है जिसके लिए \(\mathrm{f}(\mathrm{x})>0\) और \( \mathrm{f}(\mathrm{x})+\int_0^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \sqrt{1-\left(\log _{\mathrm{e}} \mathrm{f}(\mathrm{t})\right)^2} \mathrm{dt}=\mathrm{e}, \forall \mathrm{x} \in\left[0, \frac{\pi}{2}\right] \) है। तब \(\left(6 \log _{\mathrm{e}} \mathrm{f}\left(\frac{\pi}{6}\right)\right)^2\) बराबर है_____________. 

  1. A \(25\)
  2. B \(26\)
  3. C \(23\)
  4. D \(27\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(27\)

Step-by-step Solution

Detailed explanation

\(f ( x )+\int_0^{ x } f ( t ) \sqrt{1-\left(\log _{ e } f ( t )\right)^2} dt = e\) \(\Rightarrow f (0)= e\) \(f ^{\prime}( x )+ f ( x ) \sqrt{1-(\ln f ( x ))^2}=0\) \(f ( x )= y\) \(\frac{ dy }{ dx }=- y \sqrt{1-(\ln y )^2}\)…
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