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JEE Mains · Maths · STD 12 - 7.2 definite integral

Let \(f\) be a differentiable function defined on \(\left[0, \frac{\pi}{2}\right]\) such that \(f(x) > 0\) and \(f(x)+\int \limits_0^x f(t) \sqrt{1-\left(\log _e f(t)\right)^2} d t=e, \forall x \in\left[0, \frac{\pi}{2}\right]\) Then \(\left(6 \log _{ e } f \left(\frac{\pi}{6}\right)\right)^2\) is equal to \(.............\)

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