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JEE Mains · Maths · STD 12 - 7.2 definite integral

ધારોકે \(f\) એ \(\left(0, \frac{\pi}{2}\right)\) માં વિકલનીય વિધેય છે. જો \(\int_{\cos x}^{1} t^{2} f(t) d t=\sin ^{3} x+\cos x-1\) હોય, તો \(\frac{1}{\sqrt{3}} f^{\prime}\left(\frac{1}{\sqrt{3}}\right)=\dots\dots\dots\) :

  1. A \(6-9 \sqrt{2}\)
  2. B \(\frac{9}{\sqrt{2}}-6\)
  3. C \(\frac{9}{2}-6 \sqrt{2}\)
  4. D \(6-\frac{9}{\sqrt{2}}\)
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Answer & Solution

Correct Answer

(D) \(6-\frac{9}{\sqrt{2}}\)

Step-by-step Solution

Detailed explanation

\(\int\limits_{\cos x}^{1} t^{2} f(t) d t=\sin ^{3} x+\cos x-1\) Calculation for option differentiating both sides \(-\cos ^{2} x f(\cos x) \cdot(-\sin x)=3 \sin ^{2} x \cdot \cos x-\sin x\) \(\Rightarrow f (\cos x )=3 \tan x -\sec ^{2} x\)…
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