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

વિધેય \(f(x)=\tan ^{-1}(\sin x-\cos x)\) ની અંતરાલ  \([0, \pi]\) માં વૈવ્ચિક મહતમ અને વૈવ્ચિક ન્યૂનતમ કિંમતો નો સરવાળો મેળવો.

  1. A \(0\)
  2. B \(\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)-\frac{\pi}{4}\)
  3. C \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}\)
  4. D \(\frac{-\pi}{12}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\frac{\pi}{4}\)

Step-by-step Solution

Detailed explanation

\(f(x)=\tan ^{-1}(\sin x-\cos x)\) \(f^{\prime}(x)=\frac{\cos x+\sin x}{(\sin x-\cos x)^{2}+1}=0\) \(\therefore x=\frac{3 \pi}{4}\) \(x\) \(0\) \(\frac{3\,\pi}{4}\) \(\pi\) \(f(x)\) \(\frac{-\pi}{4}\) \(\tan ^{-1} \sqrt{2}\) \(\frac{\pi}{4}\)…
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