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

माना \(\mathrm{g}(\mathrm{x})=\mathrm{f}(\mathrm{x})+\mathrm{f}(1-\mathrm{x})\) तथा \(\mathrm{f}^{\prime \prime}(\mathrm{x})>0, \mathrm{x} \in(0,1)\) है। यदि \(\mathrm{g}\) अंतराल \((0, \alpha)\) में ह्रासमान है तथा अंतराल \((\alpha, 1)\) में वर्धमान है, तो \(\tan ^1(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)\) बराबर है :

  1. A \(\frac{3 \pi}{2}\)
  2. B \(\pi\)
  3. C \(\frac{5 \pi}{4}\)
  4. D \(\frac{3 \pi}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\pi\)

Step-by-step Solution

Detailed explanation

\(g(x)=f(x)+f(1-x) f^{\prime}(x) > 0, x \in(0,1)\) \(g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)=0\) \(f^{\prime}(x)=f(1-x)\) \(x=1-x\) \(x=\frac{1}{2}\) \(g^{\prime}(x)=0\) \(\text { at } x=\frac{1}{2}\) \(g^{\prime \prime}(x)=f^{\prime}(x)+f^{\prime \prime}(1-x) > 0\)…
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