JEE Mains · Maths · STD 12 - 6. Application of derivatives
Let \(g(x)=f(x)+f(1-x)\) and \(f^{\prime \prime}(x) > 0, x \in(0,1)\). If \(g\) is decreasing in the interval \((0, \alpha)\) and increasing in the interval \((\alpha, 1)\), then \(\tan ^1(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)\) is equal to :
- A \(\frac{3 \pi}{2}\)
- B \(\pi\)
- C \(\frac{5 \pi}{4}\)
- D \(\frac{3 \pi}{4}\)
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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