JEE Mains · Maths · STD 11 - 12. limits
Suppose \(f(x)=\frac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}\). Then the value of \(f^{\prime}(0)\) is equal to
- A \(\pi\)
- B \(0\)
- C \(\sqrt{\pi}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\pi}\)
Step-by-step Solution
Detailed explanation
\( f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h} \) \( =\lim _{h \rightarrow 0} \frac{\left(2^h+2^{-h}\right) \tan h \sqrt{\tan ^{-1}\left(h^2-h+1\right)}-0}{\left(7 h^2+3 h+1\right)^3 h} \) \( =\sqrt{\pi}\)
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