Verified Solution • 2025–26 Session moderate

QJEEM24Ma6196: JEE Main 2024 Mathematics - Class 12 JEEM24Ma6196 Solution: Suppose . Then the value of is equal to

Suppose f(x)=(2x+2−x)tan⁡xtan⁡−1(x2−x+1)(7x2+3x+1)3f(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′(0)f^{\prime}(0) is equal to

Multiple Choice Options

  • A π\pi
  • B π\sqrt{\pi}
  • C 0
  • D π2\frac{\pi}{2}

Step-by-Step Verified Solution

amp;f′(0)=lim⁡h→0f(h)−f(0)hamp;=lim⁡h→0(2h+2−h)tan⁡htan⁡−1(h2−h+1)−0(7h2+3h+1)3hamp;=π\begin{aligned}& 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}\end{aligned}

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