JEE Main 202429 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002Suppose f(x)=(2x+2−x)tanxtan−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}f(x)=(7x2+3x+1)3(2x+2−x)tanxtan−1(x2−x+1). Then the value of f′(0)f^{\prime}(0)f′(0) is equal toAOption A. π\piπBOption B. π\sqrt{\pi}πCOption C. 0DOption D. π2\frac{\pi}{2}2πCheckSkip · show solution
