Mathematics · 2024
JEE Main · 29 January 2024, Shift 1 · Q9
Suppose f(x)=((2^x+2^-x) tan x √(tan^-1(x^2-x+1)))/((7 x^2+3 x+1)^3). Then the value of f^′(0) is equal to
Suppose $\displaystyle f(x)=\frac{\left(2^x+2^{-x}\right) \tan x \sqrt{\tan ^{-1}\left(x^2-x+1\right)}}{\left(7 x^2+3 x+1\right)^3}$. Then the value of $\displaystyle f^{\prime}(0)$ is equal to
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \sqrt{\pi}$
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JEE Main 2024 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.