Mathematics · 2020
JEE Main · 9 January 2020, Shift 1 · Q61
If f^′(x)= tan^-1( sec x+ tan x),-(π)/2<x<(π)/2, and f(0)=0, then f(1) is equal to:
If $\displaystyle f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}<x<\frac{\pi}{2}$,
and $\displaystyle f(0)=0$, then $\displaystyle f(1)$ is equal to :
Official answer
(3)
$\displaystyle \frac{\pi+1}{4}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.