Mathematics · 2021
JEE Main · 24 February 2021, Shift 2 · Q68
Let f(x) be a differentiable function defined on [0, 2] such that f^′(x)=f^′(2-x) for all x ∈(0,2), f(0)=1 and f(2)= e^2. Then the value of ∫_0^2…
Let $\displaystyle f(x)$ be a differentiable function defined on [$\displaystyle 0$, $\displaystyle 2$] such that $\displaystyle f^{\prime}(x)=f^{\prime}(2-x)$ for all $\displaystyle x \in(0,2), f(0)=1$ and $\displaystyle f(2)=\mathrm{e}^2$. Then the value of $\displaystyle \int_0^2 f(x) \mathrm{d} x$ is :
Official answer
(2)
$\displaystyle 1+\mathrm{e}^2$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.