Mathematics · 2022
JEE Main · 28 July 2022, Shift 1 · Q20
The minimum value of the twice differentiable function f(x)=∫_0^x e^x- t f^′( t ) dt -(x^2-x+1) e^x, x ∈ R, is:
The minimum value of the twice differentiable function $\displaystyle f(x)=\int_0^x \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^2-x+1\right) \mathrm{e}^x$, $\displaystyle x \in \mathbf{R}$, is :
Official answer
(1)
$\displaystyle -\frac{2}{\sqrt{\mathrm{e}}}$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.