JEE Main 2022 · 28 Jul · Shift 1 · Mathematics Q20

MathematicsSingle correct+4 / -1Question 20 of 90

The minimum value of the twice differentiable function $f(x)=\displaystyle\int_0^xe^{x-t}f'(t)\,dt-\left(x^2-x+1\right)e^x$ , $x\in\mathbf R$ , is :

-\dfrac2{\sqrt e}

-2\sqrt e

-\sqrt e

\dfrac2{\sqrt e}

Answer and worked solution

Official answer: A

Differentiate. Writing $F(x)=e^x\int_0^xe^{-t}f'(t)dt$ , $F'=F+f'$ . Then $f=(2x-1)e^x$ .

Minimum. $f'=(2x+1)e^x=0$ at $x=-\frac12$ : $f=-2e^{-1/2}$ .

(A)

Topics:Leibniz rule Maxima and minima

JEE Main 2022 · 28 Jul · Shift 1, Mathematics question 20