JEE Main 202228 Jul, Morning shiftMathematicsMediumIntegral Calculus369× since 2002The minimum value of the twice differentiable function f(x)=∫0xex−tf′(t)dt−(x2−x+1)exf(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}f(x)=0∫xex−tf′(t)dt−(x2−x+1)ex, x∈Rx \in \mathbf{R}x∈R, is :AOption A. −2e-\frac{2}{\sqrt{\mathrm{e}}}−e2BOption B. −2e-2 \sqrt{\mathrm{e}}−2eCOption C. −e-\sqrt{\mathrm{e}}−eDOption D. 2e\frac{2}{\sqrt{\mathrm{e}}}e2CheckSkip · show solution
