JEE Math Practice Question
The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :
- A.$$-\frac{2}{\sqrt{\mathrm{e}}}$$
- B.$$-2 \sqrt{\mathrm{e}}$$
- C.$$-\sqrt{\mathrm{e}}$$
- D.$$\frac{2}{\sqrt{\mathrm{e}}}$$
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