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Definite Integration question

2022 · 28 Jul · Shift 1 · Q37
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  5. /2022 · 28 Jul · Shift 1 · Q37

Definite Integration question

2022 · 28 Jul · Shift 1 · Q37

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The 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∫x​ex−tf′(t)dt−(x2−x+1)ex, x∈Rx \in \mathbf{R}x∈R, is :
  1. A
    −2e-\frac{2}{\sqrt{\mathrm{e}}}−e​2​
  2. B
    −2e-2 \sqrt{\mathrm{e}}−2e​
  3. C
    −e-\sqrt{\mathrm{e}}−e​
  4. D
    2e\frac{2}{\sqrt{\mathrm{e}}}e​2​
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