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Joint Entrance Examination
Mathematics
Definite Integration
Question 190
Question 190 of 246
Medium
MCQ
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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Official solution
\[f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} \] \[f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} \] \[{e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} \] Differentiate on both side \[{e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1\]
f
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f(x) = {e^x}(2x - 1)
f
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f'(x) = {e^x}(2) + {e^x}(2x - 1)
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= {e^x}(2x + 1)
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\[x = - {1 \over 2}\]
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f''(x) = {e^x}(2) + (2x + 1){e^x}
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= {e^x}(2x + 3)
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For \[x = - {1 \over 2}\,\,f''(x) > 0\] \[\Rightarrow\] Maxima \[\therefore\] Max. \[ = {e^{ - {1 \over 2}}}( - 1 - 1)\] \[\therefore\] \[ - {2 \over {\sqrt e }}\]
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