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QJEEM22Ma5d06: JEE Main 2022 Mathematics - Class 12 JEEM22Ma5d06 Solution: The minimum value of the twice differentiable function , , is :

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} , x∈Rx \in \mathbf{R} , is :

Multiple Choice Options

  • A −2e-\frac{2}{\sqrt{\mathrm{e}}}
  • B −2e-2 \sqrt{\mathrm{e}}
  • C −e-\sqrt{\mathrm{e}}
  • D 2e\frac{2}{\sqrt{\mathrm{e}}}

Step-by-Step Verified Solution

f(x)=∫0xex−tf′(t)dt−(x2−x+1)exf(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}}

f(x)=ex∫0xe−tf′(t)dt−(x2−x+1)exf(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}}

e−xf(x)=∫0xe−tf′(t)dt−(x2−x+1){e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)}

Differentiate on both side

e−xf′(x)+(−f(x)e−x)=e−xf′(x)−2x+1{e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1

f(x)=ex(2x−1)f(x) = {e^x}(2x - 1)

f′(x)=ex(2)+ex(2x−1)f'(x) = {e^x}(2) + {e^x}(2x - 1)

=ex(2x+1)= {e^x}(2x + 1)

x=−12x = - {1 \over 2}

f′′(x)=ex(2)+(2x+1)exf''(x) = {e^x}(2) + (2x + 1){e^x}

=ex(2x+3)= {e^x}(2x + 3)

For x = - {1 \over 2}\,\,f''(x) > 0

⇒\Rightarrow Maxima

∴\therefore Max. =e−12(−1−1)= {e^{ - {1 \over 2}}}( - 1 - 1)

∴\therefore −2e- {2 \over {\sqrt e }}

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