Q62 / 155 · WBJEE 2022 · Mathematics

Let $f(x)=\int_{\sin x}^{\cos x} e^{-t^2} d t$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

  • A
    $\sqrt{1 / e}$
  • Correct answer
    B
    $-\sqrt{2 / e}$
  • C
    $\sqrt{2 / e}$
  • D
    $-\sqrt{1 / e}$

Solution

$f^{\prime}(x)=-e^{-\cos ^2 x} \sin x-e^{-\sin ^2 x} \cos x$ $\therefore f^{\prime}\left(\frac{\pi}{4}\right)=-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2 e}}=-\sqrt{\frac{2}{e}}$

Hence, the correct answer is option: 'B'.

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