WBJEE · Maths · Definite Integration
Let \(f(x)=\int_{\sin x}^{\cos x} e^{-t^2} d t\). Then \(f^{\prime}\left(\frac{\pi}{4}\right)\) equals
- A \(\sqrt{1 / e}\)
- B \(-\sqrt{2 / e}\)
- C \(\sqrt{2 / e}\)
- D \(-\sqrt{1 / e}\)
Answer & Solution
Correct Answer
(B) \(-\sqrt{2 / e}\)
Step-by-step Solution
Detailed explanation
\(f^{\prime}(x)=-e^{-\cos ^2 x} \sin x-e^{-\sin ^2 x} \cos x\) \(\therefore f^{\prime}\left(\frac{\pi}{4}\right)=-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2 e}}=-\sqrt{\frac{2}{e}}\)
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