A) \[\sqrt{\pi }/6\]
B) \[-\,\sqrt{(\pi /6)}\]
C) \[1/\sqrt{6}\]
D) \[\pi /\sqrt{6}\]
Correct Answer: B
Solution :
\[f(x)=\sqrt{1+{{\cos }^{2}}({{x}^{2}})}\] \[f'(x)=\frac{1}{2\sqrt{1+{{\cos }^{2}}{{(x)}^{2}}}}.(2\cos {{x}^{2}}).(-\sin {{x}^{2}}).(2x)\] \[f'(x)=\frac{-x\sin 2{{x}^{2}}}{\sqrt{1+{{\cos }^{2}}({{x}^{2}})}}\] At \[x=\frac{\sqrt{\pi }}{2},f'\left( \frac{\sqrt{\pi }}{2} \right)=\frac{-\frac{\sqrt{\pi }}{2}.\sin \frac{2\pi }{4}}{\sqrt{1+{{\cos }^{2}}\frac{\pi }{4}}}=\frac{-\frac{\sqrt{\pi }}{2}.1}{\sqrt{\frac{3}{2}}}\] \ \[f'\left( \frac{\sqrt{\pi }}{2} \right)=-\sqrt{\frac{\pi }{6}}\].
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