Differentiation Question 33
Question: If $ f(x)=\sqrt{1+{{\cos }^{2}}(x^{2})} $ , then $ f’( \frac{\sqrt{\pi }}{2} ) $ is
[Orissa JEE 2004]
Options:
A) $ \sqrt{\pi }/6 $
B) $ -\sqrt{(\pi /6)} $
C) $ 1/\sqrt{6} $
D) $ \pi /\sqrt{6} $
Show Answer
Answer:
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 2x^{2}}{\sqrt{1+{{\cos }^{2}}(x^{2})}} $
At $ x=\frac{\sqrt{\pi }}{2},f’( \frac{\sqrt{\pi }}{2} )=\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’( \frac{\sqrt{\pi }}{2} )=-\sqrt{\frac{\pi }{6}} $ .
BETA
