A) \[\frac{3}{4}\]
B) 0
C) \[\frac{1}{2}\]
D) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
\[f(x)={{\cos }^{-1}}\left[ \cos \left( \frac{\pi }{2}-\sqrt{\frac{1+x}{2}} \right) \right]+{{x}^{x}}\] \[f(x)=\frac{\pi }{2}-\sqrt{\frac{1+x}{2}}+{{x}^{x}}\] \[\therefore f'(x)=-\frac{1}{\sqrt{2}}.\frac{1}{2\sqrt{1+x}}+{{x}^{x}}(1+\log x)\] \[f'(1)=-\frac{1}{4}+1=\frac{3}{4}\].You need to login to perform this action.
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