JEE Main & Advanced Sample Paper JEE Main - Mock Test - 5

  • question_answer
    If \[f(x)={{\cos }^{-1}}\left( \sin \sqrt{\frac{1+x}{2}} \right)+{{x}^{x}},\] then at \[x=1,\]\[f(x)\] is equal to

    A)  \[0\]                            

    B) \[1/2\]        

    C) \[-1/2\]                        

    D) \[3/4\]

    Correct Answer: D

    Solution :

    Let \[y={{\cos }^{-1}}\left( \sin \sqrt{\frac{1+x}{2}} \right)+{{x}^{x}}\]
    \[={{\cos }^{-1}}\cos \left( \frac{\pi }{2}-\sqrt{\frac{1+x}{2}} \right)+{{x}^{x}}=\frac{\pi }{2}-\sqrt{\frac{1+x}{2}}+{{x}^{x}}\]
    \[f'(x)=\frac{dy}{dx}=0-\frac{1}{2\sqrt{\frac{1+x}{2}}}\times \frac{1}{2}+{{x}^{x}}(1+\log x)\]
    At \[x=1\,f'(x)=-\frac{1}{2\sqrt{\frac{1+1}{2}}}\times \frac{1}{2}+1(1+0)\]
    \[=-\frac{1}{4}+1=3/4\]


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