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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    If f is a differentiable function satisfying \[f(x)=\int\limits_{0}^{x}{\sqrt{1-{{f}^{2}}(t)}}dt+\frac{1}{2},\] then the value of \[{{\sin }^{-1}}(f(\pi ))\] is equal to

    A) \[-\frac{\pi }{3}\]                   

    B)        \[-\frac{\pi }{6}\]               

    C) \[\frac{\pi }{6}\]                     

    D)        \[\frac{\pi }{3}\]           

    Correct Answer: B

    Solution :

      [b] \[f(x)=\int\limits_{0}^{x}{\sqrt{1-{{f}^{2}}(t)}dt+\frac{1}{2}}\] \[\Rightarrow \,\,\,f'(x)=\sqrt{1-{{f}^{2}}(x)}\] \[\Rightarrow \,\,\,\frac{f'(x)}{\sqrt{1-{{f}^{2}}(x)}}=1\] \[\Rightarrow \,\,\,{{\sin }^{-1}}(f(x))=x+c\] For\[x=0,\]  \[{{\sin }^{-1}}\left( \frac{1}{2} \right)=c\]and thus, \[c=\frac{\pi }{6}\]. \[\Rightarrow \,\,\,f(x)=\sin \left( x+\frac{\pi }{6} \right)\] \[\Rightarrow \,\,\,\,\,f(\pi )=\sin \left( -\frac{\pi }{6} \right)\] \[{{\sin }^{-1}}(f(\pi ))=-\frac{\pi }{6}\]


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