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JEE Main & Advanced AIEEE Solved Paper-2012

  • question_answer
    If f : \[R\to R\] is a function defined by \[f(x)=[x]\cos \left( \frac{2x-1)}{2} \right)\pi \], where \[[x]\] denotes the greatest integer function, then f is:   AIEEE  Solved  Paper-2012

    A) continuous for every real \[x\].

    B) discontinuous only at \[x=0\].

    C) discontinuous only at non-zero integral values of \[x\].

    D)              continuous only at \[x=0\].

    Correct Answer: A

    Solution :

                 Doubtful points are \[x=n,\,\,n\in I\] L.H.L \[=\underset{x\to {{n}^{-}}}{\mathop{\lim }}\,[x]\cos \left( \frac{2x-1}{2} \right)\pi =(n-1)\cos \] \[\left( \frac{2n-1}{2} \right)\pi =0\] R.H.L. \[=\underset{x\to {{n}^{+}}}{\mathop{\lim }}\,[x]\cos \left( \frac{2n-1}{2} \right)\pi =n\,\,\,\cos \]    \[\left( \frac{2n-1}{2} \right)\pi =0\]              \[f(n)=0\]              Hence continuous


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