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 continuousYou need to login to perform this action.
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