A) \[f(x)\] is continuous at x = 0
B) \[f(x)\] is continuous in (-1,0)
C) \[f(x)\] is differentiable at x = 1
D) \[f(x)\] is differentiable in (-1,1)
Correct Answer: C
Solution :
If \[-1\le x\le 1,\]then \[0\le x\sin \pi x\le \frac{1}{2}\] \[\therefore \] \[f(x)=[x\sin \pi x]=0,\] for \[-1\le x\le 1\] If \[1<x<1+h,\]where h is a small positive real number, then \[\pi <\pi x<\pi +\pi h\] \[\Rightarrow \] \[-1<\sin \pi x<0\] \[\Rightarrow \] \[-1<x\sin \pi x<0\] \[\therefore \]\[f(x)=[x\sin \pi x]=-1\]in the right neigh bourhood Thus, \[f(x)\]is constant and equal to zero in \[[-1,1]\]and so\[f(x)\]is differentiable and hence continuous on \[(-1,1)\] At \[x=1,f(x)\]is discontinuous because \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=0\]and \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=-1\] Hence, \[f(x)\]is not differentiable at \[x=1\]
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