A) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\] does not exist
B) \[f(x)\]is continuous at\[x=0\]
C) \[f(x)\] is not differentiate at\[x=0\]
D) \[f'(0)=1\]
Correct Answer: B
Solution :
We have, \[-\frac{\pi }{4}<x<\frac{\pi }{4}\] \[\Rightarrow \] \[-1<\tan x<1\] \[\Rightarrow \] \[0\le {{\tan }^{2}}x<1\] \[\Rightarrow \] \[[{{\tan }^{2}}x]=0\] \[\therefore \] \[f(x)=[{{\tan }^{2}}x]=0,\,\,\forall x\in \left( -\frac{\pi }{4},\,\,\frac{\pi }{4} \right)\] Thus, \[f(x)\] is constant function on\[\left( -\frac{\pi }{4},\,\,\frac{\pi }{4} \right)\]. Hence, it is continuous on\[\left( -\frac{\pi }{4},\,\,\frac{\pi }{4} \right)\]and\[f'(x)=0\]\[,\forall x\in \left( -\frac{\pi }{4},\,\,\frac{\pi }{4} \right)\].
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