A) \[\underset{x\to 0}{\mathop{\lim }}\,\,f(x)\]does not exist
B) \[f(x)\] is continuous at \[x=0\]
C) \[f(x)\] is non-differentiable at \[x=0\]
D) \[f(0)=1\]
Correct Answer: B
Solution :
We have, \[Lf'(0)=\,\underset{h\to \infty }{\mathop{\lim }}\,\,\frac{f(0-h)\,-f(0)}{-h}\] \[=\,\underset{h\to \infty }{\mathop{\lim }}\,\,\frac{[{{\tan }^{2}}(-h)]-(0)}{-h}\] \[=\,\underset{h\to \infty }{\mathop{\lim }}\,\,\frac{[{{\tan }^{2}}h]}{-h}=0\] \[Rf'(0)=\underset{h\to \infty }{\mathop{\lim }}\,\,\frac{f(0+h)-f(0)}{h}\] \[=\underset{h\to \infty }{\mathop{\lim }}\,\,\frac{[{{\tan }^{2}}h]-0}{h}=0\] \[\therefore \] \[Lf'(0)=Rf'(0)\] Hence, \[f(x)\] is differentiable and hence continuous at \[x=0,\] then \[f'(0)=0\].
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