A) f is continuous and differentiable at every point
B) f is continuous at every point but not differentiable
C) f is differentiable at every point
D) f is differentiable only at the origin
Correct Answer: B
Solution :
\[\operatorname{f}({{0}^{+}})=\,\,\underset{h\to 0}{\mathop{lim}}\,\,\,f(x)\,\,=\,\,\underset{h\to 0}{\mathop{lim}}\,\,f\,\left( 0+h \right)\] \[=\,\,\underset{h\to 0}{\mathop{\lim }}\,(0+h)\frac{{{e}^{1/0+h}}-{{e}^{1/0+h}}}{{{e}^{1/0+h}}+{{e}^{-1/0+h}}}=\,\,\underset{x\to 0}{\mathop{\lim }}\,h\,\frac{{{e}^{1/h}}-{{e}^{-1/h}}}{{{e}^{1/h}}+{{e}^{-1/h}}}=0\] and \[f({{0}^{-}})=\underset{h\to 0}{\mathop{lim}}\,(0-h)=\underset{x\to 0}{\mathop{lim}}\,-h\frac{{{e}^{-1/h}}-{{e}^{1/h}}}{{{e}^{-1/h}}+{{e}^{1/h}}}=0\] and \[\operatorname{f}(0)=0.\,\,\,\,\therefore \,\,f(0{{\,}^{+}})\,\,=\,\,f(0\,-)=f(0)\] Hence f is continuous at \[\operatorname{x} = 0\] At remaining points f(x) is obviously continuous. Thus it is everywhere continuous. Again, \[\operatorname{Lf}'\,(0)\,\,=\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{f(0-h)-f(0)}{-h}\] \[=\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{h.\frac{{{e}^{-1/h}}-{{e}^{1/h}}}{{{e}^{-1/h}}+{{e}^{1/h}}}-0}{-h}=-1\] \[R\,f'(0)=\underset{h\to 0}{\mathop{lim}}\,\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\frac{h\frac{{{e}^{1/h}}-{{e}^{-1/h}}}{{{e}^{1/h}}+{{e}^{-1/h}}}}{h}=1\] \[\because \,\,\,L\,f'\,(0)\ne \,\,Rf'(0)\] \[\therefore \,\,f\,is not differentiable at x = 0\]
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