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