A) continuous as well as differentiable for all x
B) continuous for all x but not differentiable at \[x=0\]
C) neither differentiable nor continuous at \[x=0\]
D) discontinuous everywhere
Correct Answer: B
Solution :
A function is continuous in an interval [a, b], if it is continuous in an open interval and also\[f(a+0)=f(a)\] and \[f(b-0)=f(b)\]. Continuity at \[x=t\] \[f(0-0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,(0-h){{e}^{-\left( \frac{1}{\left| -h \right|}+\frac{1}{(-h)} \right)}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,(-h){{e}^{-\left( \frac{1}{h}-\frac{1}{h} \right)}}=\underset{h\to 0}{\mathop{\lim }}\,\,(-h)=0\] and \[f(0+0)=\underset{h\to 0}{\mathop{\lim }}\,\,(0+h){{e}^{-\left( \frac{1}{\left( \left| h \right| \right)}+\frac{1}{h} \right)}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,h{{e}^{-\left( \frac{1}{h}+\frac{1}{h} \right)}}=\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{{{e}^{2/h}}}=0\] \[\Rightarrow \] \[f(0-0)=f(0)=f(0+0)\] Therefore, /(x) is continuous for all x. Differentiability at \[x=0\] \[Rf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{h{{e}^{-\left( \frac{1}{h}+\frac{1}{h} \right)}}-0}{h-0}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{{{e}^{2/h}}}=0\] and \[Lf'(0)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{(-h){{e}^{-\left( \frac{1}{h}-\frac{1}{h} \right)}}-0}{(-h)-0}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,{{e}^{0}}=1\] \[Rf'(0)\ne Lf'(0)\] \[f(x)\] is not differentiable at\[x=0\].
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