JEE MAIN - Mathematics (2003 - No. 9)
If $$f(x) = \left\{ {\matrix{
{x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}} & {,x \ne 0} \cr
0 & {,x = 0} \cr
} } \right.$$
then $$f(x)$$ is
then $$f(x)$$ is
discontinuous everywhere
continuous as well as differentiable for all x
continuous for all x but not differentiable at x = 0
neither differentiable nor continuous at x = 0
Explanation
$$f\left( 0 \right) = 0;\,\,f\left( x \right) = x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}$$
$$R.H.L.\,\,$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - 2/h}}$$
$$ = \,\mathop {\lim }\limits_{h \to 0} {h \over {{e^{2/h}}}} = 0$$
$$L.H.L.$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} = 0$$
therefore, $$f(x)$$ is continuous,
$$R.H.D=$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} {{\left( {0 + h} \right){e^{ - \left( {{1 \over h} + {1 \over h}} \right)}} - 0} \over h} = 0$$
$$L.H.D.$$ $$\,\,\, = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} - 0} \over { - h}} = 1$$
therefore, $$L.H.D. \ne R.H.D.$$
$$f(x)$$ is not differentiable at $$x=0.$$
$$R.H.L.\,\,$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - 2/h}}$$
$$ = \,\mathop {\lim }\limits_{h \to 0} {h \over {{e^{2/h}}}} = 0$$
$$L.H.L.$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} = 0$$
therefore, $$f(x)$$ is continuous,
$$R.H.D=$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} {{\left( {0 + h} \right){e^{ - \left( {{1 \over h} + {1 \over h}} \right)}} - 0} \over h} = 0$$
$$L.H.D.$$ $$\,\,\, = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} - 0} \over { - h}} = 1$$
therefore, $$L.H.D. \ne R.H.D.$$
$$f(x)$$ is not differentiable at $$x=0.$$
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