A) f is discontinuous
B) f is continuous only, if \[\lambda =0\]
C) f is continuous only, whatever \[\lambda \] may be
D) None of these
Correct Answer: A
Solution :
[a] As we know. |
A function f(x) is said to be continuous as a point x = a iff |
\[\underset{x\to a}{\mathop{\lim }}\,\,\,f(x)=f(a),\] otherwise not continuous. |
Thus f(x) is continuous at x = a iff |
\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\,\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)\] |
Since, \[f(x)=\left\{ \begin{matrix} {{5}^{1/x}}, & x<0 \\ \lambda [x], & x\ge 0 \\ \end{matrix}and\,\,\lambda \in R \right.\] |
RHL \[at\,x=0:\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,f(x)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\lambda [x]\] |
\[=\underset{h\to 0}{\mathop{\lim }}\,\lambda [h]=0\] |
LHL at \[x=0:\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\,f(x)=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,{{5}^{1/x}}\] |
\[=\underset{h\to 0}{\mathop{\lim }}\,\,{{5}^{-1/h}}={{5}^{\infty }}=\infty \] |
and \[f(0)=\lambda [0]=0.\] |
Since, LHL\[\ne \]RHL |
\[\therefore \] \[f(x)\] is not continuous. |
You need to login to perform this action.
You will be redirected in
3 sec