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. |
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