Search any question & find its solution
Question:
Answered & Verified by Expert
A function is defined as follows:
$f(\mathrm{x})=\left\{\begin{array}{cl}-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}}} & , \mathrm{x} \neq 0 \\ 0, & \mathrm{x}=0\end{array}\right.$
Which one of the following is correct in respect of the above function?
Options:
$f(\mathrm{x})=\left\{\begin{array}{cl}-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}}} & , \mathrm{x} \neq 0 \\ 0, & \mathrm{x}=0\end{array}\right.$
Which one of the following is correct in respect of the above function?
Solution:
2730 Upvotes
Verified Answer
The correct answer is:
$f(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{-\mathrm{x}}{\sqrt{\mathrm{x}^{2}}}, \mathrm{x} \neq 0 \\ 0, \mathrm{x}=0\end{array}\right.$
$=\left\{\begin{array}{l}\frac{-\mathrm{x}}{|\mathrm{x}|}, \mathrm{x} \neq 0 \\ 0, \mathrm{x}=0\end{array}\right.$
$\mathrm{f}(0+\mathrm{h})=\frac{-\mathrm{h}}{\mathrm{h}}=-1 ; \mathrm{f}(0-\mathrm{h})=\frac{\mathrm{h}}{\mathrm{h}}=1$
So, $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$.
$=\left\{\begin{array}{l}\frac{-\mathrm{x}}{|\mathrm{x}|}, \mathrm{x} \neq 0 \\ 0, \mathrm{x}=0\end{array}\right.$
$\mathrm{f}(0+\mathrm{h})=\frac{-\mathrm{h}}{\mathrm{h}}=-1 ; \mathrm{f}(0-\mathrm{h})=\frac{\mathrm{h}}{\mathrm{h}}=1$
So, $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.