Question:
If \(f(x)=\begin{cases} xe^{-(\frac{1}{|x|}+\frac{1}{x})}; & \text{if } x\ne0 \text{ then} \\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ ; & \text{if }x=0\end{cases}\)
which of the following is correct ?
If \(f(x)=\begin{cases} xe^{-(\frac{1}{|x|}+\frac{1}{x})}; & \text{if } x\ne0 \text{ then} \\ 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ ; & \text{if }x=0\end{cases}\)
which of the following is correct ?
which of the following is correct ?
Updated On: Sep 15, 2024
- f(x) is continuous and f'(0) does not exist
- f(x) is not continuous
- f(x) is continuous and f'(0) also exists
- None of these
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The Correct Option is A
Solution and Explanation
The correct option is (A) : f(x) is continuous and f'(0) does not exist.
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