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 ?

Question

cdquestions Admin · Accepted Answer

The correct option is (A) : f(x) is continuous and f'(0) does not exist.

Answer

f(x) is continuous and f'(0) does not exist

Answer

f(x) is not continuous

Answer

f(x) is continuous and f'(0) also exists

Answer

None of these

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 ?

The Correct Option is A

Solution and Explanation

Questions Asked in BCECE exam

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 ?

The Correct Option is A

Solution and Explanation

Questions Asked in BCECE exam

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 ?