Question

If $f\left(x\right)=\{\begin{array}{c}x{e}^{-(\frac{1}{|x|}+\frac{1}{x})},x\ne 0\\ 0,x=0\end{array}$, then $f\left(x\right)$ is

• A. continuous for all $x$ but is not differentiable
• B. neither differentiable nor continuous
• C. discontinuous everywhere
• D. continuous as well as differentiable for all $x$

Answer 1

The correct option is A continuous for all $x$ but is not differentiable

For $x>0$, $|x|=x$
$\Rightarrow f\left(x\right)=x{e}^{\frac{-2}{x}}$

For $x<0$, $|x|=-x$
$\Rightarrow f\left(x\right)=x$

For $x=0$, $f\left(x\right)=0$

Now as $x\to 0$
R.H.L at $x=0$, $f\left(x\right)=x{e}^{\frac{-2}{x}}\Rightarrow 0$
L.H.L at $x=0$, $f\left(x\right)=x\Rightarrow 0$
And $f\left(0\right)=0$

$\because f\left(0^{-}\right)=f\left(0^{+}\right)=f\left(0\right)$
Hence, the function is continuous.

Now let us check for the differentiability at $x=0$
L.H.D at $x=0$, $f^{\prime }\left(x\right)=1$ ($\because f\left(x\right)=x$)
R.H.D at $x=0$, $f^{\prime }\left(x\right)=e^{-\frac{2}{x}}+\frac{2}{x}{e}^{-\frac{2}{x}}\Rightarrow 0$
$\because L.H.D\ne R.H.D$
Function is not differentiable at $x=0$

Hence, option A.