Question

$f\left(x\right)=\{\begin{array}{cc}x{e}^{-(\frac{1}{x}+\frac{1}{|x|})},& x\ne 0\\ a,& x=0\end{array}$.

The value of $a$, such that $f\left(x\right)$ is differentiable at $x=0$, is equal to?

• A. $1$
• B. $-1$
• C. $0$
• D. None of the above

Answer 1

The correct option is D None of the above

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

at $x=0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}x{e}^{-(\frac{1}{x}+\frac{1}{x})}$

$=\underset{x\to 0^{-}}{lim}x=0$

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}x{e}^{-(\frac{1}{x}+\frac{1}{x})}$

$=\underset{x\to 0^{+}}{lim}\frac{x}{e^{2/x}}$

$x\to 0^{+}\to \frac{1}{x}\to \infty$

$=0.0=0$

$f^{\prime }\left(x\right)=\{\begin{array}{c}1;x<0\\ 0;x=0\\ x{e}^{1/x}\left(\frac{-2}{x^2}\right)+e^{2/x};x>0\end{array}$

at $x=0$

$\underset{x\to 0^{-}}{lim}{f}^{\prime }\left(x\right)=1$

$f^{\prime }\left(0\right)=0$

LHL$\ne f^{\prime }\left(0\right)\Rightarrow$f is not differentiable

$\Rightarrow$a cannot be defined.