$f(x)=\left\{\begin{matrix}xe^{-(\frac{1}{x}+\frac{1}{|x|})},&x≠0\\a,&x=0\end{matrix}\right.$. Value of ‘a’ such that f(x) is differentiable at x = 0, is equal to

none of these

$\underset{x→0^-}{\lim}xe^{-\frac{1}{x}+\frac{1}{x}}=x×1=0=LHL$

$\underset{x→0^+}{\lim}xe^{-\frac{1}{x}-\frac{1}{x}}=x×e^{-∞}=0=RHL$

$⇒a=0$ for continuity

$LHD=f'(0^-)=\underset{h→0^-}{\lim}\frac{-he^{-(-\frac{1}{h}+\frac{1}{h})-a}}{-h}=\underset{x→0}{\lim}\frac{-h-a}{-h}=1$

$RHD=f'(0^+)=\underset{h→0}{\lim}\frac{he^{-(\frac{1}{h}+\frac{1}{h})-a}}{h}=0$

$RHD≠LHD$ ⇒ function is not differentiable at x = 0