Question

Let $g\left(x\right)=\{\begin{array}{cc}{e}^{2x}& \forall x<0\\ e^{-2x}& \forall x\ge 0\end{array}$ then g(x) does not satisfies the condition

• A. continuous $\forall x\in R$
• B. not differentiable at $x=0$
• C. continuous $\forall x\in R$ and non differentiable at $x=0$
• D. $g\left(x\right)$ is continuous & differentiable everywhere

Answer 1

Answer: (C)

continuous $\forall x\in R$ and non differentiable at $x=0$

$g\left(x\right)=\{\begin{array}{c}\begin{array}{cc}{e}^{2x},& x<0\end{array}\\ \begin{array}{cc}{e}^{-2x}.& x\ge 0\end{array}\end{array}$

then $g^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}2e^{2x},& x<0\end{array}\\ \begin{array}{cc}-2e^{-2x},& x\le 0\end{array}\end{array}$

$\therefore L.H.D$ at $x=0$ is $2e^0=2$

and $R.H.D$ at $x=0$ is $-2e^0=-2$

As $L.H.D\ne R.H.D$ at $x=0$

Again $\underset{x\to 0^{-}}{lim}g\left(x\right)=\underset{x\to 0}{lim}{e}^{2x}=1$

and $\underset{x\to 0^{+}}{lim}g\left(x\right)=\underset{x\to 0}{lim}{e}^{-2x}=1$

$\because \underset{x\to 0^{-}}{lim}g\left(x\right)=\underset{x\to 0^{+}}{lim}g\left(x\right)$

$\therefore g\left(x\right)$ is continuous at $x=0,$

Moreover $g\left(x\right)$ is continuous $\forall x\in R$