Question

Let $f\left(x\right)=\{\begin{array}{cc}-2,& -3\le x\le 0\\ x-2,& x<x\le 3\end{array}$ and $g\left(x\right)=f\left(|x|\right)+|f\left(x\right)|$

Which of the following statements are correct?
1. $g\left(x\right)$ is continuous at $x=0$.
2. $g\left(x\right)$ is continuous at $x=2$.
3. $g\left(x\right)$ is continuous at $x=-1$.
Select the correct answer using the code given below

• A. 1 and 2 only
• B. 2 and 3 only
• C. 1 and 3 only
• D. 1, 2 and 3

Answer 1

Answer: (D)

1, 2 and 3

at $x=0$,

$\underset{x\to 0^{+}}{lim}g\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(|x|\right)+|f\left(x\right)|=-2+2=0$

$\underset{x\to 0^{-}}{lim}g\left(x\right)=\underset{x\to 0^{-}}{lim}f\left(|x|\right)+|f\left(x\right)|=-2+2=0$

Hence, it is continuous at $x=0$.

At $x=2$,

$\underset{x\to 2^{+}}{lim}g\left(x\right)=\underset{x\to 2^{+}}{lim}f\left(|x|\right)+|f\left(x\right)|=0$

$\underset{x\to 2^{-}}{lim}g\left(x\right)=\underset{x\to 2^{-}}{lim}f\left(|x|\right)+|f\left(x\right)|=0$

Hence, the function is continuous at $x=2$

At $x=-1$,

$\underset{x\to -1^{+}}{lim}g\left(x\right)=\underset{x\to -1^{+}}{lim}f\left(|x|\right)+|f\left(x\right)|=-1+2=1$

$\underset{x\to -1^{-}}{lim}g\left(x\right)=\underset{x\to -1^{-}}{lim}f\left(|x|\right)+|f\left(x\right)|=-1+2=1$

Hence, the function is continuous at $x=-1$