Question

$Givenf\left(x\right)=\{\begin{array}{c}|x-1|;0\le x\le 2\\ \left[x\right];-2\le x\le 0\end{array}Whichofthefollowingfunctioniscontinuousatx=0$

• A. f(|x|)
• B. |f(x)|
• C. g(x)=|f(x)|+f(|x|)
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A f(|x|)
B |f(x)|
C g(x)=|f(x)|+f(|x|)

Here, $\underset{x\to 0^{-}}{lim}f\left(|x|\right)=[-x]=1-x=1$

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

Hence $f\left(|x|\right)$ is continuous at $x=0$

Now,

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

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

Hence $|f\left(x\right)|$ is continuous at $x=0$

Since sum of two continuous function is also continuous, hence

$g\left(x\right)=|f\left(x\right)|+f\left(|x|\right)$ is also a continuous function.