Question

Let the function $f\left(x\right)$ be defined as follows $f\left(x\right)=\{\begin{array}{c}|x|\forall x\in [-2,2]\\ \left[x\right]\forall x\in [2,4]\end{array}$, where $[\cdot ]$ represents the greatest integer function. Then

• A. $f\left(x\right)$ is continuous at $x=2$
• B. $f\left(x\right)$ is differentiable at $x=2$
• C. $f\left(x\right)$ is differentiable in $[-2,4]$
• D. $f\left(x\right)$ is continuous in $[-2,4]$

Answer 1

The correct option is A $f\left(x\right)$ is continuous at $x=2$

$f\left(x\right)=\{\begin{array}{cc}\left|x\right|& for-2\le x\le 2\\ \left[x\right]& for2<x\le 4\end{array}$
$LHL=\underset{x\to 2^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(2-h)$
$=\underset{h\to 0}{lim}|2-h|$
$=2$
$RHL=\underset{x\to 2^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(2+h)$
$=\underset{h\to 0}{lim}[2+h]$
$=2$
Also, $f\left(2\right)=|2|=2$
Hence, $LHL=RHL=f\left(2\right)$
Hence, function is continuous at $x=2$
But $f\left(x\right)$ is not continuous in $[-2,4]$ as greatest integer function is discontinuous at integers.
Now, differentiability at $x=2$
$LHD=\underset{h\to 0}{lim}\frac{f(2-h)-f\left(2\right)}{-h}$
$=\underset{h\to 0}{lim}\frac{|2-h|-2}{-h}$
$\underset{h\to 0}{lim}\frac{2-h-2}{-h}$
$=1$
$RHD=\underset{h\to 0}{lim}\frac{f(2+h)-f\left(2\right)}{h}$
$=\underset{h\to 0}{lim}\frac{[2+h]-2}{h}$
$=0$
Since, $LHD\ne RHD$
Hence, $f\left(x\right)$ is not differentiable at $x=2$.So not differentiable in $[-2,4]$