Question

The value of $x$ for which the function
$f\left(x\right)=\{\begin{array}{c}(1-x),x<1\\ (1-x)(2-x),1\le x\le 2\\ (3-x),x>2\end{array}$
fails to be continuous or differentiable, is

• A. $1$
• B. $2$
• C. $1,2$
• D. $3$

Answer 1

Answer: (B)

$2$

We have $\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}(1-x)=0$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(1-x)(2-x)=0$
and $f\left(1\right)=0$
$\therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=f\left(1\right)$
So, $f\left(x\right)$ is continuous at $x=1$
Now, (LHD at $x=1)={\left[\frac{d}{dx}\right(1-x\left)\right]}_{x=1}=-1$
(RHD at $x=1)=={\left[\frac{d}{dx}\right(1-x\left)\right(2-x\left)\right]}_{x=1}=-1$
$\therefore$ (LHD at $x=1)=$ (RHD at $x=1)$
so, $f\left(x\right)$ is differentiable at $x=1$
Now, $\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}(1-x)(2-x)=0$
and $\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}(3-x)=1$
$\therefore \underset{x\to 2^{-}}{lim}f\left(x\right)\ne \underset{x\to 2^{+}}{lim}f\left(x\right)$
Thus, $f\left(x\right)$ is discontinuous and not differentiable at $x=2$
Hence, the only point of discontinuity and non-differentiability of $f\left(x\right)$ is $x=2$