Question

A function $f(x )$ is defined as follows for real $x$
$f(x) = \begin{cases} 1-x^2, & \quad \text{for} \,\,x < 1 \\ 0, & \quad \text{for } x = 1 \\ 1+x^2, & \quad \text{for } x > 2 \end{cases} $ Then

• A. $f(x)$ is not continuous at $x = 1$
• B. $f(x)$ is continuous but not differentiable at $x = 1$
• C. $f(x )$ is both continuous and differentiable at $ x=1$
• D. None of the above

Answer 1

Answer: (A)

Since,
$f(x) =
\begin{cases}
1-x^2, & \quad \text{for} \,\,x < 1 \\
0, & \quad \text{for } x = 1 \\
1+x^2, & \quad \text{for } x > 2
\end{cases} $

$\therefore LHL = \displaystyle\lim_{x\to1^{-}} f\left(x\right) = \displaystyle \lim_{h\to0} \left[1 -\left(1-h\right)^{2}\right] = 0$

and $RHL = \displaystyle\lim_{x\to 1^{+}} f\left(x\right) = \displaystyle\lim _{h\to 0} \left\{ 1+\left(1+h\right)^{2}\right\} $

$ =2 $

Also, $f\left(1\right)= 0 $

$\Rightarrow RHL \ne LHL = f\left(1\right) $

$ \Rightarrow f\left(x\right)$ is not continuous at $x = 1$.