Question

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

$\therefore LHL=\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }\left[1-{\left(1-h\right)}^2\right]=0$

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

$=2$

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

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

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