Question

A function is defined as follows
$f(x)=\begin{cases}1, & \text { when }-\infty < x< 0 \\ 1+\sin x, & \text { when } 0 \leq x < \frac{\pi}{2} \\ 2+\left(x-\frac{\pi}{2}\right)^{2}, & \text { when } \frac{\pi}{2} \leq x < \infty\end{cases}$
continuity of $f(x)$ is

• A. $ f(x) $ is continuous at $ x=\frac{\pi }{2} $
• B. $ f(x) $ is continuous at $ x=0 $
• C. $ f(x) $ is discontinuous at $ x=0 $
• D. $ f(x) $ is continuous over the whole real number

Answer 1

Answer: (D)

Continuity at $x=0$
$LHL$ At $x=0, \displaystyle\lim _{x \rightarrow 0} f(x)=\displaystyle\lim _{x \rightarrow 0^{-}}(1)=1$
RHL At $x=0, \displaystyle\lim _{x \rightarrow 0^{+}} f(x)$
$=\displaystyle\lim _{x \rightarrow 0}(1+\sin x)=1$
$f(0)=1+\sin 0=1 $
$=L=R H L=f(0)$
so $f(x)$ is continuous at $x=0$.
Continuity at $x=\frac{\pi}{2}$
LHL At $x=\frac{\pi}{2},\displaystyle \lim _{x \rightarrow \frac{\pi}{2}} f(x)$
$=\displaystyle\lim _{x \rightarrow \frac{\pi}{2}}(1+\sin x)=1+1=2$
RHL At $x=\frac{\pi}{2},\displaystyle \lim _{x \rightarrow \frac{\pi}{2}} f(x)=2+\left(\frac{\pi}{2}-\frac{\pi}{2}\right)^{2}=2$
$f\left(\frac{\pi}{2}\right)=2+\left(\frac{\pi}{2}-\frac{\pi}{2}\right)^{2}=2$
$\therefore LHL = RHL =f\left(\frac{\pi}{2}\right)$
So, $f(x)$ is continuous at $x=\frac{\pi}{2}$.
Hence, $f(x)$ is continuous over the whole real number.