Question

A function is defined as follows
$f(x)=\{\begin{array}{ll}1,& \text{ when }-\infty <x<0\\ 1+\sin x,& \text{ when }0\le x<\frac{\pi }{2}\\ 2+{\left(x-\frac{\pi }{2}\right)}^2,& \text{ when }\frac{\pi }{2}\le x<\infty \end{array}$
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 $x=0,\underset{x\to 0}{\lim }f(x)=\underset{x\to 0^{-}}{\lim }(1)=1$
RHL At $x=0,\underset{x\to 0^{+}}{\lim }f(x)$
$=\underset{x\to 0}{\lim }(1+\sin x)=1$
$f(0)=1+\sin 0=1$
$=L=RHL=f(0)$
so $f(x)$ is continuous at $x=0$.
Continuity at $x=\frac{\pi }{2}$
LHL At $x=\frac{\pi }{2},\underset{x\to \frac{\pi }{2}}{\lim }f(x)$
$=\underset{x\to \frac{\pi }{2}}{\lim }(1+\sin x)=1+1=2$
RHL At $x=\frac{\pi }{2},\underset{x\to \frac{\pi }{2}}{\lim }f(x)=2+{\left(\frac{\pi }{2}-\frac{\pi }{2}\right)}^2=2$
$f\left(\frac{\pi }{2}\right)=2+{\left(\frac{\pi }{2}-\frac{\pi }{2}\right)}^2=2$
$\therefore LHL=RHL=f\left(\frac{\pi }{2}\right)$
So, $f(x)$ is continuous at $x=\frac{\pi }{2}$.
Hence, $f(x)$ is continuous over the whole real number.