Question

A function is defined as follows $f\left(x\right)=\{\begin{array}{cc}1,& when-\infty <x<0\\ 1+\sin x,& when0\le x<\frac{\pi }{2}\\ 2+{(x-\frac{\pi }{2})}^2& when\frac{\pi }{2}\le x<\infty \end{array}$ continuity of f(x) is

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

Answer 1

The correct option is D $f\left(x\right)$ is continuous over the whole real number

Continuity at $x=0$

LHL $=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}\left(1\right)=1$

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

$=\underset{x\to 0}{lim}(1+\sin x)=1$

And $f\left(0\right)=1+\sin 0=1$

$\therefore$ LHL $=$ RHL $=f\left(0\right)$

So, $f\left(x\right)$ is continuous at $x=0$.

Continuity at $x=\frac{\pi }{2}$

LHL $=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)$

$=\underset{x\to \frac{\pi }{2}}{lim}(1+\sin x)=1+1=2$

RHL $=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)$

$=\underset{x\to \frac{\pi }{2}}{lim}2+{(x-\frac{\pi }{2})}^2$

$=2+{(\frac{\pi }{2}-\frac{\pi }{2})}^2=2$

And $f\left(\frac{\pi }{2}\right)=2+{(\frac{\pi }{2}-\frac{\pi }{2})}^2=2$

$\therefore LHL=RHL=f\left(\frac{\pi }{2}\right)$

So, $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$.

Hence, $f\left(x\right)$ is continuous over the whole real number.