Question

Discuss the continuity of $f\left(x\right)=\{\begin{array}{c}|x-3|;0\le x<1\\ \sin x;1\le x\le \frac{\pi }{2}\\ {\log }_{\frac{\pi }{2}}x;\frac{\pi }{2}<x<3\end{array}$ in $[0,3)$

• A. Discontinuous at $x=1$
• B. Continuous at $x=1$
• C. continuous at $x=\frac{\pi }{2}$
• D. Discontinuous at $x=\frac{\pi }{2}$

Answer 1

Answer: (A), (C)

The correct options are
A Discontinuous at $x=1$
C continuous at $x=\frac{\pi }{2}$

$f\left(x\right)=\{\begin{array}{c}|x-3|0\le x<1\\ \sin x1\le x\le \frac{\pi }{2}\\ {\log }_{\frac{\pi }{2}}x\frac{\pi }{2}<x<3\end{array}$

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}|x-3|=\underset{x\to 1^{-}}{lim}3-x=2$

$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}sinx=sin1$

So $f\left(x\right)$ is discontinous at $x=1$

$\underset{x\to {\frac{\pi }{2}}^{-}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{-}}{lim}sinx=1$

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}{\log }_{\frac{\pi }{2}}x=1$

Discontinuous at $x=1$ and continuous at $x=\frac{\pi }{2}$