Question

Discuss the continuity of the function $f(x)=\sin 2x-1$ at the point $x=0,$ and $x=\pi$ .

• A. Continuous at $x=0,\pi$
• B. Discontinuous at $x=0$ but continuous at $x=\pi$
• C. Continuous at $x=0$ but discontinuous at $x=\pi$
• D. Discontinuous at $x=0,\pi$

Answer 1

Answer: (A)

Given, $f(x)=\sin 2x-1$ At $x=0,$ $\underset{x\to 0^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0-h)$
$=\underset{h\to 0}{\lim }[-\sin 2h-1]$
$=0-1=-1$ $\underset{x\to 0^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(0+h)$
$=\underset{h\to 0}{\lim }\sin 2h-1$
$=0-1=-1$ and $f(0)=sin0-1=-1$ $\because$ $\underset{x\to 0^{-}}{\lim }f(x)=\underset{x\to 0^{+}}{\lim }f(x)=f(0)$
$\therefore$ $f(x)$ is continuous at $x=0$ Now, at $x=\pi$ $\underset{x\to {\pi }^{-}}{\lim }f(x)=\underset{h\to 0}{\lim }f(\pi -h)$
$=\underset{h\to 0}{\lim }\sin 2(\pi -h)-1$
$=\underset{h\to 0}{\lim }-\sin 1h-1=-1$ $\underset{x\to {\pi }^{+}}{\lim }f(x)=\underset{h\to 0}{\lim }f(\pi +h)$
$=\underset{h\to 0}{\lim }\sin 2(\pi +h)-1$
$=\underset{h\to 0}{\lim }\sin 2h-1$
$=0-1=-1$ and $f(\pi )=\sin 2\pi -1=-1$ $\because$ $\underset{x\to {\pi }^{-}}{\lim }f(x)=\underset{x\to {\pi }^{+}}{\lim }f(x)=f(\pi )$
$\therefore$ $f(x)$ is continuous at $x=\pi$ also.