Question

The function $f\left(x\right)=\left|\sin x\right|(-2\pi \le x\le 2\pi )$ is

• A. continuous everywhere
• B. differentiable everywhere
• C. monotonic increasing
• D. invertible

Answer 1

The correct option is A continuous everywhere

$f\left(x\right)=\{\begin{array}{c}+\sin x;-2\pi \le x\le -\pi \\ -\sin x;-\pi \le x<0\\ +\sin x;0\le x<\pi \\ -\sin x;\pi \le x\le 2\pi \end{array}$

At $x=-\pi$

${lim}_{x\to {-\pi }^{-}}f\left(x\right)={lim}_{x\to {-\pi }^{-}}\left(\sin x\right)=\sin (-\pi )=0$

${lim}_{x\to {-\pi }^{+}}f\left(x\right)={lim}_{x\to {-\pi }^{+}}(-\sin x)=\sin (-\pi )=0$

LHL $=$ RHL $\Rightarrow$continuous

Similarly at $x=0$

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

$\Rightarrow$continuous

at $x=\pi$

${lim}_{x\to {\pi }^{-}}f\left(x\right)={lim}_{x\to {\pi }^{+}}f\left(x\right)=0$

$\Rightarrow$continuous

at all other points $f\left(x\right)$ is continuous too.

Since $\sin x$ & $-\sin x$ are continuous functions.