Question

Which of the following options is (are) true regarding the function $f\left(x\right)=\left|\sin (x-\frac{\pi }{3})\right|$?

• A. Period of $f\left(x\right)$ is $\frac{2\pi }{3}$
• B. Period of $f\left(x\right)$ is $\pi$
• C. Range of $f\left(x\right)$ is $[0,1]$
• D. $f\left(x\right)=0$ has only one solution in $[0,\pi ]$

Answer 1

Answer: (B), (C), (D)

The correct options are
B Period of $f\left(x\right)$ is $\pi$
C Range of $f\left(x\right)$ is $[0,1]$
D $f\left(x\right)=0$ has only one solution in $[0,\pi ]$

To obtain the graph of $\sin (x-\frac{\pi }{3}),$ shift the graph of $\sin x$ to right by $\frac{\pi }{3}$


Now, the graph of $f\left(x\right)=\left|\sin (x-\frac{\pi }{3})\right|$ is


Period does not change on shifting the graph and from the graph, we observe that graph is repeated after the interval of $\pi .$
So, period of $f\left(x\right)$ is $\pi .$
Clearly, range of $f\left(x\right)$ is $[0,1]$
$f\left(x\right)=0$ has only one solution $x=\frac{\pi }{3}$ in $[0,\pi ]$