Question

Let $f\left(x\right)=\cos (x+\frac{\pi }{3})$ and $g\left(x\right)=\sin (x-\frac{\pi }{6})$ be two functions. Which of the following is/are correct ?
( where $x\in [0,2\pi ]$ )

• A. $f\left(x\right)=g\left(x\right)$ has two solutions
• B. Period of $f\left(x\right)$ is $\frac{5\pi }{3}$
• C. Period of $g\left(x\right)$ is $\frac{11\pi }{6}$
• D. $g\left(x\right)>f\left(x\right)$ when $x\in (\frac{\pi }{6},\frac{7\pi }{6})$

Answer 1

Answer: (A), (D)

The correct options are
A $f\left(x\right)=g\left(x\right)$ has two solutions
D $g\left(x\right)>f\left(x\right)$ when $x\in (\frac{\pi }{6},\frac{7\pi }{6})$

Draw the graph of $f\left(x\right)=\cos (x+\frac{\pi }{3})$ and $g\left(x\right)=\sin (x-\frac{\pi }{6})$


From graph, the number of solutions of $f\left(x\right)=g\left(x\right)$ is $2$ in $x\in [0,2\pi ].$

Since, on shifting the graph, period does not change.
Hence, period of both $f\left(x\right)$ and $g\left(x\right)$ is $2\pi .$

From graph, $\sin (x-\frac{\pi }{6})>\cos (x+\frac{\pi }{3})$ when $x\in (\frac{\pi }{6},\frac{7\pi }{6})$