Question

Let the curves $f\left(x\right)=\sin 3x$ and $g\left(x\right)=\cos x$ and $-\frac{\pi }{2}\le x\le \frac{\pi }{2}$

• A. number of solutions of $f\left(x\right)=g\left(x\right)$ is $3$
• B. number of solutions of $f\left(x\right)=g\left(x\right)$ is $4$
• C. sum of solutions of $f\left(x\right)=g\left(x\right)$ is $0$
• D. sum of solutions of $f\left(x\right)=g\left(x\right)$ is $\frac{2\pi }{3}$

Answer 1

Answer: (A), (C)

The correct options are
A number of solutions of $f\left(x\right)=g\left(x\right)$ is $3$
C sum of solutions of $f\left(x\right)=g\left(x\right)$ is $0$

At the intersection pointof $y=\cos x$ and $y=\sin 3x,$ we have $\cos x=\sin 3x$
$\Rightarrow \cos x=\cos (\frac{\pi }{2}-3x)\Rightarrow x=2n\pi \pm \frac{\pi }{2}\mp 3x\Rightarrow x=\frac{n\pi }{2}+\frac{\pi }{8},-n\pi +\frac{\pi }{4},n\in Z$
$\Rightarrow x=-\frac{3\pi }{8},\frac{\pi }{4},\frac{\pi }{8}$
$[\therefore -\pi /2\le x\le \pi /2]$