Question

The number of real solutions of $(x,y)$, where is $\left|y\right|=\sin x,y={\cos }^{-1}\left(\cos x\right),-2\pi \le x\le 2\pi ,$ is

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$\left|y\right|=\sin x,y={\cos }^{-1}\left(\cos x\right),-2\pi \le x\le 2\pi$
$\left|y\right|=\sin x----1y={\cos }^{-1}\left(\cos x\right),-2\pi \le x\le 2\pi ----2\to y=xwhen0\le x\le \pi \to y=2\pi -x\pi \le x\le 2\pi \to y=-2\pi -x-2\pi \le x\le -\pi \to y=-x-\pi \le x\le 0$
from (1) and (2)
three different cases are $|x|=\sin x$, $|2\pi -x|=\sin x$ and $|2\pi +x|=\sin x$

For the above simultaneous equations, we get exactly 3 solutions.
For $x=0$ we get $y=0$ for both the equations,
For $x=k\left(2\pi \right)$ where $k$ is an integer, we get $y=0$ in both the cases.
Since $xϵ[-2\pi ,2\pi ]$ we get
$x=-2\pi$ and $x=2\pi$.
Hence 3 solutions,
at $x=0$
$x=-2\pi$ and
$x=2\pi$
Hence, no of real solutions is 3