Question

The number of integer $x$ satisfying $si{n}^{-1}|x-2|+co{s}^{-1}(1-|3-x|)=\frac{\pi }{2}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

If $x<2$
$si{n}^{-1}(2-x)+co{s}^{-1}(1-(3-x\left)\right)$
$=si{n}^{-1}(2-x)+co{s}^{-1}(-2+x)$
$=si{n}^{-1}(2-x)+\pi -co{s}^{-1}(2-x)$
$=\frac{\pi }{2}$
$-\frac{pi}{2}+si{n}^{-1}(2-x)+\pi +si{n}^{-1}(2-x)=\frac{\pi }{2}$
$2si{n}^{-1}(2-x)=0$
$x=2$ ...(i)
For $2<x<3$
$si{n}^{-1}(x-2)+co{s}^{-1}(-2+x)$
$=si{n}^{-1}(x-2)+co{s}^{-1}(x-2)$
$=\frac{\pi }{2}$
For $x>3$
$si{n}^{-1}(x-2)+co{s}^{-1}(4-x)=\frac{\pi }{2}$
$x-2=4-x$
$2x=6$
$x=3$
Hence there are in total 2, solutions.