Question

The number of integral values of $x$ satisfying $||x-\pi |-|\pi x-1||=(x-1)(1+\pi )$, is

Answer 1

Let $x-\pi =a$ and $\pi x-1=b$.
Then, $a+b=(x-1)(1+\pi )$
So, the equation is of the form $||a|-|b||=|a+b|,$ which is possible only if $ab\le 0$
So, $(x-\pi )(\pi x-1)\le 0$
$\Rightarrow \pi (x-\pi )(x-\frac{1}{\pi })\le 0$
$\Rightarrow x\in [\frac{1}{\pi },\pi ]$
Possible integral values of $x$ are $1,2,3$
So, number of integral values of $x$ is $3$