Question

The range of $f\left(x\right)=\left[\frac{1}{[x-\pi ]}\right]$, where $[\cdot ]$ denotes the greatest integer function $\le x$ is

• A. $(0,1)$
• B. $(-1,1)$
• C. $\{-1,+1\}$
• D. $\{-1,0,1\}$

Answer 1

The correct option is D $\{-1,0,1\}$

We will consider the following cases
1) $x\ge \pi +2$:
Then $\left[x-\pi \right]\ge 2$;
hence $\frac{1}{\left[x-\pi \right]}\le \frac{1}{2}$, i.e, $\left[\frac{1}{\left[x-\pi \right]}\right]=0$
2) $\pi +1\le x<\pi +2$
Then $1\le x-\pi <2$. Therefore $\left[x-\pi \right]=1$ and $\frac{1}{\left[x-\pi \right]}=1$, i.e, $\left[\frac{1}{\left[x-\pi \right]}\right]=1$
3) $\pi \le x<\pi +1$
Since $\left[x-\pi \right]=0$, the function is not defined in the region.
4)$\pi -1\le x<\pi$
Then $-1\le x-\pi <0$. Therefore $\left[x-\pi \right]=-1$ and $\frac{1}{\left[x-\pi \right]}=-1$, i.e, $\left[\frac{1}{\left[x-\pi \right]}\right]=-1$
5) $x<\pi -1$
Then $\left[x-\pi \right]\le -2$;
hence $0>\frac{1}{\left[x-\pi \right]}\ge -\frac{1}{2}$, i.e, $\left[\frac{1}{\left[x-\pi \right]}\right]=-1$
Therefore the only possible values the function can attend is $\left\{-1,0,1\right\}$