Find the number of points of discontinuity for $[\cot^{-1} x]$ [.] is the floor function

Question

The range of $\cot ^{-1} $ is $[-\frac{\pi}{2},\frac{\pi}{2}]$

So the range for $[\cot ^{-1} ]$ is $\{-2,-1,0,1\}$

So there must be 4 points of discontinuity, but the answer says there are only three.

Which of my deduced points are incorrect?

Answer 1

Assuming erroneously the range of $\cot^{-1}$ is $(-\pi/2,\pi/2)$: consider the following plot. (Four values = three jumps.)

(Note that, at university level, you are expected to be clear on domain/codomain when you first introduce your function.)

Really, of course, if we had $\cot^{-1}: \to (0,\pi)$, as might be considered reasonable, the same thing applies: "how many times does it jump?" is the question you should ask.

Answer 2

Firstly, the range of $\cot^{−1}$ is $(0,π)$.

$[\cot^{−1}(x)]$ is discontinuous when $\cot^{−1} x = 1$ or $\cot^{−1} x =2$ or $\cot^{−1} x = 3$