Let \(f(x) = [x]\), where [x] denotes the greatest integer less than or equal to x. The number of points in the interval \((-2, 3)\) at which f is discontinuous is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4
The greatest integer function is constant between two consecutive integers, so it is continuous at every non-integer point.
At an integer n, it jumps:
\( \displaystyle \lim_{x \to n^-} [x] = n - 1, \quad \lim_{x \to n^+} [x] = n \)
So f is discontinuous exactly at the integers. The integers inside the open interval \((-2, 3)\) are
\(-1, \ 0, \ 1, \ 2\)
The end points −2 and 3 are not in the interval. So there are 4 such points.