Let \(f(x) = [x^{2}]\), where [t] denotes the greatest integer less than or equal to t. The number of points in the interval \((1, 2)\) at which f is discontinuous is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 2
The greatest integer function jumps only when its input crosses an integer. Here the input is \(x^{2}\), which increases on (1, 2):
\(1 \lt x \lt 2 \ \Rightarrow \ 1 \lt x^{2} \lt 4\)
The integers strictly between 1 and 4 are 2 and 3, reached at
\(x^{2} = 2: \ x = \sqrt{2}\)
\(x^{2} = 3: \ x = \sqrt{3}\)
So f takes the values 1, 2, 3 on \((1, \sqrt{2})\), \([\sqrt{2}, \sqrt{3})\), \([\sqrt{3}, 2)\) and jumps at \(\sqrt{2}\) and \(\sqrt{3}\) only.
So there are 2 points of discontinuity.