JEE MAIN - Mathematics (2025 - 7th April Morning Shift - No. 23)

The number of points of discontinuity of the function $f(x)=\left[\frac{x^2}{2}\right]-[\sqrt{x}], x \in[0,4]$, where $[\cdot]$ denotes the greatest integer function, is ________.
Answer
8

Explanation

To determine the points of discontinuity of the function $ f(x) = \left[\frac{x^2}{2}\right] - [\sqrt{x}] $, where $[\cdot]$ denotes the greatest integer function, we need to identify possible values of $ x $ where discontinuities might occur within the interval $[0,4]$.

Discontinuity Analysis

For the term $\left[\frac{x^2}{2}\right]$:

The probable values of $ x $ that could cause discontinuities are the roots or specific values where the integer part changes between consecutive integers. The transitions happen when:

$ \begin{aligned} & = 1, 2, 3, 4, 5, 6, 7, 8 \\ & \implies x = \sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}, \sqrt{10}, 2\sqrt{3}, \sqrt{14}, 4 \end{aligned} $

For the term $[\sqrt{x}]$:

The values of $ x $ where $[\sqrt{x}]$ changes are straightforward. They occur at:

$ x = 1, 2 $

Discontinuity Check

By evaluating $ f(x) $ at all these potential points, we find the function is indeed discontinuous at:

$ x = 1, \sqrt{2}, 2, \sqrt{6}, 2\sqrt{2}, \sqrt{10}, 2\sqrt{3}, \sqrt{14} $

Thus, the function $ f(x) $ has 8 discontinuities on the interval $[0,4]$.

Comments (0)

Advertisement