JEE Mathematics 2025 Question with Solution

Given: $f(x) = \left\lfloor \frac{x^2}{2} \right\rfloor - \left\lfloor \sqrt{x} \right\rfloor$ on $[0,4]$.

Find: The number of points where $f(x)$ is discontinuous.

The solution checks the points where the expressions inside the greatest integer functions become integers.

For $\left\lfloor \frac{x^2}{2} \right\rfloor$, discontinuity occurs when

$$\frac{x^2}{2} = k, \quad k \in \mathbb{Z}$$

So

$$x^2 = 2k$$

For $x \in [0,4]$, the listed points are

$$x = 0, \sqrt{2}, 2, \sqrt{6}$$

For $\left\lfloor \sqrt{x} \right\rfloor$, discontinuity occurs when

$$\sqrt{x} = k, \quad k \in \mathbb{Z}$$

Hence

$$x = k^2$$

For $x \in [0,4]$, the listed points are

$$x = 0, 1, 4$$

The second approach in the solution further lists candidate points as

$$\{0, 1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$$

and states that the function is continuous at $0^+$ and at $4^-$. Therefore, excluding the interval endpoints, the function is discontinuous at the remaining $8$ points.

Therefore, the number of points of discontinuity is $8$.

Given: $f(x) = \left\lfloor \frac{x^2}{2} \right\rfloor - \left\lfloor \sqrt{x} \right\rfloor$ for $x \in [0,4]$.

Find: The total number of discontinuities.

The solution gives two observations:

1. Candidate points arise when the arguments of the floor functions hit integer values.
2. The set explicitly written in the second approach is

$$\{0, 1, \sqrt{2}, 2, \sqrt{6}, \sqrt{8}, \sqrt{10}, \sqrt{12}, \sqrt{14}, 4\}$$

It then states that the function is continuous at $0^+$ and continuous at $4^-$.

So the count of discontinuity points becomes

$$10 - 2 = 8$$

Thus, according to the solution, the function is discontinuous at $8$ points.