Q50 / 75 · JEE Main 2026 (02 April Shift 2) · Mathematics

The number of points in the interval $[2, 4]$, at which the function $f(x) = \left[x^2 - x - \dfrac{1}{2}\right]$, where $[\cdot]$ denotes the greatest integer function, is discontinuous, is _______.

Correct answer: 10

Solution

Let $g(x) = x^2 - x - \dfrac{1}{2}$.

Differentiating with respect to $x$, we get $g'(x) = 2x - 1$.

For $x \in [2, 4]$, $g'(x) > 0$, which implies that $g(x)$ is strictly increasing in the interval $[2, 4]$.

The values of $g(x)$ at the endpoints are:

$g(2) = 2^2 - 2 - \dfrac{1}{2} = 1.5$

$g(4) = 4^2 - 4 - \dfrac{1}{2} = 11.5$

The function $f(x) = [g(x)]$ is discontinuous at all points where $g(x)$ is an integer, as $g(x)$ is strictly monotonic and crosses these integer values.

The integers between $1.5$ and $11.5$ are $2, 3, 4, 5, 6, 7, 8, 9, 10, 11$.

Since $g(x)$ is strictly increasing, it attains each of these $10$ integer values exactly once in the interval $(2, 4)$.

At the endpoints $x = 2$ and $x = 4$, $g(x)$ is not an integer, so $f(x)$ is continuous from the right at $x = 2$ and continuous from the left at $x = 4$.

Thus, the number of points of discontinuity in the interval $[2, 4]$ is $10$.

Answer: $10$

Hence, the correct answer is 10.

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