JEE Main 2026 — Continuity and Differentiability Question with Solution

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

Differentiating with respect to $x$, we get $g^{\prime }(x)=2x-1$.

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

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

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

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

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: $10$