JEE Mathematics 2026 Question with Solution

Given: $f(x) = [x]^2 - [x+3] - 3$, where $[\,\cdot\,]$ denotes the greatest integer function.

Find: Which statement is correct.

Let $n = [x]$, where $n \in \mathbb{Z}$. Then $x \in [n, n+1)$.

Step 1: Express the function in terms of $n$.

$$f(x) = n^2 - (n+3) - 3 = n^2 - n - 6$$

Step 2: Determine when $f(x) < 0$.

$$n^2 - n - 6 < 0$$

Factorizing,

$$(n-3)(n+2) < 0$$

This gives

$$-2 < n < 3$$

Thus,

$$n = -1,\,0,\,1,\,2$$

Step 3: Translate back to intervals of $x$. For $n = -1,0,1,2$,

$$x \in [-1,3)$$

Hence,

$$f(x) < 0 \text{ only for } x \in [-1,3)$$

Therefore, the correct option is B.](streamdown:incomplete-link)

For greatest integer function problems, analyze the expression on intervals of the form $[n,n+1)$, where $[x]$ remains constant.

Here $[x] = n$ and also $[x+3] = n+3$ for every $x \in [n,n+1)$. So on each such interval,

$$f(x) = n^2 - n - 6$$

This depends only on the integer $n$, not on the exact value of $x$ inside the interval.

So the sign of $f(x)$ is decided by solving

$$n^2 - n - 6 < 0$$

which gives

$$(n-3)(n+2) < 0$$

and hence

$$-2 < n < 3$$

The integer values possible are $n = -1, 0, 1, 2$. Therefore $x$ must lie in the union

$$[-1,0) \cup [0,1) \cup [1,2) \cup [2,3) = [-1,3)$$

So $f(x) < 0$ only for $x \in [-1,3)$. Hence the correct option is B.](streamdown:incomplete-link)