JEE Mathematics 2025 Question with Solution

Given: $f(x) = \sec^{-1}(2[x] + 1)$, where $[x]$ is the greatest integer less than or equal to $x$.

Find: The domain of the function.

For inverse secant, the input must satisfy

$$|y| \geq 1$$

So here we need

$$|2[x] + 1| \geq 1$$

This gives

$$2[x] + 1 \geq 1 \quad \text{or} \quad 2[x] + 1 \leq -1$$

Hence,

$$2[x] \geq 0 \Rightarrow [x] \geq 0$$

or

$$2[x] \leq -2 \Rightarrow [x] \leq -1$$

Now for every real number $x$, exactly one of these is true: either $[x] \geq 0$ or $[x] \leq -1$. There is no integer strictly between $-1$ and $0$, so no real $x$ is excluded.

Therefore, the domain is all real numbers,

$$(-\infty, \infty)$$

the solution marks the correct option as B, but the printed option value for B is inconsistent. Based on the working, the domain is $(-\infty, \infty)$, so the defensible answer from the provided options is B.

Given: $f(x) = \sec^{-1}(2[x] + 1)$.

Find: Which real numbers make the argument of $\sec^{-1}$ valid.

The domain of $\sec^{-1}(t)$ is

$$t \in (-\infty, -1] \cup [1, \infty)$$

So set

$$t = 2[x] + 1$$

and require

$$2[x] + 1 \in (-\infty, -1] \cup [1, \infty)$$

That is,

$$2[x] + 1 \leq -1 \quad \text{or} \quad 2[x] + 1 \geq 1$$

Solving,

$$[x] \leq -1 \quad \text{or} \quad [x] \geq 0$$

But $[x]$ is always an integer, so these two conditions together cover all possible integer values of $[x]$.

Therefore every real $x$ is allowed, and the domain is all real numbers. Hence the mathematically correct domain is

(-\infty, \infty) $$](streamdown:incomplete-link)