JEE Main 2025 Jan 28 Shift 2, Mathematics Q15: Inverse Trigonometric Functions

JEE Main2025Jan 28, Shift 2Mathematics

Let $[x]$ denote the greatest integer less than or equal to $x$ . Then domain of $f (x) = sec^{- 1} (2 [x] + 1)$ is :

Solution

$sec^{- 1} (u)$ requires $∣ u ∣ \geq 1$ , i.e. $u \leq - 1$ or $u \geq 1$ .

Set $u = 2 [x] + 1$ . Solve:

$2 [x] + 1 \leq - 1 \Rightarrow [x] \leq - 1 \Rightarrow x < 0$ .

$2 [x] + 1 \geq 1 \Rightarrow [x] \geq 0 \Rightarrow x \geq 0$ .

The union $x < 0$ or $x \geq 0$ is all of $R$ , so the domain is $(- \infty, \infty)$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.