JEE Main 2025 · 28 January · evening · Single correct · +4 / −1
Let [x] denote the greatest integer less than or equal to x. Then the domain of f(x) = sec⁻¹(2[x] + 1) is:
- (a) (-∞, ∞)
- (b) (-∞, ∞)- {0}
- (c) (-∞, -1] ∪ [0, ∞)
- (d) (-∞, -1] ∪ [1, ∞)
Answer and solution
Answer (a) (-∞, ∞)
Solution
Maths sir · 3 steps
Since 2[x]+1 is always an odd integer, it can never fall strictly between -1 and 1.
- Condition for sec inverse. sec⁻¹ y is defined only when |y| ≥ 1, that is y ≤ -1 or y ≥ 1. The only values it rejects are those in the open interval (-1, 1).
- Possible inputs. [x] is always an integer n, so 2[x]+1 = 2n+1 is always an odd integer. Odd integers are …, -3, -1, 1, 3, …, and none of them lies in (-1, 1).
x in [x] 2[x]+1 Allowed? [-2, -1) -2 -3 yes [-1, 0) -1 -1 yes, since sec⁻¹(-1) = π [0, 1) 0 1 yes, since sec⁻¹(1) = 0 [1, 2) 1 3 yes - So. For every real x, |2[x]+1| ≥ 1, so f(x) is defined everywhere. The domain is (-∞, ∞), option (a).
The idea to keep. When the input of sec⁻¹ or csc⁻¹ contains [x], first list the integer values it can take. Then check only those values against |y| ≥ 1.
Answer from the NTA final answer key. Solution by PrepWiser.