Let the sets A and B denote the domain and range respectively of the function f(x)=1/√(⌈ x⌉-x), where ⌈ x⌉ denotes the smallest integer greater than or equal to x. Then among the statements
(S1) : A ∩ B=(1, ∞)-ℕ and
(S2) : A ∪ B=(1, ∞)

Solution

Maths sir · 3 steps

Away from integers, ⌈ x⌉ - x sits strictly between 0 and 1, and that fixes both sets.

1. Domain. We need ⌈ x⌉ - x > 0. It is 0 exactly when x is an integer and positive otherwise, so A = ℝ - ℤ.
2. Range. For non-integer x, ⌈ x⌉ - x = 1 - {x}, which takes every value in (0, 1). Its square root also lies in (0, 1), so its reciprocal covers (1, ∞). Hence B = (1, ∞).

   0 < ⌈ x⌉ - x < 1 ⇒ f(x) = 1/√(⌈ x⌉ - x) > 1

   Statement	Work it out	Verdict
   (S1) A ∩ B	(1, ∞) with the integers removed. The integers greater than 1 are 2, 3, 4, …, all natural numbers, so this is (1, ∞) - ℕ	True
   (S2) A ∪ B	A already contains non-integers such as -2.5 and 0.5, so the union is ℝ minus the integers ≤ 1, which is much bigger than (1, ∞)	False

3. So. Only (S1) is true, which is option (b).

The idea to keep. ⌈ x⌉ - x is 0 at integers and lies in (0, 1) everywhere else. So its reciprocal square root is defined only at non-integers and always exceeds 1.

Answer from the NTA final answer key. Solution by PrepWiser.