Question

Let the sets A and B denote the domain and range respectively of the function \(f(x)=\frac{1}{\sqrt[x]-x}\) 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, ∞) 

• A. (1) only (S1) is true
• B. both (S1) and (S2) are true
• C. neither (S1) nor (S2) is true
• D. only (S2) is true

Answer 1

The Correct Option is A

To determine the validity of statements \(S1\) and \(S2\) relating to the function \(f(x)=\frac{1}{\sqrt[x]-x}\), where \([\cdot]\) denotes the smallest integer greater than or equal to \(x\), we first need to analyze the function's domain and range.

1. Determining Domain (Set A):
   • The function \(f(x)\) is undefined when the denominator \([\sqrt{x}] - x = 0\). This occurs when \([\sqrt{x}] = x\).
   • The condition \([\sqrt{x}] = x\) implies \(x\) must be an integer since \([\cdot]\) is the ceiling function.
   • For \(x\) to be an integer and satisfy the condition \([\sqrt{x}] = x\), \(x\) must equal to \(n^2\) for \(n \in \mathbb{N}\), resulting in discontinuity at perfect squares.
   • Therefore, the domain \(A\) of the function is \((1, \infty) - N\), where \(N\) is the set of all perfect squares.
2. Determining Range (Set B):
   • The function output, \(f(x)=\frac{1}{[\sqrt{x}] - x}\), restricts output values such that as \(x\) approaches any perfect square, the output approaches undefined values, but over the non-perfect square values of \(x\), \(f(x)\) takes on a complete set of real numbers.
   • This implies the range \(B\) is all real numbers \((1, \infty)\).
3. Analyzing Statement (S1): \(A \cap B = (1, \infty) - N\):
   • The intersection of the domain \(A\) and range \(B\) essentially reflects back to set \(A\), implying all real numbers except some specific integer values (perfect squares), ensuring that \((1, \infty) - N\) indeed represents both correctly.
   • Thus, statement \(S1\) is true.
4. Analyzing Statement (S2): \(A \cup B = (1, \infty)\):
   • The union of domain \(A\) and range \(B\) covers from just greater than 1 to infinity, incorporating every real number in between there, with the feature gaps at perfect squares.
   • Since the perfect squares still contribute to areas where \(f(x)\) is undefined, \(A \cup B\) cannot smoothly cover \((1, \infty)\).
   • Thus, statement \(S2\) is not true.

In conclusion, the correct answer is that only statement (S1) is true.