Question

If \( f \) is a real valued function from \( A \) onto \( B \) defined by \( f(x) = \frac{1}{\sqrt{|x| - |x|}} \), then \( A \cap B \) is:}

• A. \( \emptyset \)
• B. \( (-\infty, 0) \)
• C. \( (0, \infty) \)
• D. \( (-\infty, \infty) \)

Hint:

Always carefully check the domain of the function and look for any points where it becomes undefined or contradictory.

Answer 1

The Correct Option is A

Given the function: \[ f(x) = \frac{1}{\sqrt{|x| - |x|}} \] 

Step 1: Checking the Denominator The denominator of the function is: \[ \sqrt{|x| - |x|} \] To analyze this expression, consider different cases for \( x \): 

Case 1: \( x \geq 0 \) For \( x \geq 0 \), we have \( |x| = x \), so: \[ |x| - |x| = x - x = 0 \] Thus, the denominator becomes \( \sqrt{0} = 0 \), making the function undefined. 

Case 2: \( x<0 \) For \( x<0 \), we have \( |x| = -x \), so: \[ |x| - |x| = (-x) - (-x) = 0 \] Again, the denominator becomes \( \sqrt{0} = 0 \), making the function undefined. 

Step 2: Conclusion Since the function is undefined for all \( x \in \mathbb{R} \), it does not map any values from set \( A \) to set \( B \). This implies that \( A \) and \( B \) must be disjoint, meaning: \[ A \cap B = \emptyset \] 

Final Answer: \(\boxed{\emptyset}\)