Question

If \(A\) and \(B\) are the domain and range of the real valued function, \[ f(x)=\dfrac{|x|}{\sqrt{1-|x|}} \] then \(A \cup B =\ ?\) 

• A. \((1,\infty)\)
• B. \([0,\infty)\)
• C. \((-1,\infty)\)
• D. \(\mathbb{R}\) \bigskip

Hint:

For functions containing square roots in the denominator, always use the condition \[ \text{Radicand}>0 \] instead of merely \(\ge0\). This helps determine the exact domain correctly.

Answer 1

The Correct Option is C

Concept: To determine the domain and range of a function involving a square root in the denominator, we must ensure:
• The expression inside the square root is positive.
• The denominator is not equal to zero.
• The function remains real valued. After finding the domain and range separately, we compute their union.

Step 1: Find the domain of the function.
Given \[ f(x)=\frac{|x|}{\sqrt{1-|x|}} \] For the square root to be defined, \[ 1-|x|\ge 0 \] which gives \[ |x|\le 1. \] Since the square root appears in the denominator, it cannot become zero. Therefore, \[ 1-|x|>0. \] Hence, \[ |x|<1. \] Thus the domain is \[ A=(-1,1). \]

Step 2: Find the range of the function.
Let \[ t=|x|. \] Since \(x\in(-1,1)\), \[ 0\le t<1. \] Then \[ f(x)=\frac{t}{\sqrt{1-t}}. \]

Step 3: Analyze the behavior of the function.
When \[ t=0, \] we obtain \[ f(0)=0. \] As \[ t\rightarrow 1^{-}, \] the denominator approaches zero from the positive side. Hence \[ f(t)\rightarrow \infty. \] Since the function is continuous on \(0\le t<1\), every non-negative value is attained. Therefore, \[ B=[0,\infty). \]

Step 4: Find \(A\cup B\).
We have \[ A=(-1,1) \] and \[ B=[0,\infty). \] Therefore, \[ A\cup B=(-1,\infty). \] Since \[ [0,\infty)\subset (-1,\infty), \] the union becomes \[ \boxed{(-1,\infty)}. \] Hence, \[ \boxed{A\cup B=(-1,\infty)}. \] Therefore the correct option is \[ \boxed{(C)}. \]