Question

The domain of the real valued function \[ f(x)=\sqrt{\frac{2-|x|}{3-|x|}} \] is:

• A. \((-\infty,\infty)\)
• B. \((-\infty,-3)\cup(2,\infty)\)
• C. \((-\infty,-3]\cup(-2,2)\cup[3,\infty)\)
• D. \((-\infty,-3)\cup[-2,2]\cup(3,\infty)\)

Hint:

For square-root functions involving rational expressions, first make the expression inside the square root non-negative and separately exclude values that make the denominator zero.

Answer 1

The Correct Option is D

Step 1: Condition for the square root to be real.
For \[ f(x)=\sqrt{\frac{2-|x|}{3-|x|}}, \] the expression inside the square root must be non-negative. Hence, \[ \frac{2-|x|}{3-|x|}\geq 0 \] Also, \[ 3-|x|\neq 0. \]

Step 2: Substitute \(t=|x|\).
Let \[ t=|x|,\qquad t\geq 0. \] Then the inequality becomes \[ \frac{2-t}{3-t}\geq 0. \] The critical points are \[ t=2,\qquad t=3. \]

Step 3: Sign analysis.
For \(0\leq t<2\), \[ 2-t>0,\qquad 3-t>0, \] so the fraction is positive.
For \(2<t<3\), \[ 2-t0, \] so the fraction is negative.
For \(t>3\), \[ 2-t<0,\qquad 3-t<0, \] so the fraction is positive.
At \[ t=2, \] the fraction is \(0\), which is allowed.
At \[ t=3, \] the denominator becomes zero, which is not allowed.
Therefore, \[ t\in [0,2]\cup(3,\infty). \]

Step 4: Convert back to \(x\).
Since \(t=|x|\), \[ |x|\leq 2 \] gives \[ -2\leq x\leq 2. \] Also, \[ |x|>3 \] gives \[ x3. \] Hence the domain is \[ (-\infty,-3)\cup[-2,2]\cup(3,\infty). \]

Step 5: Final conclusion.
Therefore, \[ \boxed{(-\infty,-3)\cup[-2,2]\cup(3,\infty)} \]