Question:medium

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

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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.
Updated On: Jun 18, 2026
  • \((-\infty,\infty)\)
  • \((-\infty,-3)\cup(2,\infty)\)
  • \((-\infty,-3]\cup(-2,2)\cup[3,\infty)\)
  • \((-\infty,-3)\cup[-2,2]\cup(3,\infty)\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Impose the reality condition for the square root.
For f(x) = √((2 - |x|)/(3 - |x|)), the radicand must be non-negative. Hence we require (2 - |x|)/(3 - |x|) ≥ 0, along with the restriction 3 - |x| ≠ 0 to avoid division by zero.

Step 2: Introduce the substitution t = |x|.

Set t = |x| where t ≥ 0. The inequality transforms to (2 - t)/(3 - t) ≥ 0. The critical boundary points are t = 2 (numerator zero) and t = 3 (denominator zero).

Step 3: Perform sign analysis on the intervals.

For 0 ≤ t<2: both numerator and denominator are positive, yielding a positive fraction. For 2<t<3: numerator is negative while denominator remains positive, making the fraction negative. For t>3: both are negative, producing a positive fraction. At t = 2 the fraction equals zero, which is acceptable. At t = 3 the denominator vanishes, which is forbidden. The solution in t is therefore [0, 2] ∪ (3, ∞).

Step 4: Translate back to the variable x.

Since t = |x|, the condition |x| ≤ 2 implies -2 ≤ x ≤ 2. The condition |x|>3 splits into x<-3 or x>3. Combining these yields the domain (-∞, -3) ∪ [-2, 2] ∪ (3, ∞).

Step 5: Final conclusion.

The natural domain of the function is (-∞, -3) ∪ [-2, 2] ∪ (3, ∞).
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