Question

The set of all real values of \(x\) for which \(f(x) = \sqrt{\frac{|x|-2}{|x|-3}}\) is a well defined function is

• A. \( (-3,-2] \cup (2,3] \)
• B. \( \mathbb{R} - ([-3,-2) \cup (2,3]) \)
• C. \( \mathbb{R} - [-3,3] \)
• D. \( (-3,3) \)

Hint:

For \( \sqrt{g(x)} \), require \(g(x) \ge 0\). For \(N(x)/D(x)\), require \(D(x) \neq 0\). Let \(y=|x|\). Solve \( \frac{y-2}{y-3} \ge 0 \). This implies \(y \le 2\) or \(y>3\). So \(|x| \le 2\) or \(|x|>3\). Convert these back to intervals for \(x\).

Answer 1

The Correct Option is B

For \(f(x)\) to be well-defined, we require:

1. The term under the square root must be non-negative: \[ \frac{|x|-2}{|x|-3} \ge 0 \]
2. The denominator must not be zero: \[ |x|-3 \neq 0 \implies |x| \neq 3 \]

Let \(y = |x|\). Then \(y \ge 0\). The inequality becomes \( \frac{y-2}{y-3} \ge 0 \).
The critical points for this inequality are \(y=2\) and \(y=3\).
We analyze the sign of the expression \( \frac{y-2}{y-3} \) in different intervals:

• If \(y < 2\) (and \(y \ge 0\)): \(y-2 < 0\), \(y-3 < 0\). So \( \frac{y-2}{y-3} > 0 \). This implies \( 0 \le |x| < 2 \).
• If \(y = 2\): \( \frac{2-2}{2-3} = \frac{0}{-1} = 0 \). So \( \frac{y-2}{y-3} \ge 0 \) is satisfied. This implies \(|x| = 2\).
• If \(2 < y < 3\): \(y-2 > 0\), \(y-3 < 0\). So \( \frac{y-2}{y-3} < 0 \). Not satisfied. This implies \(2 < |x| < 3\).
• If \(y > 3\): \(y-2 > 0\), \(y-3 > 0\). So \( \frac{y-2}{y-3} > 0 \). Satisfied. This implies \(|x| > 3\).
• If \(y = 3\): Denominator is zero, so undefined. This implies \(|x| \neq 3\).

Combining the conditions where \( \frac{y-2}{y-3} \ge 0 \), we have \(y \le 2\) or \(y > 3\).
Substituting back \(y = |x|\):

• \(|x| \le 2 \implies -2 \le x \le 2 \implies x \in [-2, 2]\).
• \(|x| > 3 \implies x < -3 \text{ or } x > 3 \implies x \in (-\infty, -3) \cup (3, \infty)\).

The set of all real values of \(x\) is the union of these intervals: \[ (-\infty, -3) \cup [-2, 2] \cup (3, \infty) \] This can be written as: \[ \mathbb{R} - ([-3, -2) \cup (2, 3]) \] \[ \boxed{\mathbb{R} - ([-3, -2) \cup (2, 3])} \]