Sets, Relations & Functions
Domain of functions involving modulus
Grade 11
<p>Find the values of \(x\) for which the following function is defined: \[f(x) = \sqrt{\dfrac{1}{|x-2| - (x-2)}}\]</p>
Step-by-Step Solution
Key Concept: The expression |x-2| - (x-2) equals 0 when x ≥ 2 and equals -2(x-2) when x < 2; for the square root to be defined, we need the denominator positive, which forces x < 2.
<p><strong>Step 1:</strong> Analyze |x-2| - (x-2) by cases.</p><p>When x ≥ 2: |x-2| = x-2, so |x-2| - (x-2) = (x-2) - (x-2) = 0</p><p>When x < 2: |x-2| = -(x-2), so |x-2| - (x-2) = -(x-2) - (x-2) = -2(x-2) = 2(2-x)</p><p><strong>Step 2:</strong> For f(x) to be defined, we need: (i) denominator ≠ 0, and (ii) expression under square root ≥ 0.</p><p>This requires: $\frac{1}{|x-2| - (x-2)} \geq 0$ with denominator ≠ 0.</p><p><strong>Step 3:</strong> When x ≥ 2, denominator = 0, so function is undefined.</p><p>When x < 2, denominator = 2(2-x) > 0, so $\frac{1}{2(2-x)} > 0$ ✓</p><p><strong>Step 4:</strong> The square root condition is satisfied when x < 2.</p><p>∴ Answer: (-∞, 2)</p>
Correct Answer: (-∞, 2)