<p>Domain of definition of the function \(f(x) = \dfrac{3}{4 - x^2} + \log_{10}(x^3 - x)\) is</p>
<p>\((1, 2)\)</p>
<p>\((-1, 0) \cup (1, 2)\)</p>
<p>\((1, 2) \cup (2, \infty)\)</p>
<p>\((-1, 0) \cup (1, 2) \cup (2, \infty)\)</p>
Step-by-Step Solution
Key Concept: Find the intersection of two constraint sets: the denominator must be non-zero (4 - x² ≠ 0) AND the logarithm argument must be positive (x³ - x > 0). The domain is where BOTH conditions are satisfied simultaneously.
<p><strong>Step 1:</strong> For the fraction 3/(4-x²), we need 4 - x² ≠ 0, so x ≠ ±2</p><p><strong>Step 2:</strong> For log₁₀(x³ - x), we need x³ - x > 0</p><p>Factor: x(x² - 1) > 0 → x(x-1)(x+1) > 0</p><p>Sign analysis on intervals (-∞,-1), (-1,0), (0,1), (1,∞):</p><p>• x < -1: (-)(-)(-) = negative ✗</p><p>• -1 < x < 0: (-)(-)( +) = positive ✓</p><p>• 0 < x < 1: (+)(-)(+) = negative ✗</p><p>• x > 1: (+)(+)(+) = positive ✓</p><p>So x³ - x > 0 gives x ∈ (-1, 0) ∪ (1, ∞)</p><p><strong>Step 3:</strong> Find intersection: [(-1, 0) ∪ (1, ∞)] ∩ [ℝ \ {-2, 2}]</p><p>Since -2 ∉ (-1,0) ∪ (1,∞) and 2 ∉ (-1,0) ∪ (1,∞), no additional exclusions needed.</p><p>∴ <strong>Answer: (-1, 0) ∪ (1, ∞)</strong></p>
Correct Answer: D