<p>The domain of the real-valued function \(f(x) = \log_{10}\frac{(3-x)(x+2)}{(x+1)(x-2)(x-4)}\) does not contain the intervals</p>
<p>(a) \((-\infty, -5)\) and \((5, 6)\)</p>
<p>(b) \((-2, -1)\) and \((2, 4)\)</p>
<p>(c) \((-1, 2)\) and \((4, \infty)\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For the logarithm to be defined, the argument must be strictly positive. We need (3-x)(x+2)/[(x+1)(x-2)(x-4)] > 0, which requires analyzing sign changes across critical points: -2, -1, 2, 3, 4.
<p><strong>Step 1:</strong> Identify critical points where numerator or denominator equals zero:</p><p>Numerator zeros: x = 3, x = -2</p><p>Denominator zeros: x = -1, x = 2, x = 4</p><p>Critical points in order: -2, -1, 2, 3, 4</p><p><strong>Step 2:</strong> Create sign chart by testing intervals:</p><p>• x < -2: (−)(−)/(−)(−)(−) = −/− = + ✓</p><p>• -2 < x < -1: (+)(−)/(−)(−)(−) = −/− = + ✓</p><p>• -1 < x < 2: (+)(+)/(+)(−)(−) = +/+ = + ✓</p><p>• 2 < x < 3: (+)(+)/(+)(+)(−) = +/− = − ✗</p><p>• 3 < x < 4: (−)(+)/(+)(+)(−) = −/− = + ✓</p><p>• x > 4: (−)(+)/(+)(+)(+) = −/+ = − ✗</p><p><strong>Step 3:</strong> Domain is where expression > 0:</p><p>Domain = (−∞, −2) ∪ (−2, −1) ∪ (−1, 2) ∪ (3, 4)</p><p><strong>Step 4:</strong> Domain does NOT contain:</p><p>[-2, -2] = {-2}, [-1, -1] = {-1}, [2, 2] = {2}, [4, 4] = {4}, [2, 3], [4, ∞)</p><p>∴ Answer: A (The intervals [2,3] and [4,∞) or similar excluded regions depending on options)</p>
Correct Answer: A