<p><strong>49.</strong> Consider the function \(f(x) = |x^2 - 7x + 12|(x^2 - 7x + 10)(x^2 - 4x + 3)\). Then Rolle's theorem for \(f(x)\) is not applicable to which of the following range?</p>
Step-by-Step Solution
Key Concept: Rolle's theorem requires f to be continuous on [a,b], differentiable on (a,b), and f(a)=f(b). The critical step is factoring each quadratic expression, identifying zeros, and determining where f(x)=0 at endpoints while checking differentiability at points where the absolute value expression changes sign.
<p><strong>Step 1:</strong> Factor the expressions:</p><ul><li>x² - 7x + 12 = (x-3)(x-4)</li><li>x² - 7x + 10 = (x-2)(x-5)</li><li>x² - 4x + 3 = (x-1)(x-3)</li></ul><p><strong>Step 2:</strong> Rewrite: f(x) = |(x-3)(x-4)| · (x-2)(x-5)(x-1)(x-3)</p><p><strong>Step 3:</strong> Identify zeros of f(x): x = 1, 2, 3, 4, 5</p><p><strong>Step 4:</strong> Check Rolle's applicability on intervals [a,b] where f(a)=f(b)=0:</p><ul><li>[1,2], [2,3], [3,4], [4,5], [1,3], [2,4], [3,5], [1,4], [2,5], [1,5]: f is continuous and f(endpoints)=0</li><li>At x=3 and x=4: The absolute value |x²-7x+12| has corners (non-differentiable) because the argument changes sign there AND (x-3) appears as a factor</li></ul><p><strong>Step 5:</strong> Any interval containing x=3 or x=4 in its interior fails the differentiability condition. Intervals like [1,4], [3,5], [2,4], [1,5] contain these corner points and therefore Rolle's theorem is NOT applicable.</p><p>∴ Answer: B (typically [1,5] or any interval with interior corner points)</p>
Correct Answer: B