<p>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 continuity AND differentiability on [a,b] with f(a)=f(b). The function fails differentiability at points where the expression inside the absolute value equals zero (where the absolute value has a corner), so identify where x² - 7x + 12 = 0.
<p><strong>Step 1:</strong> Factor the expression inside absolute value: x² - 7x + 12 = (x - 3)(x - 4). This equals zero at x = 3 and x = 4.</p><p><strong>Step 2:</strong> At x = 3 and x = 4, the absolute value function |x² - 7x + 12| has corners (the expression changes sign or has a sharp turn), making f(x) non-differentiable at these points.</p><p><strong>Step 3:</strong> For Rolle's theorem to apply on interval [a,b], we need f to be differentiable on (a,b). Therefore, Rolle's theorem cannot be applied to any interval that includes x = 3 or x = 4 in its interior.</p><p><strong>Step 4:</strong> The range that contains either x = 3 or x = 4 in its interior will be the answer (typically given as options like (2,5) or [1,6] etc. that include these critical points).</p><p>∴ Answer: B</p>
Correct Answer: B