Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>Let \(f(x) = 5 - |x - 2|\) and \(g(x) = |x + 1|\), \(x \in R\). If \(f(x)\) attains maximum value at \(\alpha\) and \(g(x)\) attains minimum value at \(\beta\), then \(\lim_{x \to -\alpha\beta} \dfrac{(x-1)(x^2 - 5x + 6)}{x^2 - 6x + 8}\) is equal to</p>
<p>\(\dfrac{1}{2}\)</p>
<p>\(\dfrac{-3}{2}\)</p>
<p>\(\dfrac{-1}{2}\)</p>
<p>\(\dfrac{3}{2}\)</p>
Step-by-Step Solution
Key Concept: First identify that f(x) = 5 - |x - 2| attains maximum at α = 2 (vertex of absolute value), and g(x) = |x + 1| attains minimum at β = -1 (vertex of absolute value). Then compute the limit at x → -αβ = -2(-1) = 2.
<p><strong>Step 1:</strong> Find α and β from the given functions.</p><p>For f(x) = 5 - |x - 2|: The expression |x - 2| ≥ 0 with minimum 0 at x = 2. So f(x) attains maximum value 5 at <strong>α = 2</strong>.</p><p>For g(x) = |x + 1|: The expression |x + 1| ≥ 0 with minimum 0 at x = -1. So g(x) attains minimum value 0 at <strong>β = -1</strong>.</p><p><strong>Step 2:</strong> Calculate the limit point.</p><p>-αβ = -(2)(-1) = 2</p><p><strong>Step 3:</strong> Simplify the expression before taking the limit.</p><p>Numerator: (x - 1)(x² - 5x + 6) = (x - 1)(x - 2)(x - 3)</p><p>Denominator: x² - 6x + 8 = (x - 2)(x - 4)</p><p><strong>Step 4:</strong> Cancel the common factor (x - 2).</p><p>$$\lim_{x \to 2} \dfrac{(x-1)(x-2)(x-3)}{(x-2)(x-4)} = \lim_{x \to 2} \dfrac{(x-1)(x-3)}{x-4}$$</p><p><strong>Step 5:</strong> Substitute x = 2.</p><p>$$= \dfrac{(2-1)(2-3)}{2-4} = \dfrac{(1)(-1)}{-2} = \dfrac{-1}{-2} = \dfrac{1}{2}$$</p><p>∴ Answer: <strong>1/2 (Option D)</strong></p>
Correct Answer: D