Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>If <em>x</em> = −1 and <em>x</em> = 2 are extreme points of <em>f</em>(<em>x</em>) = α log |<em>x</em>| + β<em>x</em><sup>2</sup> + <em>x</em>, then</p>
<p>\(\alpha = 2, \beta = -\dfrac{1}{2}\)</p>
<p>\(\alpha = 2, \beta = \dfrac{1}{2}\)</p>
<p>\(\alpha = -6, \beta = \dfrac{1}{2}\)</p>
<p>\(\alpha = -6, \beta = -\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: At extreme points, f'(x) = 0. Calculate f'(x), apply the condition at both x = -1 and x = 2 to get a system of equations in α and β, then solve for their values.
<p><strong>Step 1:</strong> Find f'(x).</p><p>f(x) = α log|x| + βx² + x</p><p>f'(x) = α/x + 2βx + 1</p><p><strong>Step 2:</strong> Apply f'(-1) = 0.</p><p>α/(-1) + 2β(-1) + 1 = 0</p><p>-α - 2β + 1 = 0</p><p>α + 2β = 1 ... (i)</p><p><strong>Step 3:</strong> Apply f'(2) = 0.</p><p>α/2 + 2β(2) + 1 = 0</p><p>α/2 + 4β + 1 = 0</p><p>α + 8β + 2 = 0</p><p>α + 8β = -2 ... (ii)</p><p><strong>Step 4:</strong> Solve equations (i) and (ii).</p><p>Subtract (i) from (ii):</p><p>(α + 8β) - (α + 2β) = -2 - 1</p><p>6β = -3</p><p>β = -1/2</p><p><strong>Step 5:</strong> Substitute back into (i).</p><p>α + 2(-1/2) = 1</p><p>α - 1 = 1</p><p>α = 2</p><p>∴ Answer: C (α = 2, β = -1/2)</p>
Correct Answer: C