Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let \(a, b \in \mathbb{R}\) be such that the function \(f\) given by \(f(x) = \ln|x| + bx^2 + ax,\, x \neq 0\) has extreme values at \(x = -1\) and \(x = 2\).<br><b>Statement-1:</b> \(f\) has local maximum at \(x = -1\) and at \(x = 2\).<br><b>Statement-2:</b> \(a = \dfrac{1}{2}\) and \(b = \dfrac{-1}{4}\)</p>
<p>Statement-1 is true, Statement-2 is false</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: For extreme values to exist at x = -1 and x = 2, we need f'(-1) = 0 and f'(2) = 0. Computing f'(x) = 1/x + 2bx + a and substituting these conditions gives us two linear equations in a and b that uniquely determine them. Then check the second derivative to classify the nature of extrema.
<p><strong>Step 1:</strong> Find f'(x).</p><p>f(x) = ln|x| + bx² + ax, so f'(x) = 1/x + 2bx + a</p><p><strong>Step 2:</strong> Use extreme value conditions f'(-1) = 0 and f'(2) = 0.</p><p>At x = -1: f'(-1) = -1 - 2b + a = 0 ⟹ a - 2b = 1 ... (i)</p><p>At x = 2: f'(2) = 1/2 + 4b + a = 0 ⟹ a + 4b = -1/2 ... (ii)</p><p><strong>Step 3:</strong> Solve the system of equations.</p><p>From (ii) - (i): 6b = -3/2 ⟹ b = -1/4</p><p>Substituting into (i): a - 2(-1/4) = 1 ⟹ a + 1/2 = 1 ⟹ a = 1/2</p><p><strong>Step 4:</strong> Verify Statement-2: a = 1/2 and b = -1/4 ✓ (TRUE)</p><p><strong>Step 5:</strong> Classify extrema using f''(x) = -1/x² + 2b = -1/x² - 1/2.</p><p>At x = -1: f''(-1) = -1 - 1/2 = -3/2 < 0 ⟹ local maximum</p><p>At x = 2: f''(2) = -1/4 - 1/2 = -3/4 < 0 ⟹ local maximum</p><p><strong>Step 6:</strong> Verify Statement-1: f has local maximum at x = -1 AND x = 2 ✓ (TRUE)</p><p>∴ Answer: D (Both statements are true; Statement-2 is the correct explanation of Statement-1)</p>
Correct Answer: D