Applications of Derivatives
Second Derivative Test
Grade 12
Question:
<p>Let <i>a</i>, <i>b</i> ∈ ℝ be such that the function <i>f</i> given by \[f(x) = \log|x| + bx^2 + ax, \quad x \neq 0\] has extreme values at <i>x</i> = −1 and <i>x</i> = 2.</p><p><strong>Statement I:</strong> <i>f</i> has local maximum at <i>x</i> = −1 and at <i>x</i> = 2.</p><p><strong>Statement II:</strong> \(a = -\frac{1}{2}\) and \(b = -\frac{1}{4}\)</p>
<p>(a) Statement I is false, Statement II is true</p>
<p>(b) Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I</p>
<p>(c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I</p>
<p>(d) Statement I is true, Statement II is false</p>
Step-by-Step Solution
Key Concept: Use the conditions f'(−1) = 0 and f'(2) = 0 to find a and b, then use the second derivative test to verify the nature of extrema.
<p>From $f'(x) = \frac{1}{x} + 2bx + a = 0$ at $x = -1$ and $x = 2$:</p><p>At $x = -1$: $-1 - 2b + a = 0 \Rightarrow a - 2b = 1$</p><p>At $x = 2$: $\frac{1}{2} + 4b + a = 0 \Rightarrow a + 4b = -\frac{1}{2}$</p><p>Solving: $6b = -\frac{3}{2} \Rightarrow b = -\frac{1}{4}$ and $a = 1 + 2(-\frac{1}{4}) = \frac{1}{2}$.</p><p>So Statement II values don't match exactly. For Statement I, check $f''(x) = -\frac{1}{x^2} + 2b$. At $x = -1$: $f''(-1) = -1 + 2(-\frac{1}{4}) = -\frac{3}{2} < 0$, so local maximum. At $x = 2$: $f''(2) = -\frac{1}{4} + 2(-\frac{1}{4}) = -\frac{3}{4} < 0$, so local maximum. Statement I is true.</p>
Correct Answer: c