If $y = a \ln x + b x^2 + x$ has extreme values at $x = 1$ and $x = 2$, then $(a, b)$ is:
- $(-2/3, -1/6)$
- $(2, 1)$
- $(-1, 1)$
- $(2/3, 1/6)$
Show answer and solution
Correct answer: A. $(-2/3, -1/6)$
$y' = a/x + 2bx + 1 = 0$.
At $x = 1$: $a + 2b + 1 = 0$.
At $x = 2$: $a/2 + 4b + 1 = 0$.
**Solving:** from first, $a = -2b - 1$.
**Sub into second:** $(-2b - 1)/2 + 4b + 1 = 0$
⇒ $-b - 0.5 + 4b + 1 = 0$
⇒ $3b = -0.5$ ⇒ $b = -1/6$.
Then $a = -2(-1/6) - 1 = 1/3 - 1 = -2/3$.