alleryIf $y = a \ln x + b x^2 + x$ has extreme values at $x = 1$ and $x = 2$, then $(a, b)$ is:
A$(-2/3, -1/6)$
B$(2, 1)$
C$(-1, 1)$
D$(2/3, 1/6)$
Answer & 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 $a = -2(-1/6) - 1 = 1/3 - 1 = -2/3$.
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