If $x = -1$ and $x = -2$ are the extreme points of $f(x) = a\log|x| + βx^2 +x$ then

$α=\frac{2}{3},β=\frac{1}{6}$

The correct answer is Option (3) → $α=\frac{2}{3},β=\frac{1}{6}$

Given:

$f(x)=a\log|x|+\beta x^{2}+x$

Extreme points occur where $f'(x)=0$.

Differentiate:

$f'(x)=a\frac{1}{x}+2\beta x+1$

Given that $x=-1$ and $x=-2$ are extreme points, so:

$a\frac{1}{-1}+2\beta(-1)+1=0$

$-a-2\beta+1=0$

(1) $a+2\beta=1$

Next condition:

$a\frac{1}{-2}+2\beta(-2)+1=0$

$-\frac{a}{2}-4\beta+1=0$

(2) $\frac{a}{2}+4\beta=1$

Multiply (1) by 1 and (2) by 1 to solve:

(1) $a+2\beta=1$

(2) $\frac{a}{2}+4\beta=1$

Multiply (2) by 2:

$a+8\beta=2$

Subtract (1) from this:

$a+8\beta - (a+2\beta)=2-1$

$6\beta=1$

$\beta=\frac{1}{6}$

Substitute into (1):

$a+2\left(\frac{1}{6}\right)=1$

$a+\frac{1}{3}=1$

$a=\frac{2}{3}$

The required values are:

$a=\frac{2}{3}$ and $\beta=\frac{1}{6}$