If $f(x) = a\log_e|x| + bx^2+x$ has critical points at $x = -2$ and $x = 1$, then

$a + 4b = 0$

The correct answer is Option (3) → $a + 4b = 0$

The function is

$f(x) = a \log_e |x| + b x^2 + x$

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

At $x = -2$: $\frac{a}{-2} + 2b(-2) + 1 = 0$

$\Rightarrow -\frac{a}{2} - 4b + 1 = 0$      ...(1)

At $x = 1$: $\frac{a}{1} + 2b(1) + 1 = 0$

$\Rightarrow a + 2b + 1 = 0$      ...(2)

From (1): $- \frac{a}{2} - 4b + 1 = 0$

$\Rightarrow -a - 8b + 2 = 0$   (Multiplying by 2)

$\Rightarrow -a - 8b = -2$    ...(3)

Equation (2): $a + 2b = -1$    ...(4)

Add (3) and (4):

$(-a - 8b) + (a + 2b) = -2 -1$

$-6b = -3 \Rightarrow b = \frac{1}{2}$

Substitute $b = \frac{1}{2}$ into (4):

$a + 2\cdot \frac{1}{2} = -1 \Rightarrow a + 1 = -1 \Rightarrow a = -2$

$a = -2$, $b = \frac{1}{2}$