If $f(x) = a\log_e|x| + bx^2+x$ has critical points at $x = -2$ and $x = 1$, then
Answer & explanation
Correct answer: option 3
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):
$a + 2\cdot \frac{1}{2} = -1 \Rightarrow a + 1 = -1 \Rightarrow a = -2$
$a = -2$, $b = \frac{1}{2}$