If \(x = -1\) and \(x = -2\) are the extreme points of \(f(x) = \alpha\log|x| + \beta x^2 +x\) then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\alpha=\displaystyle\frac{2}{3},\ \beta=\frac{1}{6}\)
At an extreme point of a differentiable function, \(f'(x) = 0\).
\( \displaystyle f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \)
Putting x = −1:
\(-\alpha - 2\beta + 1 = 0\)
\(\alpha + 2\beta = 1 \quad \text{(1)}\)
Putting x = −2:
\( \displaystyle -\frac{\alpha}{2} - 4\beta + 1 = 0 \)
\(\alpha + 8\beta = 2 \quad \text{(2)}\)
Subtracting (1) from (2):
\(6\beta = 1\)
\( \displaystyle \beta = \frac{1}{6} \)
From (1):
\( \displaystyle \alpha = 1 - \frac{2}{6} = \frac{2}{3} \)