Let f(x) = x e^ a x^2 + b x , where a, b R . If f(x) is strictly increasing on the interval (-1, 2) and strictly decreasing on (- , -1) (2, ) , then the ordered pair (a, b) is:

Step-by-step solution

Given f(x) = x e^ a x^2 + b x . Differentiating with respect to x using the product rule, we get: f'(x) = 1 e^ a x^2 + b x + x e^ a x^2 + b x (2ax + b) f'(x) = e^ a x^2 + b x (2ax^2 + bx + 1) Since e^ a x^2 + b x > 0 for all x R , the sign of f'(x) depends only on the quadratic expression Q(x) = 2ax^2 + bx + 1 . It is given that f(x) is increasing on (-1, 2) and decreasing elsewhere. This means f'(x) > 0 for x (-1, 2) and f'(x) Thus, Q(x) must be a downward-opening parabola (which requires 2a Using the relationship