Let f: R [-5, 4] be a surjective function defined by f(x) = x^2 + ax + b x^2 + 2x + 3 . If a > 0 , then the value of a + b is

Step-by-step solution

Let y = x^2 + ax + b x^2 + 2x + 3 yx^2 + 2yx + 3y = x^2 + ax + b (y-1)x^2 + (2y-a)x + (3y-b) = 0 Since x R , the discriminant D 0 for all y in the range. (2y-a)^2 - 4(y-1)(3y-b) 0 4y^2 - 4ay + a^2 - 4(3y^2 - (b+3)y + b) 0 -8y^2 + 4(b - a + 3)y + (a^2 - 4b) 0 8y^2 - 4(b - a + 3)y - (a^2 - 4b) 0 Since f(x) is surjective, its range is exactly [-5, 4] . Thus, the roots of the equation 8y^2 - 4(b - a + 3)y - (a^2 - 4b) = 0 must be -5 and 4 . Sum of roots = -5 + 4 = -1 4(b - a + 3) 8 = -1 b - a + 3 = -2 b - a = -5 b = a