At x = 1, substituting x = 1 in the equation: ax2 + bx + c = 5; f(1) => a + b + c = 5 .....(1) Taking the first derivative of f(x) in the original equation gives dy/dx = 2ax + b = 2x + 1 (given)....(2) From (2),=> b = 1, and 2ax = 2x, => a = 1. ( comparing coefficients) substituting into (1) 1 + 1 + c = 5, => c = 5 - 2 = 3 Thus a = 1, b = 1 and c = 3