1.Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)=(2+3a)x2+(a+2)x+bf(x) = (2+3a)x^2 + (a+2)x + bf(x)=(2+3a)x2+(a+2)x+b, a≠1a \neq 1a=1. If f(x+y)=f(x)+f(y)+1−12xyf(x+y) = f(x) + f(y) + 1 - \frac{1}{2}xyf(x+y)=f(x)+f(y)+1−21xy, then the value of 28∑r=110f(r)28\sum_{r=1}^{10} f(r)28∑r=110f(r) is:Login to continueOnly logged in users canattempt or see the solution.
