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Polynomial satisfying f(x + y) = f(x) + f(y) plus extra terms

This type of question, asked 3 times: 29 Jun 2022, 28 Jan 2025, 4 Apr 2026 S2. All 3 questions below, oldest first.

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  1. Question 1

    29 Jun 2022

    87
    Let c,k∈Rc,k\in R. If f(x)=(c+1)x2+(1−c2)x+2kf(x) = (c+ 1)x^{2}+\left( 1 -c^{2} \right)x+ 2k and f(x+y)=f(x)+f(y)−xyf(x+y) =f(x) +f(y) -xy, for all x,y∈Rx,y\in R, then the value of ∣2(f(1)+f(2)+f(3)+⋯+f(20))∣|2(f(1) +f(2) +f(3) + \cdots + f(20))| is equal to

  2. Question 2

    28 Jan 2025

    55
    Let f:R→Rf:R \rightarrow R be a function defined by f(x)=(2+3a)x2+(a+2a−1)x+b,a≠1f(x) = (2 + 3a)x^{2}+\left( \frac{a + 2}{a - 1} \right)x + b,a \neq 1. If f(x+y)=f(x)+f(y)+1−27xyf(x + y) = f(x) + f(y) + 1 -\frac{2}{7}xy, then the value of 28∑i=15∣f(i)∣28\sum_{i = 1}^{5} |f(i)| is:

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  3. Question 3

    4 Apr 2026 S2

    66
    Let for some α∈R,f:R→R\alpha\in R,f:R\rightarrow R be a function satisfying f(x+y)=f(x)+2y2+y+αxyf(x + y) =f(x) + 2y^{2}+ y +\alpha xy for all x,y∈Rx,y \in R. If f(0)=−1f(0) = - 1 and f(1)=2f(1) = 2, then the value of ∑n=15(α+f(n))\sum_{n = 1}^{5} (\alpha+f(n)) is :

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