JEE Main 2026 — Mathematics Algebra

Given $f(x+y) = f(x) + 2y^2 + y + \alpha xy$f(x+y)=f(x)+2y2+y+αxy

Substituting $x = 0$x=0 and replacing $y$y with $x$x, we get:

$$f(x) = f(0) + 2x^2 + x + \alpha(0)x$$f(x)=f(0)+2x2+x+α(0)x

Since $f(0) = -1$f(0)=−1, we have:

$$f(x) = 2x^2 + x - 1$$f(x)=2x2+x−1

To find $\alpha$α, substitute $f(x)$f(x) into the original functional equation:

$$2(x+y)^2 + (x+y) - 1 = (2x^2 + x - 1) + 2y^2 + y + \alpha xy$$2(x+y)2+(x+y)−1=(2x2+x−1)+2y2+y+αxy

$$2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + \alpha xy$$2x2+4xy+2y2+x+y−1=2x2+x−1+2y2+y+αxy

Comparing both sides, we get $\alpha = 4$α=4.

We need to find the value of $\displaystyle \sum_{n=1}^{5}(\alpha+f(n))$n=1∑5​(α+f(n)):

$$\sum_{n=1}^{5}(4 + 2n^2 + n - 1) = \sum_{n=1}^{5}(2n^2 + n + 3)$$n=1∑5​(4+2n2+n−1)=n=1∑5​(2n2+n+3)

$$= 2\sum_{n=1}^{5}n^2 + \sum_{n=1}^{5}n + \sum_{n=1}^{5}3$$=2n=1∑5​n2+n=1∑5​n+n=1∑5​3

$$= 2\left(\dfrac{5 \times 6 \times 11}{6}\right) + \dfrac{5 \times 6}{2} + 3 \times 5$$=2(65×6×11​)+25×6​+3×5

$$= 2(55) + 15 + 15$$=2(55)+15+15

$$= 110 + 30 = 140$$=110+30=140

Answer: $140$140