Let for some $\alpha \in \mathbb{R}, f: \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x})+2 \mathrm{y}^{2}+\mathrm{y}+\alpha \mathrm{xy}$ for all $\mathrm{x}, \mathrm{y} \in \mathbb{R}$. If $f(0)=-1$ and $f(1)=2$, then the value of $\sum_{\mathrm{n}=1}^{5}(\alpha+f(\mathrm{n}))$ is :

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Sets Relations and Functions

Let for some

α \in R, f : R \to R

be a function satisfying

f (x + y) = f (x) + 2 y^{2} + y + α xy

for all

x, y \in R

. If

f (0) = - 1

and

f (1) = 2

, then the value of

\sum_{n = 1}^{5} (α + f (n))

is :

Answer: (B)

Put

x = 0

f (x + y) = f (x) + 2 y^{2} + y + α x y

: with

f (0) = - 1

f (y) = - 1 + 2 y^{2} + y

, which also gives

f (1) = - 1 + 2 + 1 = 2

as required. To fix

α

, expand both sides using

f (t) = 2 t^{2} + t - 1

: the left side

f (x + y) = 2 (x + y)^{2} + (x + y) - 1

contains the cross term

4 x y

, while the right side contributes

α x y

, so

α = 4

. Then

\sum_{n = 1}^{5} (α + f (n)) = \sum_{n = 1}^{5} (4 + 2 n^{2} + n - 1) = \sum_{n = 1}^{5} (2 n^{2} + n + 3)

. Using

\sum_{1}^{5} n^{2} = 55

and

\sum_{1}^{5} n = 15

, this equals

2 (55) + 15 + 15 = 140

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option