Let for some α ∈ ℝ , f:ℝarrowℝ be a function satisfying f(x+y)=f(x)+2y^2+y+α xy for all x,y ∈ ℝ . If f(0)=-1 and f(1)=2 , then the value of ∑_ n=1 ^ 5 (α+f(n)) is:

Correct answer: B. 140 Given f(x+y) = f(x) + 2y^2 + y + α xy Substituting x = 0 and replacing y with x , we get: f(x) = f(0) + 2x^2 + x + α(0)x Since f(0) = -1 , we have: f(x) = 2x^2 + x - 1 To find α , substitute f(x) into the original functional equation: 2(x+y)^2 + (x+y) - 1 = (2x^2 + x - 1) + 2y^2 + y + α xy 2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + α xy Comparing both sides, we get α = 4 . We need to find the value of ∑_ n=1 ^ 5 (α+f(n)) : ∑_ n=1 ^ 5 (4 + 2n^2 + n - 1) = ∑_ n=1 ^ 5 (2n^2 + n + 3) = 2∑_ n=1 ^ 5 n^2 + ∑_ n=1 ^ 5 n + ∑_ n=1 ^ 5 3 = 2((5 × 6 × 11)/(6)) + (5 × 6)/(2) + 3 × 5 = 2(55) + 15 + 15 = 110 + 30 = 140 Answer: 140

JEE Main 20264 AprilEvening Shift

Exam: JEE Main
Year: 2026
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Functions

Let for some α ∈ ℝ , f:ℝarrowℝ be a function satisfying...

Let for some

\alpha \in \mathbb{R}

f:\mathbb{R}\rightarrow\mathbb{R}

be a function satisfying

f(x+y)=f(x)+2y^2+y+\alpha xy

for all

x,y \in \mathbb{R}

. If

f(0)=-1

and

f(1)=2

, then the value of

\sum_{n=1}^{5}(\alpha+f(n))

is:

Options

A $110$
B $140$
C $150$
D $170$

Correct answer

B. $140$

Step-by-step solution

Given

f(x+y) = f(x) + 2y^2 + y + \alpha xy

Substituting

x = 0

and replacing

y

with

x

, we get:

f(x) = f(0) + 2x^2 + x + \alpha(0)x

Since

f(0) = -1

, we have:

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

To find

\alpha

, substitute

f(x)

into the original functional equation:

2(x+y)^2 + (x+y) - 1 = (2x^2 + x - 1) + 2y^2 + y + \alpha xy

2x^2 + 4xy + 2y^2 + x + y - 1 = 2x^2 + x - 1 + 2y^2 + y + \alpha xy

Comparing both sides, we get

\alpha = 4

.

We need to find the value of

\sum_{n=1}^{5}(\alpha+f(n))

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

= 2\sum_{n=1}^{5}n^2 + \sum_{n=1}^{5}n + \sum_{n=1}^{5}3

= 2\left(\dfrac{5 \times 6 \times 11}{6}\right) + \dfrac{5 \times 6}{2} + 3 \times 5

= 2(55) + 15 + 15

= 110 + 30 = 140

Answer:

140

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