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

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2026

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Home Maths Relations and FunctionsQuestion #364

Let for some

α \in R

f : R \to R

be a function satisfying

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

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:

Clue 1 of 3: Functional Equation Form

The given functional equation suggests that the function $f (x)$ might be a simple polynomial. Try to determine its form.

🥷

Ninja Strategy•Dominant Term Estimation

Calculate the largest part of the sum first ( $2 \sum_{n = 1}^{5} n^{2} = 110$ ) and note that since all other terms are positive, the final answer must be strictly greater than 110.

🧠 Full Solution Path

Step 1: Determine the Form of the Function and the Value of

α

✦ Active

Given the functional equation $f (x + y) = f (x) + 2 y^{2} + y + α x y$ . By swapping $x$ and $y$ , we get $f (y + x) = f (y) + 2 x^{2} + x + α y x$ . Equating the two expressions for $f (x + y)$ gives:

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

Rearranging this, we find that $f (x) - 2 x^{2} - x = f (y) - 2 y^{2} - y$ . This implies that $f (x) - 2 x^{2} - x$ must be a constant, let's call it $C$ . Therefore, the function must be of the form $f (x) = 2 x^{2} + x + C$ .

Substituting this form back into the original functional equation to find $α$ :

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

Expanding and simplifying the equation leads to $4 x y = α x y$ , which implies $α = 4$ .

Step 2: Find the Specific Function

f (x)

○ Expand

We now have $f (x) = 2 x^{2} + x + C$ and $α = 4$ . Using the given condition $f (0) = - 1$ :

f (0) = 2 (0)^{2} + 0 + C = - 1 ⟹ C = - 1

So, the function is $f (x) = 2 x^{2} + x - 1$ . We can verify this with the other condition, $f (1) = 2 (1)^{2} + 1 - 1 = 2$ , which is correct.

Step 3: Calculate the Required Summation○ Expand

We need to compute the value of $S = \sum_{n = 1}^{5} (α + f (n))$ .

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

Using the standard formulas for sum of first $n$ integers and sum of first $n$ squares:

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

S = 2 (\frac{5 (5 + 1) (2 \cdot 5 + 1)}{6}) + \frac{5 (5 + 1)}{2} + (5 \times 3)

S = 2 (55) + 15 + 15 = 110 + 30 = 140

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Maths Question 16 – JEE-MAIN 2026

Clue 1 of 3: Functional Equation Form

🧠 Full Solution Path