JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

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 $n = 1 \sum 5 (α + f (n))$ is:

A
110
B
140
C
150
D
170

Solution

Put $x = 0$ : $f (y) = f (0) + 2 y^{2} + y = - 1 + 2 y^{2} + y$ .

Check: $f (1) = - 1 + 2 + 1 = 2$ . ✓

Find $α$ using $x = y = 1$ : $f (2) = f (1) + 2 + 1 + α = 2 + 3 + α = 5 + α$ .

Also from the closed form: $f (2) = - 1 + 8 + 2 = 9$ . So $5 + α = 9 \Rightarrow α = 4$ .

Now $α + f (n) = 4 + (- 1 + 2 n^{2} + n) = 2 n^{2} + n + 3$ .

$n = 1 \sum 5 (2 n^{2} + n + 3) = 2 \cdot \frac{5 \cdot 6 \cdot 11}{6} + \frac{5 \cdot 6}{2} + 15 = 110 + 15 + 15 = 140$ .

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