Q41 / 75 · JEE Main 2026 (04 April Shift 2) · Mathematics
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:
- A$110$
- Correct answerB$140$
- C$150$
- D$170$
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: $140$
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: $140$
Hence, the correct answer is option: 'B'.