JEE Mains · Maths · STD 12 - 1. relation and function
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 \(110\)
- B \(140\)
- C \(150\)
- D \(170\)
Answer & Solution
Correct Answer
(B) \(140\)
Step-by-step Solution
Detailed explanation
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…
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