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Mathematics · 2022

JEE Main · 29 June 2022, Shift 1 · Q27

Let c, k ∈ R. If f(x)=( c +1) x^2+(1- c^2) x+2 k and f(x+y)=f(x)+f(y)-x y, for all x, y ∈ R, then the value of |2(f(1)+f(2)+f(3)+…+f(20))| is equal…

Let $\displaystyle \mathrm{c}, \mathrm{k} \in \mathbf{R}$. If $\displaystyle f(x)=(\mathrm{c}+1) x^2+\left(1-\mathrm{c}^2\right) x+2 \mathrm{k}$ and $\displaystyle f(x+y)=f(x)+f(y)-x y$, for all $\displaystyle x, y \in \mathbf{R}$, then the value of $\displaystyle |2(f(1)+f(2)+f(3)+\ldots+f(20))|$ is equal to \_\_\_\_.
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.