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Mathematics · 2025

JEE Main · 28 January 2025, Shift 1 · Q19

Let f: R → R be a function defined by f(x)=(2+3 a) x^2+((a+2)/(a-1)) x+ b, a ≠ 1. If; f(x+ y )=f(x)+f( y )+1-2/7 x y, then the value of 28…

Let $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $$\begin{aligned} & f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+\mathrm{b}, a \neq 1 . \text { If } \\ & f(x+\mathrm{y})=f(x)+f(\mathrm{y})+1-\frac{2}{7} x \mathrm{y}, \text { then the value of } 28 \sum_{i=1}^5|f(i)| \text { is } \end{aligned} $$
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JEE Main 2025 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.