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JEE Main 28 Jan 2025 Shift 1 Paper

Section: Mathematics
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Question : 16 of 75
 
Marks: +1, -0
Let f:ℝ⟶ℝ be a function defined by f(x)=(2+3a)x2+(‌
a+2
a−1
)x+b,a≠1. If f(x+y)= f(x)+f(y)+1−‌
2
7
xy, then the value of 28
5
∑
i−1
|f(i)| is
[28 Jan 2025 Shift 1]
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