JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumSets, Relations and Functions168× since 2002Let f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R be a function defined by f(x)=(2+3a)x2+(a+2a−1)x+b,a≠1f(x) = (2 + 3a) x^2 + \left( \frac{a + 2}{a - 1} \right) x + b, a \neq 1f(x)=(2+3a)x2+(a−1a+2)x+b,a=1. If f(x+y)=f(x)+f(y)+1−27xyf(x + y) = f(x) + f(y) + 1 - \frac{2}{7} xyf(x+y)=f(x)+f(y)+1−72xy, then the value of 28∑i=15∣f(i)∣28 \sum_{i=1}^{5} |f(i)|28∑i=15∣f(i)∣ isAOption A. 545BOption B. 715COption C. 735DOption D. 675CheckSkip · show solution
