JEE Main 2025 (Online) 28th January Morning ShiftQuestion 1
Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined by
$f(x)=(2+3a)x^2+\left(\frac{a+2}{a-1}\right)x+b$, $a\neq 1$. If
$f(x+y)=f(x)+f(y)+1-\frac{2}{7}xy$, then the value of $28\sum_{i=1}^{5}|f(i)|$ is
Functions
Solution
Putting $x=y=0$ gives $b=-1$; the $xy$ coefficient gives $2(2+3a)=-\frac27\Rightarrow a=-\frac57$, so $f(x)=-\frac{x^2}{7}-\frac{3x}{4}-1$, always negative. $28\sum_{i=1}^5|f(i)|=28\left(\frac{55}{7}+\frac{45}{4}+5\right)=220+315+140=675$.