QJEEM25Ma96fb: JEE Main 2025 Mathematics - Class 11 JEEM25Ma96fb Solution: Let be a function defined by . If , then the value of is

$\begin{aligned}& f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}\right) x+b \\& f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7} x y\quad\text{.... (1)}\end{aligned}$

In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$

So, $\mathrm{f}(0)=0+0+\mathrm{b}=-1 \Rightarrow \mathrm{~b}=-1$

$\begin{aligned}& \text { In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2 \\& -1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2\end{aligned}$

$\begin{aligned}& -1=\left(2(3 a+2)+\frac{2}{7}\right) x^2+1-2 \\& \Rightarrow 6 a+4+\frac{2}{7}=0 \\& a=-\frac{5}{7}\end{aligned}$

So $\mathrm{f}(\mathrm{x})=-\frac{1}{7} \mathrm{x}^2-\frac{3}{4} \mathrm{x}-1$

$\Rightarrow|\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4 \mathrm{x}^2+21 \mathrm{x}+28\right|$

Now, $28 \sum_{i=1}^5|f(i)|=28(|f(1)|+|f(2)|+\ldots+|f(5)|)$

$28 \cdot \frac{1}{28} \cdot 675=675$