Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+b, a \neq 1$. If $f(x+y)=f(x)+f(\mathrm{y})+1-\frac{2}{7} x \mathrm{y}$, then the value of $28 \sum\limits_{i=1}^5|f(i)|$ is

<p>$$\begin{aligned} & f(x)=(3 a+2) x^2+\left(\frac{a+2}{a-1}\right) x+b \\ & f\left(x+\frac{1}{2}\right)=f(x)+f(y)+1-\frac{2}{7} x y\quad\text{.... (1)} \end{aligned}$$</p> <p>In (1) Put $x=y=0 \Rightarrow f(0)=2 f(0)+1 \Rightarrow f(0)=-1$</p> <p>So, $\mathrm{f}(0)=0+0+\mathrm{b}=-1 \Rightarrow \mathrm{~b}=-1$</p> <p>$$\begin{aligned} & \text { In (1) Put } y=-x \Rightarrow f(0)=f(x)+f(-x)+1+\frac{2}{7} x^2 \\ & -1=2(3 a+2) x^2+2 b+1+\frac{2}{7} x^2 \end{aligned}$$</p> <p>$$\begin{aligned} & -1=\left(2(3 a+2)+\frac{2}{7}\right) x^2+1-2 \\ & \Rightarrow 6 a+4+\frac{2}{7}=0 \\ & a=-\frac{5}{7} \end{aligned}$$</p> <p>So $\mathrm{f}(\mathrm{x})=-\frac{1}{7} \mathrm{x}^2-\frac{3}{4} \mathrm{x}-1$</p> <p>$$\Rightarrow|\mathrm{f}(\mathrm{x})|=\frac{1}{28}\left|4 \mathrm{x}^2+21 \mathrm{x}+28\right|$$</p> <p>Now, $28 \sum_{i=1}^5|f(i)|=28(|f(1)|+|f(2)|+\ldots+|f(5)|)$</p> <p>$28 \cdot \frac{1}{28} \cdot 675=675$</p>