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

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

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Put } y=0 \ & f(x)=f(0)+f(x)+1-0 \ & f(0)=-1 \ & f(0)=0+0+b \ & \Rightarrow b=-1 \ & f(-1+1)=f(-1)+f(1)+1+\frac{2}{7} \ & f(0)=f(-1)+f(1)+\frac{9}{7}\end{aligned}$ $\begin{aligned}-1=(2+3 a)+\left(\frac{a+2}{a-1}ight)(-1)+b+(2+3 a) & \ & +\frac{a+2}{a-1}+b+\frac{9}{7}\end{aligned}$ $\begin{aligned} & -1=4+6 a-2+\frac{9}{7} \ & -1=2+\frac{9}{7}+6 a \ & 6 a=-1-2-\frac{9}{7} \ & a=\frac{-5}{7} \ & f(x)=\frac{-x^2}{7}+\frac{\frac{9}{7}}{\frac{-12}{7}} x-1 \ & f(x)=\frac{-x^2}{7}-\frac{3}{4} x-1 \ & \sum_{i=1}^5 f(i)=-\frac{1}{7}\left(\frac{5 	imes 6 	imes 11}{6}ight)-\frac{3}{4}\left(\frac{5 	imes 6}{2}ight)-5 \ & =\frac{-55}{7}-\frac{45}{4}-5 \ & =\frac{675}{28} \ & \Rightarrow 28\left|\sum_{i-1}^5 f(i)ight|=675\end{aligned}$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x)=(2+3 a)…

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