Let f: R R be a function defined by f(x)=(2+3 a) x^2+((a+2)/(a-1)) x+b, a 1.

Question

Let $f: \mathbb{R} ightarrow \mathbb{R}$ be a function defined by $f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}ight) x+b, a 
eq 1$. 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

Accepted Answer

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

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

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

Solutions · 1