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

Question

Let $f:R	o R$ be a function defined by $f(x)=(2+3a){x}^{2}+(\frac{a+2}{a-1})x+b,a
e 1$. 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

Accepted Answer

$\begin{gathered} f(x)=(3a+2){x}^{2}+(\frac{a+2}{a-1})x+b \ f(x+\frac{1}{2})=f(x)+f(y)+1-\frac{2}{7}xy\, \end{gathered}$

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

So, $f(0)=0+0+b=-1\Rightarrow b=-1$

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

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

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

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

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

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

Solutions · 1