If $|ax^2+bx+c|\le 1\ \forall |x|\le 1$, then what is the maximum possible value of $\frac 83a^2+2b^2$? Let $f(x) = ax^2 + bx + c$ ; $a,b,c\in\mathbb R$
It is given that $|f(x)| \le 1$  $\forall |x| \le 1$
Q1) The possible value of $|a+c|$, if $\displaystyle \frac{8}{3} a^2 + 2b^2$ is maximum, is given by:
a) $0$
b) $1$
c) $2$
d) $3$
Q2) The possible value of $|a+b|$, if $\displaystyle\frac{8}{3} a^2 + 2b^2$ is maximum, is given by:
a) $0$ 
b) $1$
c) $2$
d) $3$
Q3) The maximum possible value of $\displaystyle\frac{8}{3} a^2 + 2b^2$ is given by:
a) $32$
b) $\displaystyle\frac{32}{3}$
c) $\displaystyle\frac{2}{3}$
d) $\displaystyle\frac{16}{3}$
I have no idea how to go about this question and any help will be appreciated.
 A: I'm going to write an answer because the given answer seems to have some errors. 
First of all, noting that $a,b,c$ can be written as
$$a=\frac 12\left(f(-1)+f(1)-2f(0)\right),\quad b=\frac 12\left(f(1)-f(-1)\right),\quad c=f(0)$$
might make things easy.
Let us consider first Q3).
$$\begin{align}\frac 83a^2+2b^2&=\frac 43\left(2a^2+\frac 32b^2\right)\\&=\frac 43\left(a^2+2ab+b^2+a^2-2ab+b^2-\frac 12b^2\right)\\&=\frac 43\left((a+b)^2+(a-b)^2-\frac 12b^2\right)\\&=\frac 43\left(\left(f(1)-f(0)\right)^2+\left(f(-1)-f(0)\right)^2-\frac 12b^2\right)\\&=\frac 43\left(|f(1)-f(0)|^2+|f(-1)-f(0)|^2-\frac 12b^2\right)\\&\le \frac 43\left(\left(|f(1)|+|f(0)|\right)^2+\left(|f(-1)|+|f(0)|\right)^2-\frac 12 b^2\right)\\&\le \frac 43\left(\left(1+1\right)^2+\left(1+1\right)^2-\frac 12\cdot 0^2\right)\\&=\frac{32}{3}\end{align}$$
This is attained if and only if
$$(f(-1),f(0),f(1))=(1,-1,1),(-1,1,-1),$$
i.e.
$$(a,b,c)=(2,0,-1),(-2,0,1).$$
So, the maximum possible value of $\frac 83a^2+2b^2$ is $\color{red}{\frac{32}{3}}$.
Q1) $|a+c|=|\pm 2\mp 1|=\color{red}{1}$.
Q2) $|a+b|=|\pm 2+0|=\color{red}{2}$.
