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.

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) $|a+b|=|\pm 2+0|=\color{red}{2}$.
