If $|ax^2+bx+c|\le 1\ \forall |x|\le 1$, then what is the maximum possible value of $\frac 83a^2+2b^2$?

Question

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.

Answer 1

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