Finding the relation between coefficients of quadratic equation 
My attempt:
I could prove  $|c| < 1$.
As it given that $f\lt 1$, so $f(0)\lt 1$.
I have solved using triangular inequality. Is there any other way?
ans is ABCD
 A: On the boundaries $x=0$ and $x=1$ the inequality corresponds to


*

*$x=0 \implies |c|\le 1$

*$x=1 \implies |a+b+c|\le 1 \implies 0\le|a+b|\le 2$
Let wlog $a \ge 0$, since the extrema values are also reached at $x=-\frac{b}{2a}$ we need to consider $3$ cases:
$1) \quad 0\le-\frac{b}{2a}\le1 \iff -2a\le b\le 0$


*

*$x=1 \implies |a+b+c|\le 1 \implies 0\le|a+b|\le 2 \implies 0\le a \le 2 \land -4\le b \le 0$

*$x=-\frac{b}{2a} \implies \left|\frac {b^2} {4a}-\frac{b^2}{2a}+c\right|\le 1\implies 0\le \left|-\frac{b^2}{4a}\right|\le 2\implies 0\le b^2 \le 8a\implies -2\sqrt {2a}\le b \le 0$
and since


*

*$2\sqrt {2a}=2a \iff \sqrt {2a}=a\iff a=2$


we have


*

*$0\le a \le 2$

*$-4\le b \le 0$


$2) \quad b\ge 0$


*

*$x=1 \implies |a+b+c|\le 1 \implies 0\le a+b \le 2 $


and since $a \ge 0$ we have


*

*$0\le a \le 2$

*$-2\le b \le  2$


$3) \quad b\le -2a $


*

*$x=1 \implies |a+b+c|\le 1 \implies 0\le |a+b| \le 2 \implies -2\le a+b \le 0$


and we have


*

*$0\le a \le 2$

*$-4 \le b \le 0$


Therefore all options are correct.
