If the curve $y=a x^{2}+b x+c, x \in R,$ passes through the point (1,2) and the tangent line to this curve at origin is $y = x ,$ then the possible values of $a , b , c$ are :
$a+b+c=2$...(1)
and $\left.\frac{ dy }{ dx }\right|_{(0,0)}=1$
$2 ax +\left. b \right|_{(0,0)}=1$
$b =1$
Curve passes through origin So,
$c=0$ and $a=1$