comparision...

f(x) lies completely below or above the x-axis as it has no roots. But as f(1)<0 f(x) lies completely below the x-axis. Therefore f(0) will also be less than 0, which means c<0.

$a+b+c=f(1)<0$

If $c=f(0)\geq0$ , $f(x)$ must have a real root (either at $0$ for $c=0$ or between $0$ and $1$ for $c>0$ .

Thus, the correct answer must be $\boxed{c<0}$

If the value of c becomes zero , the equation becomes pure quadratic. Therefore the answer is c <0