If a, b, c ϵ R and a ≠ 0, c > 0, the graph of f(x) = ax2+bx+c for which f(x)=0 has only imaginary roots, will look like
Given f(x) = ax2+bx+c has imaginary roots
⇒ D < 0
b2−4ac < 0
b2< 4ac
As b ϵ R, b2 is always > 0 and for 4ac > b2, 4ac should be > 0
⇒ a & c should have same sign
As c > 0, a> 0
Hence as a > 0 and D < 0, graph is bowl shaped and don't touch x-axis ⇒ f(x) is always positive.
So, A is right option