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Interesting question!

If there is at all an intersection, it would be a single point (obvious)

In that case

=> aX - b = X2 + b

=> X2 – aX + 2b = 0


For a real solution of X,

i.è. B2 – 4AC > 0



i.e. (-a)2 – 4 * 2b >0

i.e. b < a2/8



Irrespective of any sign of a, if b is less than zero, that would do the needful, therefore B should be the right answer.


Regards,
Mayank

mayank,

let go parabola but whenever a single degree equation cuts an n degree equation there will be a n point cut.

since I cannot see the options, I don't know the final response but here is a way to solve it:

x^2 + B = Ax - B
<=> x^2 - Ax + 2B = 0 <=>
<=> x=[A +- sqrt(A^2 - 8B)]/2

therefore we have to analyse de root:
if A^2 - 8B < 0 - No intersection
if A^2 - 8B = 0 - One intersection
if A^2 - 8B > 0 - two intersections

To anwser the question we need to know if there is at least one inters, i.e., A^2 - 8B >=0,

Thus,
[A >= sqrt(8B) or A < -sqrt(8B) ] and b>0

Now we can only anwser depending on the info given..