if I remember correctly we had y=x^2 and y=ln(x)+k. I just differentiated both functions and made them equal to one another and solved for x. Because the tangent at A is the same gradient for both curves. With the x co-ord you can sub back into x^2 and you have the (x,y) of point A and it all flows from there. can find k as such. the 2nd part with the integral was simple once you had k.had anyone figured the answer to the last question in the maths adv cssa? or did someone already did and i just couldn't be bothered to read the history of this discussion.
