MM CAS question from NH exam 1

[mano91]

Hi All,
Would anyone be able to help me with C) ii)?
I understand where the c=1 solution comes from however the solution also includes c>2.
I have tested this on the calculator by graphing the function and its' inverse using c=3 and c=4 etc. and I do see how it works.

Also, this is aside but, for two real solutions- I tried setting the discriminant to >0 and solve, which is c>1 however this conflicts with the solution in c) ii) for "one real solution". Thus the two real solutions must come from 1<c<2 however where does this come from?

Thank you kindly for your help.

[VanillaRice]

I have answered this question in the VCE Methods Question Thread: https://atarnotes.com/forum/index.php?topic=128232.msg995651#msg995651

This question requires some careful thought about what happens to both g and its inverse as c changes.
Consider your answer of c = 1. In part a, you had c < 1. What happens when c < 1? The graphs do not intersect each other. What happens as c approaches 1 from the negative direction (i.e. as we move from c = -2 to -1 to 0....). The graphs will get closer and closer to each other (g(x) moves to the left, while its inverse moves downwards). The first time that they touch is c = 1 (have a quick sketch of the two graphs to see what this looks like). However, what happens when we keep going? If we go slightly above c = 1 (say, c = 1.1), the graphs will intersect at two points. Let's move to the case where c = 2. Here, the two graphs intersect twice, one of which is at their turning points (at the point (-1, -1), on the line y=x).  Now, we must consider the domain restriction. Since c = 2 represents where the two graphs intersect at their 'turning points', and we only have half of a 'parabola' anything beyond c = 2 will only have 1 solution (intersection) between the two graphs, since the second half of the 'parabola' doesn't actually exist (due to domain restrictions)!

I hope this makes sense, please let me know if there's any problems

Please also see Yueni's post in response to the same question for an alternative approach: https://atarnotes.com/forum/index.php?topic=128232.msg996446#msg996446

Hope this helps, please post if you would like any clarification

EDIT:

Also, this is aside but, for two real solutions- I tried setting the discriminant to >0 and solve, which is c>1 however this conflicts with the solution in c) ii) for "one real solution". Thus the two real solutions must come from 1<c<2 however where does this come from?

Think about what is happening when you square a positive square root - you are adding extra solutions. Why? Consider that g is only half a 'parabola' (turned 90 degrees) - when you square it, you are now considering both the positive and negative square roots i.e. the entire parabola.