hi, i'm having a bit of trouble with this question, especially cii. i got c=1 but im not sure as the answer is c=1 and c>2
i've had a look at the solutions but im still a bit confused
thank you!
The second I saw this, I recalled...oh the 2017 NHT exam! The worst part is, VCAA never provide decent solutions for the NHT exams
For c)i), you just need to make the RHS = x and solve for c, so it isn't too bad and you get c < 1. As is the case with inverses, they only (in almost all circumstances unless you get into the ridiculous powers...) intersect when x=y, thus making the RHS = x is an acceptable way to solve and successfully get the two marks.
For c)ii), it's a little more difficult. I too got c=1, and upon seeing the solutions, went to my teacher (I did this exam about 3 weeks before the real one. She couldn't explain it either other than trying to do it by inspection! Again, you have to use g(x) = x (as this substitutes for the inverse), work through and you would get to x^2 + 1 - c = 0, and this would get you the c = 1 solution.
However, if you chose not to find the solution by discriminant, and instead found x, you would find x = root(c-1) and -root(c-1). As c had to fit into the domains of both g and g^-1, As the domain of g^-1 is [-1,infinity), and thus x must fit into it, you substitute x = -1 into the equation x < -root(c-1), making it 1 < root (c-1) (as subbing it into the positive would not get any real numbers due to the square root and negative).
thus root(c-1) > 1, and c >2.
Sorry for the awful explanation!
why are there two real solutions for those values? wouldnt a value like 2.54 have 2 real solutions as well?
No, as x > -1 or x = 1. If c = 2.54, the smaller x-value will be less than -1, thus out of the domain of the inverse.