Re: How did people go in the maths comp today?
I thought it was easier than that. I looked at f1(x) = x. It's a quadratic with solutions of -2 and 3. Then I looked at f2(x) = x. It's a quadratic with solutions of -2 and 3. Then I looked at f3(x) = x. It's a quadratic with solutions of -2 and 3. So on a hunch .... set of real solutions = S = {-2,3} and n(S) = 2.I had no idea what it was either, and it looked pretty daunting at first sight, but take a look at the examples, is a fraction with the variable x in both the numerator and denominator. , after simplifying is still a fraction with the variable still a power of 1. Following this pattern, will still be a fraction with the variable x in both the numerator and denominator with a power of 1.
Since it asked how many solutions there were to this would result in a quadratic if the denominator is multiplied through. Quadratics have 2 solutions and I think (not completely sure) but the set S means either real or complex solutions, i.e. it doesn't matter if the quadratic doesn't have any real roots. - if this isn't what it means, please elaborate how to distinguish between the answer 1 and 2.
Thus, the answer would be 2, even though I chose 2n in the midst of all the mayhem