3411) If you let t = \frac{y}{x}, then we can reformulate the problem as, under what conditions will the roots of the quadratic, at^2+2ht+b=0 have roots, m and -1/m.
Since product of roots is -1, we must have \frac{b}{a}=-1\Rightarrow a+b=0
2) Now, you can try this one
3) WLOG a<b<cf = \frac{(x-a)(x-c)}{(x-b)}; f: (-\infty, b) \rightarrow \mathbb{R}
It is continuous everywhere in its domain.
Now, as x \rightarrow -\infty , f(x) \rightarrow -\infty and asx \rightarrow b , f(x) \rightarrow \infty
Since its continuous, from Mean Value Theorem f takes all real values.
A similar analysis for (b, \infty) shows us that the given expression takes on every real value at least twice
