This problem is from a Sri Lankan grade 12 term test paper.
Let $y=\frac{1-kx}{(x-1)(x-2)}$; $x \not=1,2$. Show that $1/2<k<1$, such that $y$ can get any real value for two distinct real values of $x$.
I solved this one in a way. But, I am still not satisfied with my methods. Anyone please explain what those methods do. I would like to know what are the applications of these problems.
My effort:
I transformed this equation to: $$yx^2+(k-3y)x+(2y-1)=0$$
Since there are two distict values for $x$, discriminant>0. By simplifying that inequality, $$y^2+(4-6k)y+k^2>0$$
Since $y^2$ term coefficient>0, last discriminent must be negative. That gives: $(2k-1)(k-1)<0$ which implies the asked solution.
I think that you are meant to prove that the function $$ f(x) = \frac{1 - kx}{(x-1)(x-2)} $$ is a $2$-to-$1$ function with range all of $\mathbb{R}$. In other words, for any real number $y$, there are precisely $2$ distinct $x$-values such that $f(x) = y$. It turns out that this is almost true. Each $y \neq 0$ has $2$ preimages, but $y = 0$ has only the preimage $x = \tfrac1k$.
EDIT: I made a silly algebra error, actually looking at solutions to $y = \frac{1}{f(x)}$. But this is equivalent to $Y = f(x)$ for $Y = \frac1y$ (since $y \neq 0$ must be assumed), so the analysis that follows still works!
Your approach will work. When we solve $Y = f(x)$ for $x$, we obtain $$ y(1 - kx) = x^2 - 3x + 2 $$ or $$ x^2 + (ky - 3)x + (2 - y) = 0, $$ where $y = \frac1Y$, as remarked upon above.
As you note, this is a quadratic equation in $x$ with discriminant $$ \Delta = (ky - 3)^2 - 4(2 - y) = k^2y^2 + (4 - 6k)y + 1. $$ We would like to show that for any $\frac12 < k < 1$ and any $y \neq 0$, this quantity is positive, which amounts to showing that
$$ (2 - 3k)^2 < k^2, $$ i.e. that $$ (2k - 1)(k - 1) < 0, $$ which is equivalent to the hypotheses about $k$.