If $x \in R $ and $\frac{(x-a)(x-c)}{(x-b)}$ can assume any real value, then which of the following may be correct If $x \in R $ and $\frac{(x-a)(x-c)}{(x-b)}$ can assume any real value, then which of the following may be correct 


*

*$a>b>c$

*$a>c>b$

*$b>c>a$

*$b>a>c$


I have tried simplifying this equation like - 
$$\frac{(x-a)(x-c)}{(x-b)} = y$$
$$x^2 -xc -ax + ac = xy - by$$
$$x^2 -x(c+a-y) + (ac+by) = 0$$
Since this is always real, $D\ge 0$
$$(c+a-y)^2 - 4(ac+by) \ge  0$$
$$c^2 + a^2 + y^2 + 2(ac -cy-ay) - 4ac - 4by \ge 0$$
However, I have no idea how to continue from here. Can someone possibly advice on how to continue?
 A: You have started in the right way but you have made a small slip with the algebra. You should have $$y^2+y(2c+2a-4b)+(a-c)^2\geq0$$
For this quadratic to be non-negative, it must have a discriminant which is negative or zero so
$$(2c+2a-4b)^2-4(a-c)^2\leq0$$
This simplifies to $$(b-a)(b-c)\leq0$$
So we can conclude, ignoring the equal signs, that $b$ must lie between $a$ and $c$ So the first option could be correct.
A: Intuitively speaking, you can get all positive values, independent of the value of $b$ (since the leading term of the quadratic is always positive). Now, to receive all negative values, we have to have that $b$ lies between the roots of the polynomial in the denominator (so we can get the remaining negative values).
With that 'magical' guess, we can then formulate a simple proof:
Let's assume that $a < c$, and by the above 'guess' we should have $a<b<c$. Then Picking $x \ge c$, we have that
$$
\frac{(x-a)(x-c)}{x-b} \ge 0
$$
as expected. It remains to be proven that this function achieves all values in this range, but I leave this as an exercise for the reader (hint: consider the fact that the function grows unboundedly and is everywhere continuous in the interval $x \ge c$).
Now, allow $b < x \le c$, with $x\downarrow b$, then we know that
$$
\frac{(x-a)(x-c)}{x-b} \to -\infty
$$
since $(x-a)(x-c) < 0$ when $a< x < c$, while
$$
\frac{(x-a)(x-c)}{x-b} = 0
$$
at $x=c$. Again, by continuity, we have that the function achieves all negative real values. Putting both statements together completes the proof.

I was not able to find an elementary proof for necessity of the condition (which I totally forgot to prove above) that did not make use of calculus. Attempting to construct an element that couldn't be found in the range via using only simple bounds was somewhat convoluted and not particularly pretty.
Anyways, taking $a \le c < b$, wlog, it's not hard to note that we will have exactly two stationary points, one at $x< b$ and another at $x > b$ where the function is coercive (i.e., diverges to $\pm \infty$ at the boundary, diverging with the same sign at all points of the boundary) in each domain, respectively. Evaluating the function at each stationary point is enough to show that some values are never achieved.
Any simpler proof of the above statement would be great!
