Range of an equation $y=\frac{(x-\alpha)(x^3-3x+1)}{x-\alpha}$

Question

If the range of $y=\frac{(x-\alpha)(x^3-3x+1)}{x-\alpha}$ is all real numbers, then number of integers in the range of $\alpha$ is

(1) 2

(2) 3

(3) 5

(4)Infinite

I have no idea of how to do these type of problem.

Answer 1

Okay, so range of $x^3 - 3x +1$ is all real numbers clearly. But in the given function $$y = \frac{(x-\alpha)(x^3 - 3x +1)}{(x-\alpha)}$$ when we put $x = \alpha$, we get the indeterminate form $\frac{0}{0}$. For other values of of $x$, we can just cancel out the $(x-\alpha)$ factor as it is not equal to $0$.

So $y = x^3 - 3x + 1$ for all values of $x$, except for $x = \alpha$. Now if $y$ has to achieve all real values, $y(\alpha)$ must be achieved by some other value of $x$.

Now we need to think about the graph of a cubic. We know that some cubics take some values more than once, i.e., they are not strictly increasing/decreasing. As you can see in the pic below, $x^3 - 3x +1$ takes all the values from $-1$ to $+3$ more than once.

And since we've already established that $y(\alpha)$ needs to be one of these values in order for our function to be able to achieve all the real values.

$$\therefore y(\alpha) \in [-1,3] \\ \implies \alpha_{min} = x_{min}: x^3 - 3x + 1 = -1 \\ \implies (x-1)^2(x+2) = 0 \\ \implies \alpha_{min} = -2$$ $$\&$$ $$\alpha_{max} = x_{max}: x^3 - 3x + 1 = 3 \\ \implies (x+1)^2(x-2) = 0 \\ \implies \alpha_{max} = 2$$

$\therefore \alpha \in [-2, 2]$ which has $5$ integers in it.