Problem 10.1. a) Find the image of the function $\frac{x^{2}}{x-1}$.

b) Find all real numbers $a$ such that the equation

$$
x^{4}-a x^{3}+(a+1) x^{2}-2 x+1=0
$$
10.1. a) Set $t=\frac{x^{2}}{x-1}$. Then $x^{2}-t x+t=0$. This quadratic equation has a real root if its discriminant is non-negative. Then $t^{2}-4 t \geq 0$ which shows that the desired range is $(-\infty ; 0] \cup[4 ;+\infty)$.

b) Write the equation in the form $f(x)=x^{4}-a x^{2}(x-1)+(x-1)^{2}=0$. Obviously $x=1$ is not a solution. Then dividing both sides by $(x-1)^{2}$ and setting $t=\frac{x^{2}}{x-1}$, we get the equation $t^{2}-a t+1=0$. If its discriminant $a^{2}-4$ is negative, i.e. $a \in(-2 ; 2)$, the equation $f(x)=0$ has no real solutions. If $a \in(-\infty ;-2] \cup[2 ;+\infty)$, denote by $t_{1}$ and $t_{2}$ the roots of the above equation.

Now a) implies that $f(x)=0$ has no real roots if and only if $t_{1}, t_{2} \in(0 ; 4)$.

Setting $g(t)=t^{2}-a t+1$, this is equivalent to

$$
\begin{aligned}
& g(0)>0 \\
& g(4)>0 \\
& 0<\frac{a}{2}<4
\end{aligned} \quad \text { which gives }
$$

$a \in\left(0 ; \frac{17}{4}\right)$. Having in mind that $a \in(-\infty ;-2] \cup[2 ;+\infty)$, we get that $a \in\left[2 ; \frac{17}{4}\right)$.