Răspuns:
[tex]m \in \boxed{1}\,\cup\,\boxed{2} \Rightarrow m\in \varnothing\,\cup\,(1,+\infty) \Rightarrow \boxed{m \in (1,+\infty)}[/tex]
Explicație pas cu pas:
[tex]f:\mathbb{R}\to\mathbb{R},\quad f(x) = (m-1)x^2-2mx+m+1[/tex]
[tex]\\\\\text{Cazul }\boxed{1}:\\ \text{Conditiile sunt: }\begin{cases} m-1 < 0\\ \Delta < 0\end{cases} \Rightarrow \begin{cases} m<1\\ 4m^2-4(m-1)(m+1) < 0\end{cases} \Rightarrow \\ \\ \\ \Rightarrow \begin{cases}m<1 \\4m^2-4(m^2-1)<0\end{cases} \Rightarrow \begin{cases}m<1 \\4<0\quad (F)\end{cases} \Rightarrow\\\\\\\Rightarrow\begin{cases}m<1\\m\in\varnothing\end{cases}\Bigg|\Rightarrow m\in (-\infty, 1)\,\cap\,\varnothing \Rightarrow m\in \varnothing[/tex]
[tex]\\\\\text{Cazul }\boxed{2}: \\ \text{Conditiile sunt: }\begin{cases}m-1 > 0\\ \Delta > 0\end{cases} \Rightarrow \begin{cases}m>1\\ 4 > 0\quad (A)\end{cases} \Rightarrow \\ \\ \Rightarrow \begin{cases}m>1\\ m\in \mathbb{R}\,\backslash\,\{1\}\end{cases}\Bigg| \Rightarrow m\in (1,+\infty)\,\cap\, \mathbb{R}\,\backslash\,\{1\} \Rightarrow\\ \\ \\\Rightarrow m \in (1,+\infty)[/tex]
