Răspuns:
[tex]f(x)=x^2-mx+m-1; a=1;b=-m;c=m-1[/tex]
Primul caz, cand f(x) are doua solutii reale de semn contrar:
[tex]x1 =[/tex] prima solutie
[tex]x2=[/tex] a doua solutie
[tex]x1=\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex] (1)
[tex]x2=\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex] (2)
Din (1) si (2) => [tex]\sqrt{b^2-4ac}>b[/tex] si [tex]-\sqrt{b^2-4ac}>b[/tex]
[tex]Delta=b^2-4ac=(-m)^2-4*1*(m-1)=m^2-4m+4[/tex]
[tex]\sqrt{m^2-4m+1}<|-m|<=>m^2-4m+1<m^2<=>\\-4m+1<0=> -4m<-1=>m<\frac{1}{4}[/tex]
[tex]m[/tex]∈[tex](-inf;\frac{1}{4})[/tex]
Al doilea caz, cand f(x) are două soluții reale, una inversă celeilalte:
[tex]x1=\frac{1}{x2}[/tex]
[tex]P=\frac{c}{a}=\frac{m-1}{1}=m-1[/tex][tex]=> m-1=x1*x2 <=>m-1=\frac{1}{x2}*x2 =>m-1=1=>m=2[/tex]
