Question

Problema 1.23 și 1.24! Va rog!!

Answer 1

$1.24\\ \\ f:R->R~si~f(x)=\sqrt{mx^2-(m-1)x+(m-1)}.\\ \\ Conditie:mx^2-(m-1)x+(m-1)\geq 0.\\ \\ Aceasta~are~loc~daca~m>0~si~\Delta\leq 0.\\ \\ La~tine,~\Delta=(m-1)^2-4m(m-1)=m^2-2m+1-4m^2+4m=-3m^2+2m+1.\\ \\ Deci,~-3m^2+2m+1\leq 0.\\ \\ Asociem~ecuatia~de~gradul~2~cu~radacinile~m_1=1~si~m_2=-\frac{1}{3}.\\ \\ Dupa~ce~faci~tabelul,~vei~avea~ca~m\in~(-\infty,-\frac{1}{3}]~\cup~[1,\infty).\\ \\ Dar,~m>0.\\ \\ In~final,~conform~axei,~m\in~[1,\infty). \\ \\ 1.23\\ \\ Egalezi~f(x)=g(x)~\Leftrightarrow~mx^2-2x(m+2)-1=x^2-2x+m\\ \\ x^2(m-1)-2x(m+1)-(m+1)=0\\ \\ Cu~conditia~\Delta>0.\\ \\ Tu~ai~\Delta=4(m+1)^2+4(m+1)(m-1)=8m^2+8m.\\ \\ Deci,~8m(m+1)>0~\Leftrightarrow~m^2+m>0\\ \\Asociezi~ecuatia~de~gradul~2~cu~radacinile~m_1=-1~si~m_2=0\\ \\ Faci~tabel~si~ai~m \in~(-\infty,-1)~\cup~(0,+\infty).$