Question

Am nevoie de ajutor!!!!

Answer 1

Răspuns:

 $m \in (\frac{1}{2} ,1) \cup(1,2)$

Explicație pas cu pas:

Matricea $A$ este inversabila dacă $det (A) \ne 0$

a)

$det(A)=-2m+6x+x^2+3-4x-mx^2=(1-m)x^2+2x+3-2m\\(1-m)x^2+2x+3-2m \ne 0 \ \forall x \in \mathbb{R} \\1-m \ne 0 \implies m \ne 1$

$(1-m)x^2+2x+3-2m =0\\\Delta = 4-4(1-m)(3-2m)=-8m^2+20m-8$

$\Delta < 0$

$2m^2-5m+2=0\\\Delta_m=25-16=9\\m_{1,2}=\frac{5 \pm3}{4} =2,\frac{1}{2}$

$2m^2-5m+2 < 0 \implies m \in (\frac{1}{2} ,2)$

$m \in (\frac{1}{2} ,1) \cup(1,2)$