A) \[\frac{1}{3}\le y\le 3\]
B) \[3<y\le 5\]
C) \[5<y\le 7\]
D) \[7<y\le 9\]
Correct Answer: A
Solution :
\[y=\frac{{{x}^{2}}-2x+4}{{{x}^{2}}+2x+4}\] or \[y({{x}^{2}}+2x+4)={{x}^{2}}-2x+4\] or \[(y-1){{x}^{2}}+2(y+1)x+4(y-1)=0\] Since \[x\] is real, the discriminant \[D\] satisfies. \[D\ge 0\] or \[4{{(y+1)}^{2}}-16{{(y-1)}^{2}}\ge 0\] or \[({{y}^{2}}+2y+1)-4({{y}^{2}}-2y+1)\ge 0\] or \[-3{{y}^{2}}+10y-3\ge 0\] or \[3{{y}^{2}}-10y+3\le 0\] or \[(3y-1)(y-3)\le 0\] or \[\frac{1}{3}\le y\le 3\]
You need to login to perform this action.
You will be redirected in
3 sec
