A) \[[1/3,\,\,3]\]
B) \[(1/3,\,\,3)\]
C) \[(3,\,\,3)\]
D) \[(-1/3,\,\,3)\]
Correct Answer: A
Solution :
Let \[y=\frac{{{x}^{2}}-2x+4}{{{x}^{2}}+2x+4}\] \[\Rightarrow \] \[{{x}^{2}}(y-1)+x(2y+2)+4y-4=0\] Since, \[x\] is real therefore its discriminant \[{{b}^{2}}-4ac\ge 0\] \[\therefore \] \[{{(2y+2)}^{2}}-4(y-1)4(y-1)\ge 0\] \[\Rightarrow \] \[4{{y}^{2}}+8y+4-16{{y}^{2}}-16+32y\ge 0\] \[\Rightarrow \] \[-12{{y}^{2}}+40y-12\ge 0\] \[\Rightarrow \] \[3{{y}^{2}}-10y+3\le 0\] \[\Rightarrow \] \[(3y-1)(y-3)\le 0\] \[\Rightarrow \] \[\frac{1}{3}y\le 3\]
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