A) 2, 1
B) \[5,\frac{1}{5}\]
C) \[7,\frac{1}{7}\]
D) None of these
Correct Answer: C
Solution :
Let \[y=\frac{{{x}^{2}}-3x+4}{{{x}^{2}}+3x+4}\] Þ \[(y-1){{x}^{2}}+3(y+1)x+4(y-1)=0\] For x is real \[D\ge 0\] Þ \[9{{(y+1)}^{2}}-16{{(y-1)}^{2}}\ge 0\] Þ \[-7{{y}^{2}}+50y-7\ge 0\]Þ\[7{{y}^{2}}-50y+7\le 0\] Þ \[(y-7)(7y-1)\le 0\] Now, the product of two factors is negative if one in and one in . Case I : \[(y-7)\ge 0\] and \[(7y-1)\le 0\] Þ \[y\ge 7\]and \[y\le \frac{1}{7}\]. But it is impossible Case II : \[(y-7)\le 0\]and \[(7y-1)\ge 0\] Þ \[y\le 7\]and \[y\ge \frac{1}{7}\Rightarrow \frac{1}{7}\le y\le 7\] Hence maximum value is 7 and minimum value is \[\frac{1}{7}\]You need to login to perform this action.
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