question_answer

If \[x\] is real, then the maximum and minimum values of the expression \[\frac{{{x}^{2}}-3x+4}{{{x}^{2}}+3x+4}\] will be [IIT 1984]

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 \[\frac{1}{7}\]