A) 4, - 5
B) 5, - 4
C) - 4, 5
D) - 4, - 5
Correct Answer: A
Solution :
Let \[y=\frac{{{x}^{2}}+14x+9}{{{x}^{2}}+2x+3}\] Þ \[y({{x}^{2}}+2x+3)-{{x}^{2}}-14x-9=0\] \[\Rightarrow \] \[(y-1){{x}^{2}}+(2y-14)x+3y-9=0\] For real x, its discriminant \[\ge 0\] i.e. \[4{{(y-7)}^{2}}-4(y-1)3(y-3)\ge 0\] Þ \[{{y}^{2}}+y-20\le \]0 or \[(y-4)(y+5)\le 0\] Now, the product of two factors is negative if these are of opposite signs. So following two cases arise: Case I: \[y-4\ge 0\]or \[y\ge 4\] and \[y+5\le 0\]or \[y\le -5\] This is not possible. Case II: \[y-4\le 0\] or \[y\le 4\]and \[y+5\ge 0\]or \[y\ge -5\] Both of these are satisfied if \[-5\le y\le 4\] Hence maximum value of y is 4 and minimum value is - 5.
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