A) 41
B) 1
C) 17/7
D) 1/4
Correct Answer: A
Solution :
Now, \[\frac{3{{x}^{2}}+9x+17}{3{{x}^{2}}+9x+7}=1+\frac{10}{3\left( {{x}^{2}}+3x+\frac{7}{3} \right)}\] \[=1+\frac{10}{3\left( {{x}^{2}}+3x+\frac{9}{4}-\frac{9}{4}+\frac{7}{3} \right)}\] \[=1+\frac{10}{3\left[ {{\left( x+\frac{3}{2} \right)}^{2}}+\frac{1}{12} \right]}\] Maximum value of\[\frac{3{{x}^{2}}+9x+17}{3{{x}^{2}}+9x+7}\]occurs at \[x=-\frac{3}{2}\] \[\therefore \]Maximum value of\[\frac{3{{x}^{2}}+9x+17}{3{{x}^{2}}+9x+7}\] \[=1+\frac{10}{3\left( \frac{1}{12} \right)}=1+40=41\] Alternate Solution Let \[y=\frac{3{{x}^{2}}+9x+17}{3{{x}^{2}}+9x+7}\] \[\Rightarrow \]\[3(y-1){{x}^{2}}+9(y-1)x+7y-17=0\] Since, x is real. \[\therefore \]\[{{\{9y-1\}}^{2}}-4.3(y-1)(7y-17)\ge 0\] \[\Rightarrow \]\[-3{{y}^{2}}+126y-123\ge 0\] \[\Rightarrow \]\[(y-41)(y-1)\le 0\] \[\Rightarrow \]\[1\le y\le 41\]
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