• # question_answer If $x$is real, the maximum value of$\frac{3{{x}^{2}}+9x+17}{3{{x}^{2}}+9x+7}$is     AIEEE  Solved  Paper-2006 A) 41    B)                                        1                              C)        17/7               D)                        1/4

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$