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Q.
If x is real, the maximum value of $\frac{3x^2 + 9x + 17}{3x^2 + 9x + 7}$ is :
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Solution:
$\frac{3x^{2}+9x+17}{3x^{2}+9x+7}$
$=1+\frac{10}{3\left(x^{2}+3x+\frac{7}{3}\right)}$
$=1+\frac{10}{3\left[\left(x+\frac{3}{2}\right)^{2}+\frac{1}{12}\right]}$
$\Rightarrow $ Maximum value of $\frac{3x^{2}+9x+17}{3x^{2}+9x+7}$ occurs at $x=-\frac{3}{2}\cdot$
$\therefore $ Maximum value of $\frac{3x^{2}+9x+17}{3x^{2}+9x+7}$
$=1+\frac{10}{3\left(\frac{1}{12}\right)}$
= 1 + 40 = 41