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If $x$ is real, the maximum value of $\frac{3 x^2+9 x+17}{3 x^2+9 x+7}$ is
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The correct answer is:
41
41
$y=\frac{3 x^2+9 x+17}{3 x^2+9 x+7}$
$3 x^2(y-1)+9 x(y-1)+7 y-17=0$
$D \geq 0 \quad \because x$ is real
$81(y-1)^2-4 x 3(y-1)(7 y-17) \geq 0$
$\Rightarrow(y-1)(y-41) \leq 0 \Rightarrow 1 \leq y \leq 41$
$3 x^2(y-1)+9 x(y-1)+7 y-17=0$
$D \geq 0 \quad \because x$ is real
$81(y-1)^2-4 x 3(y-1)(7 y-17) \geq 0$
$\Rightarrow(y-1)(y-41) \leq 0 \Rightarrow 1 \leq y \leq 41$
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