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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

1) $1 / 4$
2) 41
3) 1
4) $17 / 7$
in Mathematics by Gold Status (10,261 points) | 8 views

1 Answer

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Answer

(2)

Explanation

$y=\frac{3 x^{2}+9 x+17}{3 x^{2}=9 x+7}$
$3 x^{2}(y-1)+9 x(y-1)+7 y-17=0$
$D \geq 0 \quad \because x$ is real
$81(y-1)^{2}-4 x 3(y-1)(7 y-17) \geq 0$
$\Rightarrow(y-1)(y-41) \leq 0 \Rightarrow 1 \leq \cdot y \leq 41$

by Gold Status (10,261 points)

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