If x is real , the maximum value of frac{{3{x | vedclass

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Question

$y=\frac{3 x^{2}+9 x+17}{3 x^{2}+9 x+7}$

$3 x^{2}(y-1)+9 x(y-1)+7 y-17=0$

$D \geq 0$ as $x$ is real

$81(y-1)^{2}-4 	imes 3(y-1)(7 y-17) \geq

Vedclass · Accepted Answer

$41$

Vedclass · Answer

$y=\frac{3 x^{2}+9 x+17}{3 x^{2}+9 x+7}$

$3 x^{2}(y-1)+9 x(y-1)+7 y-17=0$

$D \geq 0$ as $x$ is real

$81(y-1)^{2}-4 	imes 3(y-1)(7 y-17) \geq 0$

$\Rightarrow(y-1)(y-41) \leq 0$

$\Rightarrow 1 \leq y \leq 41$

$	herefore$ Max value of $y$ is $41$

If $x$ is real , the maximum value of $\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}$ is

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