4-2.Quadratic Equations and Inequations
hard

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

A

$\frac{1}{4}$

B

$1$

C

$41$

D

$\frac{{17}}{7}$

(AIEEE-2006)

Solution

$y=\frac{3 x^{2}+9 x+17}{3 x^{2}+9 x+7}$

$3 x^{2}(y-1)+9 x(y-1)+7 y-17=0$

$D \geq 0$ as $x$ is real

$81(y-1)^{2}-4 \times 3(y-1)(7 y-17) \geq 0$

$\Rightarrow(y-1)(y-41) \leq 0$

$\Rightarrow 1 \leq y \leq 41$

$\therefore$ Max value of $y$ is $41$

Standard 11
Mathematics

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