If x is real, then the greatest and least values of the expression $\frac{x^{2} - 2 x + 2}{x^{2} + 3 x + 9}$

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2 and $- \frac{2}{27}$

2 and $\frac{2}{27}$

2 and $\frac{1}{27}$

2 or 3

answer is B.

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Detailed Solution

$\begin{array}{l} L e t y = \frac{x^{2} - 2 x + 2}{x^{2} + 3 x + 9} \\ \Rightarrow x^{2} y + 3 x y + 9 y = x^{2} - 2 x + 2 \\ \Rightarrow (y - 1) x^{2} + (3 y + 2) x + 9 y - 2 = 0 \\ \forall x \in R, D \geq 0 \\ \Rightarrow {(3 y + 2)}^{2} - 4 (y - 1) (9 y - 2) \geq 0 \\ \Rightarrow 9 y^{2} + 4 + 12 y - 4 (9 y^{2} - 2 y - 9 y + 2) \geq 0 \\ \Rightarrow 9 y^{2} + 4 + 12 y - 36 y^{2} + 8 y + 36 - 8 \geq 0 \\ \Rightarrow - 27 y^{2} + 56 y - 4 \geq 0 \\ \Rightarrow 27 y^{2} - 56 y + 4 \leq 0 \\ \Rightarrow 27 y^{2} - 54 y - 2 y + 4 \leq 0 \\ \Rightarrow 27 y (y - 2) - 2 (y - 2) \leq 0 \\ \Rightarrow (y - 2) (27 y - 2) \leq 0 \\ \Rightarrow \frac{2}{27} \leq y \leq 2 \end{array}$