If x is real then the range of x2−2x+9x2+2x+9 is

x2−2x+9x2+2x+9=y⇒(1−y)x2−(1+y)2x+9(1−y)=0;Δ≥0⇒2y2−5y+2≤0⇒y∈[12,2]

If x is real then the range of x2−2x+9x2+2x+9 is

x2−2x+9x2+2x+9=y⇒(1−y)x2−(1+y)2x+9(1−y)=0;Δ≥0⇒2y2−5y+2≤0⇒y∈[12,2]

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If $x$ is real then the range of $\frac{x^{2} - 2 x + 9}{x^{2} + 2 x + 9}$ is

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$[\frac{1}{2}, 2]$

$(- \infty, - \frac{2}{9}] \cup (1, \infty)$

$(- \infty, 0] \cup (1, \infty)$

$(- \infty, - 6] \cup [- 2, \infty)$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{x^{2} - 2 x + 9}{x^{2} + 2 x + 9} = y \Rightarrow (1 - y) x^{2} - (1 + y) 2 x + 9 (1 - y) = 0; \\ Δ \geq 0 \Rightarrow 2 y^{2} - 5 y + 2 \leq 0 \Rightarrow y \in [\frac{1}{2}, 2] \end{array}$

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If x is real then the range of x2−2x+9x2+2x+9 is

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If $x$ is real then the range of $\frac{x^{2} - 2 x + 9}{x^{2} + 2 x + 9}$ is