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

( − ∞ , − 6 ] ∪ [ − 2 , ∞ )

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

( − ∞ , − 6 ] ∪ [ − 2 , ∞ )

( − ∞ , 0 ] ∪ ( 1 , ∞ )

( − ∞ , − 6 ] ∪ [ − 2 , ∞ )

− ∞ , − 2 9 ∪ ( 1 , ∞ )

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

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

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

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

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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