The range of x2−2x+9x2+2x+9 is

y=y2−2x+9x2+2x+9⇒(y−1)x2+(2y+2)x+(9y−9)=0X is real ⇒(2y+2)2−4(y−1)(9y−9)≥0⇒4y2+4+8y−4(9y2−18y+9)≥0⇒−32y2+80y−32≥0⇒2y2−5y+2≤0⇒(2y−1)(y−2)≤0⇒1/2≤y≤2⇒ Range [1/2,2]

The range of x2−2x+9x2+2x+9 is

y=y2−2x+9x2+2x+9⇒(y−1)x2+(2y+2)x+(9y−9)=0X is real ⇒(2y+2)2−4(y−1)(9y−9)≥0⇒4y2+4+8y−4(9y2−18y+9)≥0⇒−32y2+80y−32≥0⇒2y2−5y+2≤0⇒(2y−1)(y−2)≤0⇒1/2≤y≤2⇒ Range [1/2,2]

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The range of $\frac{x^{2} - 2 x + 9}{x^{2} + 2 x + 9}$ is

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$[1 / 2, 2]$

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

$(- \infty, - 2 / 9] \cup (1, \infty)$

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

answer is B.

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Detailed Solution

$y = \frac{y^{2} - 2 x + 9}{x^{2} + 2 x + 9} \Rightarrow (y - 1) x^{2} + (2 y + 2) x + (9 y - 9) = 0$

X is real $\Rightarrow {(2 y + 2)}^{2} - 4 (y - 1) (9 y - 9) \geq 0 \Rightarrow 4 y^{2} + 4 + 8 y - 4 (9 y^{2} - 18 y + 9) \geq 0$

$\Rightarrow - 32 y^{2} + 80 y - 32 \geq 0 \Rightarrow 2 y^{2} - 5 y + 2 \leq 0 \Rightarrow (2 y - 1) (y - 2) \leq 0 \Rightarrow 1 / 2 \leq y \leq 2$ $\Rightarrow$ Range $[1 / 2, 2]$