If x∈R then the range of xx2-5x+9 is

Let y=xx2-5x+9⇒yx2-5xy+9y=x ⇒yx2-5y+1x+9y=0 ∀ x∈R ,∆≥0 ⇒b2-4ac≥0 ⇒5y+12-4y9y≥0 ⇒25y2+1+10y-36y2≥0 ⇒-11y2+10y+1≥0 ⇒11y2-10y-1≤0 ⇒11y2-11y+y-1≤0 ⇒11yy-1+1y+1≤0 ⇒y-111y+1≤0 ⇒y∈-111 , 1

If x∈R then the range of xx2-5x+9 is

Let y=xx2-5x+9⇒yx2-5xy+9y=x ⇒yx2-5y+1x+9y=0 ∀ x∈R ,∆≥0 ⇒b2-4ac≥0 ⇒5y+12-4y9y≥0 ⇒25y2+1+10y-36y2≥0 ⇒-11y2+10y+1≥0 ⇒11y2-10y-1≤0 ⇒11y2-11y+y-1≤0 ⇒11yy-1+1y+1≤0 ⇒y-111y+1≤0 ⇒y∈-111 , 1

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If $x \in R$ then the range of $\frac{x}{x^{2} - 5 x + 9}$ is

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$(- \frac{1}{11}, 1)$

$[\frac{- 1}{11}, 1]$

$(- 1_{1} \frac{1}{1}))$

$(- \infty \frac{- 1}{11}) \cup (1, \infty)$

answer is C.

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Detailed Solution

Let $y = \frac{x}{x^{2} - 5 x + 9}$

$\Rightarrow y x^{2} - 5 x y + 9 y = x \Rightarrow y x^{2} - (5 y + 1) x + 9 y = 0 \forall x \in R, ∆ \geq 0 \Rightarrow b^{2} - 4 a c \geq 0 \Rightarrow {(5 y + 1)}^{2} - 4 y (9 y) \geq 0 \Rightarrow 25 y^{2} + 1 + 10 y - 36 y^{2} \geq 0 \Rightarrow - 11 y^{2} + 10 y + 1 \geq 0 \Rightarrow 11 y^{2} - 10 y - 1 \leq 0 \Rightarrow 11 y^{2} - 11 y + y - 1 \leq 0 \Rightarrow 11 y (y - 1) + 1 (y + 1) \leq 0 \Rightarrow (y - 1) (11 y + 1) \leq 0 \Rightarrow y \in [\frac{- 1}{11}, 1]$