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

If $x \in R$, then the range of $\frac{x}{x^2-5 x+9}$ is
  1. $\left(-\frac{1}{11}, 1\right)$
  2. $\left(-\infty, \frac{-1}{11}\right) \cup(1, \infty)$
  3. $\left[\frac{-1}{11}, 1\right]$
  4. $\left[-1, \frac{1}{11}\right]$

Solution

$ \begin{aligned} & \text {Let } \frac{x}{x^2-5 x+9}=y \\ & \Rightarrow \quad y x^2-(5 y+1) x+9 y=0 \\ & \because \quad x \in R \Rightarrow D \geq 0 \Rightarrow(5 y+1)^2-36 y^2 \geq 0 \\ & \Rightarrow \quad 25 y^2+10 y+1-36 y^2 \geq 0 \\ & \Rightarrow \quad 11 y^2-10 y-1 \leq 0 \\ & \Rightarrow \quad 11 y^2-11 y+y-1 \leq 0 \\ & \Rightarrow \quad(11 y+1)(y-1) \leq 0 \Rightarrow y \in\left[-\frac{1}{11}, 1\right] \end{aligned} $ Hence, option (c) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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