Q52 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Mathematics

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

  • A
    $\left(-\frac{1}{11}, 1\right)$
  • B
    $\left(-\infty, \frac{-1}{11}\right) \cup(1, \infty)$
  • Correct answer
    C
    $\left[\frac{-1}{11}, 1\right]$
  • D
    $\left[-1, \frac{1}{11}\right]$

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.

Hence, the correct answer is option: 'C'.

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