AP EAMCET · Maths · Functions
If \(x \in R\), then the range of \(\frac{x}{x^2-5 x+9}\) is
- A \(\left(-\frac{1}{11}, 1\right)\)
- B \(\left(-\infty, \frac{-1}{11}\right) \cup(1, \infty)\)
- C \(\left[\frac{-1}{11}, 1\right]\)
- D \(\left[-1, \frac{1}{11}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\frac{-1}{11}, 1\right]\)
Step-by-step Solution
Detailed explanation
\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 \\…
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