TS EAMCET · Maths · Functions
The set of all real values of the expression \(\frac{x^2-x+2}{x^2+x-2}\) for all \(x \in \mathbb{R}-\{-2,1\}\) is
- A \((-2,3)\)
- B \(\left[\frac{7}{9}, \infty\right)\)
- C \((-\infty,-1] \cup\left[\frac{7}{9}, \infty\right)\)
- D \((-\infty,-1]\)
Answer & Solution
Correct Answer
(C) \((-\infty,-1] \cup\left[\frac{7}{9}, \infty\right)\)
Step-by-step Solution
Detailed explanation
Given \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^2-\mathrm{x}+2}{\mathrm{x}^2+\mathrm{x}-2}=\forall \in \mathbf{R}-\{-2,1\}\) Take \(y=\frac{x^2-x+2}{x^2+x-2}\) \[ \begin{aligned} & y^2+y x-2 y=x^2-x+2 \\ & x^2(y-1)+x(y+1)+(-2 y-2)=0 \end{aligned} \] Here, \(\Delta=0\)…
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