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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
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Verified Answer
The correct answer is:
$(-\infty,-1] \cup\left[\frac{7}{9}, \infty\right)$
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, $\Delta=0$
$$
\begin{aligned}
& (\mathrm{y}+1)^2-4(\mathrm{y}-1)(-2 \mathrm{y}-2)=0 \\
& \mathrm{y}^2+1+2 \mathrm{y}+8\left(\mathrm{y}^2-1\right)=0 \\
& \mathrm{y}^2+2 \mathrm{y}+1+8 \mathrm{y}^2-8=0 \\
& 9 \mathrm{y}^2+2 \mathrm{y}-7=0 \\
& 9 \mathrm{y}^2+9 \mathrm{y}-7 \mathrm{y}-7=0 \\
& 9 \mathrm{y}(\mathrm{y}+1)-7(\mathrm{y}+1)=0 \\
& (9 \mathrm{y}-7)(\mathrm{y}+1)=0 \\
& \Rightarrow \mathrm{y}=\frac{7}{9},-1
\end{aligned}
$$
So, required set of values is $[-\infty,-1] \cup\left[\frac{7}{9}, \infty\right]$
So, option (c) is correct.
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, $\Delta=0$
$$
\begin{aligned}
& (\mathrm{y}+1)^2-4(\mathrm{y}-1)(-2 \mathrm{y}-2)=0 \\
& \mathrm{y}^2+1+2 \mathrm{y}+8\left(\mathrm{y}^2-1\right)=0 \\
& \mathrm{y}^2+2 \mathrm{y}+1+8 \mathrm{y}^2-8=0 \\
& 9 \mathrm{y}^2+2 \mathrm{y}-7=0 \\
& 9 \mathrm{y}^2+9 \mathrm{y}-7 \mathrm{y}-7=0 \\
& 9 \mathrm{y}(\mathrm{y}+1)-7(\mathrm{y}+1)=0 \\
& (9 \mathrm{y}-7)(\mathrm{y}+1)=0 \\
& \Rightarrow \mathrm{y}=\frac{7}{9},-1
\end{aligned}
$$
So, required set of values is $[-\infty,-1] \cup\left[\frac{7}{9}, \infty\right]$
So, option (c) is correct.
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