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Range of $f(x)=\frac{x^2+34 x-71}{x^2+2 x-7}$ is
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Verified Answer
The correct answer is:
$(-\infty, 5] \cup[9, \infty)$
$\begin{aligned}
& \text { Let } \frac{x^2+34 x-71}{x^2+2 x-7}=y \\
& \Rightarrow x^2(1-y)+2(17-y) x+(7 y-71)=0
\end{aligned}$
For real value of $x, B^2-4 A C \geq 0$
$\Rightarrow y^2-14 y+45 \geq 0 \Rightarrow y \geq 9, y \leq 5$
& \text { Let } \frac{x^2+34 x-71}{x^2+2 x-7}=y \\
& \Rightarrow x^2(1-y)+2(17-y) x+(7 y-71)=0
\end{aligned}$
For real value of $x, B^2-4 A C \geq 0$
$\Rightarrow y^2-14 y+45 \geq 0 \Rightarrow y \geq 9, y \leq 5$
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