JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 2)
Let $f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function
defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.
Then range of $f$ is
defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.
Then range of $f$ is
$ \left(-\infty,-\frac{21}{4}\right] \cup[1, \infty) $
$\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty) $
$\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty) $
$\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$
Explanation
$y=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$
$\Rightarrow(y-1) x^{2}-(8 y+2) x+12 y-1=0$
Let $y \neq 1$, then $D \geq 0$
$$ 4(4 y+1)^{2}-4(y-1)(12 y-1) \geq 0 $$
$\Rightarrow 16 y^{2}+1+8 y-\left(12 y^{2}-13 y+1\right) \geq 0$
$\Rightarrow 4 y^{2}+21 y \geq 0$
$\Rightarrow y \in\left(-\infty,-\frac{21}{4}\right) \cup[0, \infty)-\{1\}$
$$ \text { for } y=1 \text {, } $$
$$ \begin{aligned} & -8 x+12=2 x+1 \\\\ & x=\frac{11}{10} \quad \therefore I \in R \end{aligned} $$
$\therefore $ Range $=\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
$\Rightarrow(y-1) x^{2}-(8 y+2) x+12 y-1=0$
Let $y \neq 1$, then $D \geq 0$
$$ 4(4 y+1)^{2}-4(y-1)(12 y-1) \geq 0 $$
$\Rightarrow 16 y^{2}+1+8 y-\left(12 y^{2}-13 y+1\right) \geq 0$
$\Rightarrow 4 y^{2}+21 y \geq 0$
$\Rightarrow y \in\left(-\infty,-\frac{21}{4}\right) \cup[0, \infty)-\{1\}$
$$ \text { for } y=1 \text {, } $$
$$ \begin{aligned} & -8 x+12=2 x+1 \\\\ & x=\frac{11}{10} \quad \therefore I \in R \end{aligned} $$
$\therefore $ Range $=\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
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