Let f: R-2,6 R be real valued function defined as f(x)=(x^2+2 x+1)/(x^2-8 x+12).

Question

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

Accepted Answer

$y=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$

$\Rightarrow(y-1) x^{2}-(8 y+2) x+12 y-1=0$

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 $=\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$

Solutions · 1