Question

Let $f: R -\{2,6\} \rightarrow R$ be real valued function defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$. Then range of $f$ is

• A. $\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$
• B. $\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
• C. $\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty)$
• D. $\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)$

Answer 1

Answer: (B)

Sol. Let $y=\frac{x^2+2 x+1}{x^2-8 x+12}$
By cross multiplying
$y x^2-8 x y+12 y-x^2-2 x-1=0 $
$x^2(y-1)-x(8 y+2)+(12 y-1)=0$
Case $1, y \neq 1$
$ D \geq 0$
$ \Rightarrow(8 y+2)^2-4(y-1)(12 y-1) \geq 0$
$ \Rightarrow y(4 y+21) \geq 0 $

$ y \in\left(-\infty, \frac{-21}{4}\right] \cup[0, \infty)-\{1\}$
Case $2, y =1$
$ x^2+2 x+1=x^2-8 x+12 $
$ 10 x=11$
$x =\frac{11}{10} $ So, $y$ can be 1
Hence $y \in\left(-\infty, \frac{-21}{4}\right] \cup[0, \infty)$