Question

Let $f:R-\{2,6\}\to R$ be real valued function defined as $f(x)=\frac{x^2+2x+1}{x^2-8x+12}$. 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 $y=\frac{x^2+2x+1}{x^2-8x+12}$
By cross multiplying
$y{x}^2-8xy+12y-x^2-2x-1=0$
$x^2(y-1)-x(8y+2)+(12y-1)=0$
Case $1,y\ne 1$
$D\ge 0$
$\Rightarrow (8y+2{)}^2-4(y-1)(12y-1)\ge 0$
$\Rightarrow y(4y+21)\ge 0$

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