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. $\left(-\infty,-\frac{21}{4}\right] \cup[1, \infty)$

Hint:

To find the range of a rational function, check the discriminant of the quadratic obtained by cross-multiplying and ensure the solutions satisfy the domain restrictions.

Answer 1

The Correct Option is B

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 $y\in \left(-\infty ,\frac{-21}{4}\right]\cup [0,\infty )$
So, the correct option is (B) : $\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$

Answer 2

Let \( y = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}. \) By cross-multiplying: \[ y(x^2 - 8x + 12) = x^2 + 2x + 1. \] Simplifying the equation: \[ yx^2 - 8xy + 12y = x^2 + 2x + 1, \] \[ yx^2 - x^2 - 8xy + 12y - 2x - 1 = 0. \] Case 1: Assume \( y \neq 1 \). \[ x^2(y - 1) - x(8y + 2) + (12y - 1) = 0. \] The discriminant condition for real solutions is \( D \geq 0 \). Simplifying: \[ (8y + 2)^2 - 4(y - 1)(12y - 1) \geq 0. \] Step 1: Solving this inequality results in the range for \( y \), which is \[ y \in \left( -\infty, \frac{-21}{4} \right] \cup [0, \infty). \] Case 2: Assume \( y = 1 \). Substitute into the equation: \[ x^2 + 2x + 1 = x^2 - 8x + 12. \] Simplifying: \[ 10x = 11 \quad \Rightarrow \quad x = \frac{11}{10}. \] Thus, \( y \) can be 1.
Step 2: Combining the solutions, the range of \( f(x) \) is \[ \left( -\infty, \frac{-21}{4} \right] \cup [0, \infty). \]