Sets, Relations & Functions
Range of Rational Functions with Excluded Points
nta_pyq_2023_jan
Grade None
Question:
Let $f: \mathbb{R} - \{2,6\} \to \mathbb{R}$ be real valued function defined as $f(x) = \dfrac{x^2+2x+1}{x^2-8x+12}$. Then range of $f$ is
$\left(-\infty, -\dfrac{21}{4}\right] \cup [0,\infty)$
$\left(-\infty, -\dfrac{21}{4}\right) \cup (0,\infty)$
$\left(-\infty, -\dfrac{21}{4}\right] \cup \left[\dfrac{21}{4}, \infty\right)$
$\left(-\infty, -\dfrac{21}{4}\right] \cup [1,\infty)$
Step-by-Step Solution
Key Concept: Set $y=f(x)$, rearrange to quadratic in $x$, require discriminant $\ge 0$ and exclude values giving $x=2$ or $x=6$.
Setting $f(x)=y$ gives $x^2(1-y)+x(2+8y)+(1-12y)=0$. Discriminant $\ge 0$ gives range $\left(-\infty,-\frac{21}{4}\right]\cup[0,\infty)$.
Correct Answer: 1