Mathematics · 2023
JEE Main · 31 January 2023, Shift 2 · Q72
Let f: R -{2,6} → R be real valued function defined as f(x)=(x^2+2 x+1)/(x^2-8 x+12). Then range of f is
Let $\displaystyle f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function defined as $\displaystyle f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.
Then range of $\displaystyle f$ is
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle \left(-\infty,-\frac{21}{4}\right] \cup[0, \infty)$
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JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.