Let f: \mathbb{R}-\{2,6\} \to \mathbb{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 [31-Jan-2023 Shift 2]

Correct Answer is : (−∞,−214]∪[0,∞)

(-\infty,-\frac{21}{4}] \cup[0, \infty)

(-\infty,-\frac{21}{4}) \cup(0, \infty)

(-\infty,-\frac{21}{4}] \cup [\frac{21}{4}, \infty)

(-\infty,-\frac{21}{4}] \cup[1, \infty)

JEE Mains 31-Jan-2023 Shift 2 Solved Paper

Section: Mathematics

Question : 70 of 90

Marks: +1, -0

Let f:ℝ−{2,6}→ℝ be real valued function defined as f(x)=‌

x2+2x+1

x2−8x+12

. Then range of f is

[31-Jan-2023 Shift 2]

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