Let $f: [2, \infty) \to \mathbb{R}$ be the function defined by $f(x) = x^2 - 4x + 5$, then the range of $f$ is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $[1, \infty)$ ##
Given, $f(x) = x^2 - 4x + 5$
Let $y = x^2 - 4x + 5 \Rightarrow y = x^2 - 4x + 4 + 1$
$\Rightarrow y = (x - 2)^2 + 1 \Rightarrow (x - 2)^2 = y - 1$
$\Rightarrow x - 2 = \sqrt{y - 1} \Rightarrow x = \sqrt{y - 1} + 2$
Since, $y - 1 \ge 0 \Rightarrow y \ge 1 \quad ∴\text{Range} = [1, \infty)$