If $f : [2, \infty) \to R$ be the function defined by $f(x) = x^2 - 4x + 5$, then the range of $f$ is

$[1, \infty)$

The correct answer is Option (2) → $[1, \infty)$ ##

Given that, $f(x) = x^2 - 4x + 5$

Let $y = x^2 - 4x + 5$

$\Rightarrow y = x^2 - 4x + 4 + 1 = (x - 2)^2 + 1$

$\Rightarrow (x - 2)^2 = y - 1 \Rightarrow x - 2 = \sqrt{y - 1}$

$\Rightarrow x = 2 + \sqrt{y - 1}$

$∴y - 1 \ge 0, y \ge 1$

Range $= [1, \infty)$