If $f : [2, \infty) \to R$ 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 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 $= [1, \infty)$