A function f: ℝ → ℝ defined as f(x) = x^2 - 4x + 5 is:

The correct answer is Option (4) → neither injective nor surjective ##

Given, $f: \mathbb{R} \rightarrow \mathbb{R}$

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

One-to-one (injective)

$f(x_1) = f(x_2)$

$⇒x_1^2 - 4x_1 + 5 = x_2^2 - 4x_2 + 5$

$⇒x_1^2 - x_2^2 = 4(x_1 - x_2)$

$⇒(x_1 - x_2)(x_1 + x_2) = 4(x_1 - x_2)$

$⇒x_1 + x_2 = 4$

$⇒x_1 = 4 - x_2$

Thus, $f(x)$ is not an injective mapping.

Onto (surjective)

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

$⇒y = (x - 2)^2 + 1$

$⇒y - 1 = (x - 2)^2$

$⇒x = \sqrt{(y - 1)} + 2$

Thus, for any value of $y < 1$, $x \notin \mathbb{R}$. So, we don't have a pre-image for all $y \in \mathbb{R}$ in $x \in \mathbb{R}$.

Thus, $f(x) = x^2 - 4x + 5$ is not surjective.