Let f: |2, ∞) → R be the function defined by f(x) – x² – 4x + 5, then the range of f is

The correct option is [1, ∞).

To find the range of the function $f\left(x\right)=x^2-4x+5$ defined on the domain $[2,\infty )$, we can rewrite the quadratic expression by completing the square.

Let us express the function as:
$$f\left(x\right)=x^2-4x+4+1$$
This simplifies to:
$$f\left(x\right)={(x-2)}^2+1$$

Now, we analyze the behavior of this function over the given domain $x\ge 2$.
Since $x\ge 2$, subtracting 2 from both sides gives:
$$x-2\ge 0$$

Squaring both sides of this inequality (since both sides are non-negative), we get:
$${(x-2)}^2\ge 0$$

Adding 1 to both sides of the inequality yields:
$${(x-2)}^2+1\ge 1$$

Therefore, we have:
$$f\left(x\right)\ge 1$$
This shows that the minimum value of the function is 1 (occurring at the boundary $x=2$), and as $x$ approaches infinity, $f\left(x\right)$ also grows without bound. Thus, the range of the function is $[1,\infty )$.