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(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.To find the range $B$ for the function to be a bijection, we need to

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$[1, \infty)$

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(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.To find the range $B$ for the function to be a bijection, we need to determine the set of all possible values of $f(x)$.First, we rewrite the function by completing the square:$f(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$.Since the domain is $x \in [2, \infty)$, we have $x - 2 \geq 0$.Therefore, $(x - 2)^2 \geq 0$.Adding $1$ to both sides, we get $(x - 2)^2 + 1 \geq 1$.Thus, $f(x) \geq 1$.The range of the function is $[1, \infty)$.Since the function is strictly increasing on $[2, \infty)$, it is injective. For it to be a bijection, the codomain $B$ must be equal to the range.Therefore, $B = [1, \infty)$.

If $f:[2, \infty) \rightarrow B$ defined by $f(x)=x^2-4x+5$ is a bijection, then $B$ is equal to

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