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(B) Given the function $f(x) = x^2 - 4x + 5$ defined on the domain $[2, \infty)$.To be a bijection,the codomain $B$ must be equal to the range of the

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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 be a bijection,the codomain $B$ must be equal to the range of the function.We find the range by analyzing the function's behavior.First,find the derivative: $f'(x) = 2x - 4$.Setting $f'(x) = 0$ gives $x = 2$.For all $x \in [2, \infty)$,$f'(x) \geq 0$,which means the function is monotonically increasing on this interval.The minimum value occurs at the boundary $x = 2$.$f(2) = (2)^2 - 4(2) + 5 = 4 - 8 + 5 = 1$.As $x 	o \infty$,$f(x) 	o \infty$.Since the function is continuous and increasing,the range is $[f(2), \infty) = [1, \infty)$.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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