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If $f:[2, \infty) \rightarrow \mathbb{R}$ is defined by $f(x)=x^2-4 x+5$, then the range of $f$ is
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$[1, \infty)$
$f:[2, \infty) \rightarrow R$
$\begin{aligned} & f(x)=x^2-4 x+5 \\ & \because \quad D < 0 \\ & \therefore \quad f(x)>0, \forall x \in R \\ & \text { Vertex }\left(\frac{-b}{2 a}, \frac{-D}{4 a}\right) \equiv\left(\frac{4}{2}, \frac{4}{4}\right) \equiv(2,1) \\ & \because \quad \text { for } \quad x \in[2, \infty) \\ & \quad y \in[1, \infty)\end{aligned}$
$\begin{aligned} & f(x)=x^2-4 x+5 \\ & \because \quad D < 0 \\ & \therefore \quad f(x)>0, \forall x \in R \\ & \text { Vertex }\left(\frac{-b}{2 a}, \frac{-D}{4 a}\right) \equiv\left(\frac{4}{2}, \frac{4}{4}\right) \equiv(2,1) \\ & \because \quad \text { for } \quad x \in[2, \infty) \\ & \quad y \in[1, \infty)\end{aligned}$
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