Let $f : [2, \infty) \to R$ be the function defined by $f (x) = x^{2} - 4 x + 5,$
then the range of $f$ is

[4, \infty)

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[1, \infty)

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R

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[5, \infty)

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Solution

The correct option is B $[1, \infty)$
Convert the given quadratic into perfect square form

Given: $f : [2, \infty) \to R$ be the function

defined by $f (x) = x^{2} - 4 x + 5$

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

$\Rightarrow f (x) = (x - 2)^{2} - 4 + 5$

$\Rightarrow f (x) = (x - 2)^{2} + 1$

Find the range

Now, $(x - 2)^{2} \geq 0$

$(x - 2)^{2} + 1 \geq 1$

$\Rightarrow f (x) \geq 1$

$∴ Range of f = [1, \infty)$

Hence, Option (B) is correct

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

Let $f : [2, \infty) \to R$ be the function defined by $f (x) = x^{2} - 4 x + 5,$
then the range of $f$ is