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Q.

$I f f : [2, \infty) \to B d e f i n e d b y f (x) = x^{2} - 4 x + 5 i s a b i j e c t i o n, t h e n B =$

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a

$[0, \infty)$

b

$[1, \infty)$

c

$[4, \infty)$

d

$[5, \infty)$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = x^{2} - 4 x + 4 + 1 \\ = {(x - 2)}^{2} + 1 \\ ∵ 2 \leq x < \infty \\ \Rightarrow 0 \leq x - 2 < \infty \\ \Rightarrow 0 \leq {(x - 2)}^{2} < \infty \\ \Rightarrow 1 \leq {(x - 2)}^{2} + 1 < \infty \\ ∴ B = R a n g e = [1, \infty) \end{array}$

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