If a function


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

(a) $R$

(b) $[1, \infty)$

(C) $[4, \infty)$

(d) $[5, \infty)$


Since f is a bijection, co-domain of f = range of f

$\Rightarrow B=$ range of $f$

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

Let $f(x)=y$

$\Rightarrow y=x^{2}-4 x+5$

$\Rightarrow x^{2}-4 x+(5-y)=0$

$\because$ Discrimant, $D=b^{2}-4 a c \geq 0$,

$(-4)^{2}-4 \times 1 \times(5-y) \geq 0$

$\Rightarrow 16-20+4 y \geq 0$

$\Rightarrow 4 y \geq 4$

$\Rightarrow y \geq 1$

$\Rightarrow y \in[1, \infty)$

$\Rightarrow$ Range of $f=[1, \infty)$

$\Rightarrow B=[1, \infty)$

So, the answer is (b).

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