Question:
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) $[5, \infty)$
Solution:
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).