If $\quad f:[2, \infty) \rightarrow B \quad$ defined by $f(x)=x^2-4 x+5$ is a bijection, then $B$ is equal to
If $\quad f:[2, \infty) \rightarrow B \quad$ defined by $f(x)=x^2-4 x+5$ is a bijection, then $B$ is equal to
$[0, \infty)$
$[1, \infty)$
$[4, \infty)$
$[5, \infty)$
Solution
Given, $f(x)=x^2-4 x+5$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=2 x-4$
Put $f^{\prime}(x)=0 \Rightarrow x=2$
For $x>2, f^{\prime}(x)>0$, increasing
$\therefore$ Minimum value is $f(2)=4-8+5=1$
$\therefore \quad B \in[1, \infty)$