Question:
Let $f:[2, \infty) \rightarrow X$ be defined by $f(x)=4 x-x^{2}$. Then, $f$ is invertible if $X=$
(a) $[2, \infty)$
(b) $(-\infty, 2]$
(c) $(-\infty, 4]$
(d) $[4, \infty)$
Solution:
Since f is invertible, range of f = co domain of f = X
So, we need to find the range of f to find X.
For finding the range, let
$f(x)=y$
$\Rightarrow 4 x-x^{2}=y$
$\Rightarrow x^{2}-4 x=-y$
$\Rightarrow x^{2}-4 x+4=4-y$
$\Rightarrow(x-2)^{2}=4-y$
$\Rightarrow x-2=\pm \sqrt{4-y}$
$\Rightarrow x=2 \pm \sqrt{4-y}$
This is defined only when
$4-y \geq 0$
$\Rightarrow y \leq 4$
$X=$ Range of $f=(-\infty, 4]$
So, the answer is (c).
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