Question

Let $f:[2,\infty )\to X$ be defined by $f\left(x\right)=4x-x^2$. Then, f is invertible if X =
(a) $[2,\infty )$
(b) $(-\infty ,2]$
(c) $(-\infty ,4]$
(d) $[4,\infty )$

Answer 1

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
$$\begin{gathered} f\left(x\right)=y \\ \Rightarrow 4x-x^2=y \\ \Rightarrow x^2-4x=-y \\ \Rightarrow x^2-4x+4=4-y \\ \Rightarrow {\left(x-2\right)}^2=4-y \\ \Rightarrow x-2=\pm \sqrt{4-y} \\ \Rightarrow x=2\pm \sqrt{4-y} \\ \text{This is defined only when} \\ 4-y\ge 0 \\ \Rightarrow y\le 4 \\ X=\text{Range of}f=(-\infty ,4] \end{gathered}$$
So, the answer is (c).