Let f : [2, ∞)→X be defined by f(x)


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) $[4, \infty)$


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


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

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

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


$\Rightarrow x-2=\pm \sqrt{4-y}$

$\Rightarrow x=2 \pm \sqrt{4-y}$

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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