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Question
Let
\[f : [2, \infty ) \to X\] be defined by
\[f\left( x \right) = 4x - x^2\] Then, f is invertible if X =
Options
\[[2, \infty )\]
\[( - \infty , 2]\]
\[( - \infty , 4]\]
\[[4, \infty )\]
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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
\[f\left( x \right) = y\]
\[ \Rightarrow 4x - x^2 = y\]
\[ \Rightarrow x^2 - 4x = - y\]
\[ \Rightarrow x^2 - 4x + 4 = 4 - y\]
\[ \Rightarrow \left( x - 2 \right)^2 = 4 - y\]
\[ \Rightarrow x - 2 = \pm \sqrt{4 - y}\]
\[ \Rightarrow x = 2 \pm \sqrt{4 - y}\]
\[\text{This is defined only when}\]
\[4 - y \geq 0\]
\[ \Rightarrow y \leq 4\]
\[X = \text{Range of f} = ( - \infty , 4]\]
So, the answer is (c).
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