Question
Let $$f:\left( { - \infty ,\,1} \right] \to \left( { - \infty ,\,1} \right]$$ such that $$f\left( x \right) = x\left( {2 - x} \right).$$ Then $${f^{ - 1}}\left( x \right)$$ is :
A.
$$1 + \sqrt {1 - x} $$
B.
$$1 - \sqrt {1 - x} $$
C.
$$\sqrt {1 - x} $$
D.
none of these
Answer :
$$1 - \sqrt {1 - x} $$
Solution :
$$\eqalign{
& y = x\left( {2 - x} \right) \cr
& \Rightarrow {x^2} - 2x + y = 0 \cr
& \Rightarrow x = \frac{{2 \pm 2\sqrt {1 - y} }}{2} = 1 \pm \sqrt {1 - y} \cr
& {\text{But }}x \leqslant 1.\,\,{\text{So, }}x = 1 - \sqrt {1 - y} \cr
& {\text{Therefore,}}\,\,{f^{ - 1}}\left( x \right) = 1 - \sqrt {1 - x} \cr} $$