Question
Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \geqslant - 1.$$ Then the set $$\left\{ {x:f\left( x \right) = {f^{ - 1}}\left( x \right)} \right\}$$ is
A.
$$\left\{ {0, - 1,\frac{{ - 3 + i\sqrt 3 }}{2},\frac{{ - 3 - i\sqrt 3 }}{2}} \right\}$$
B.
$$\left\{ {0,1, - 1} \right\}$$
C.
$$\left\{ {0, - 1} \right\}$$
D.
empty
Answer :
$$\left\{ {0, - 1} \right\}$$
Solution :
$$\eqalign{
& f\left( x \right) = {f^{ - 1}}\left( x \right) \Rightarrow fof\left( x \right) = x \cr
& {\left[ {{{\left( {x + 1} \right)}^2} - 1 + 1} \right]^2} - 1 = x \cr
& \Rightarrow {\left( {x + 1} \right)^4} = x + 1 \cr
& \therefore x = 0\,{\text{or}}\, - 1 \cr
& \therefore {\text{Req}}{\text{.}}\,{\text{set}}\,{\text{is}}\,\left\{ {0, - 1} \right\} \cr} $$