Let $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \geqslant - 1$$
Statement -1 : The set $$\left\{ x \right.:f\left( x \right) = {f^{ - 1}}\left( x \right) = \left\{ {0, - 1} \right\}$$
Statement-2 : $$f$$ is a bijection.

Solution :
Given that $$f\left( x \right) = {\left( {x + 1} \right)^2} - 1,x \geqslant - 1$$
Clearly $${D_f} = \left[ { - ,\infty } \right)$$   but co-domain is not given. Therefore $$f\left( x \right)$$  need not be necessarily onto.
But if $$f\left( x \right)$$  is onto then as $$f\left( x \right)$$  is one one also, $$\left( {x + 1} \right)$$   being something $$+ve,$$  $${f^{ - 1}}\left( x \right)$$  will exist where $${\left( {x + 1} \right)^2} - 1 = y$$
$$\eqalign{ & \Rightarrow x + 1 = \sqrt {y + 1} \,\left( {{\text{ + ve}}\,{\text{square root as}}\,x + 1 \geqslant 0} \right) \cr & \Rightarrow x = - 1 + \sqrt {y + 1} \Rightarrow {f^{ - 1}}\left( x \right) = \sqrt {x + 1} - 1 \cr & {\text{Then}}\,f\left( x \right) = {f^{ - 1}}\left( x \right) \Rightarrow {\left( {x + 1} \right)^2} - 1 = \sqrt {x + 1} - 1 \cr & \Rightarrow {\left( {x + 1} \right)^2} = \sqrt {x + 1} \Rightarrow {\left( {x + 1} \right)^4} = \left( {x + 1} \right) \cr & \Rightarrow \left( {x + 1} \right)\left[ {{{\left( {x + 1} \right)}^3} - 1} \right] = 0 \Rightarrow x = - 1,0 \cr} $$
$$\therefore $$ The statement-1 is correct but statement-2 is false.