Let f(x) = `(x+1)^(2) - 1, (x ge - 1)`. Then, the set S = {x : f(x) = `f^(-1)(x)`} is

Let f(x) = (x+1)^2 -1, x ge -1 ). Then the set {x: f(x) = f^(-1) (x)} is

Let f(x)=(x+1)^(2)-1, x ge -1 . Then the set {x:f(x)=f^(-1)(x)} is

Let f(x) =(x+1)^(2) - 1, x ge -1 Statement 1: The set {x : f(x) =f^(-1)(x)}= {0,-1} Statement-2: f is a bijection.

If f(x)=(x+1)^(2)-1, x ge -1 then the set S={x, f(x)=f^(-1)x} is

Let f(x) = (x+2)^(2) - 4, x ge 2 . Let S = {x : f(x) =f^(-1)(x)} , Then S is equals to

Let, f: R rarr R be given by f(x)=(x+1)^(2)-1 , x ge-1 . Then the set of values of x for which f(x)=f^(-1)(x) is given by

Let f:R rarr R be given by f(x)=(x-1)^(2)+1,xge1 . Then the set of values of x for which f(x)=f^(-1)(x) is given by