Let `f:[-1,oo] in [-1,oo]` be a function given `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.

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.

Let f be a function defined by f(x)=(x-1)^(2)+x, (x ge 1) . Statement-1: The set [x:f(x)=f^(-1)(x)]={1,2} Statement-2: f is a bijectioon and f^(-1)(x)=1+sqrt(x-1), x ge 1

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

Let f be a function defined by f(x)=(x-1)^(2)+1,(xge1) . Statement 1: The set (x:f(x)=f^(-1)(x)}={1,2} Statement 2: f is a bijection and f^(-1)(x)=1+sqrt(x-1),xge1 .

Let f be a function defined by f(x)=(x-1)^(2)+1,(xge1) . Statement 1: The set (x:f(x)=f^(-1)(x)}={1,2} Statement 2: f is a bijection and f^(-1)(x)=1+sqrt(x-1),xge1 .

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 S = {x : f(x) = f^(-1)(x)} is

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