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.A. Statement 1 is ture, statement 2 is true, statement 2 is a correct explanation for statement 1.B. Statement 1 is ture, statement 2 is true, statement 2 is not a correct explanation for statement 1.C. Statement 1 is ture, statement 2 is false.D. Statement 1 is false, statement 2 is true.

Let `f:[-1,oo] in [-1,oo]` be a function given `f(x)=(x+1)^(2)-1, x ge

1.	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.A. Statement 1 is ture, statement 2 is true, statement 2 is a correct explanation for statement 1.B. Statement 1 is ture, statement 2 is true, statement 2 is not a correct explanation for statement 1.C. Statement 1 is ture, statement 2 is false.D. Statement 1 is false, statement 2 is true.
Answer» Correct Answer - C There is no information about co-domain therefore `f(x)` is not necessarily onto. Therefore statement 2 is false. However for `x ge -1, f(x) =(x+1)^(2)-1` is one-one. Assume that `f(x)` is onto. Now roots of `f(x)=f^(-1)(x)` lies on line `y=x`. So we have to find roots of `f(x)=x or (x+1)^(2)-1=x` ` :. x^(2) +x=0` ` :. x=0, -1` Hence, statement 1 is true.