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.

Options

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

C

Explanation

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.