Let $f(x)=(x+1)^2-1,x≥-1$
Statement 1 : The set $\{x:f(x)=f^{-1}(x)\}=\{0,-1\}$
Statement 2 : f is a bijection
Answer & explanation
Correct answer: option 3
There is no information about co-domain therefore f (x) is not necessarily onto.
Let $f(x)=(x+1)^2-1,x≥-1$
Statement 1 : The set $\{x:f(x)=f^{-1}(x)\}=\{0,-1\}$
Statement 2 : f is a bijection
Correct answer: option 3
There is no information about co-domain therefore f (x) is not necessarily onto.