Given that f(x)=(x+1)2−1,x≥−1 Clearly Df=[−1,∞) but co-domain is not given ∴f(x) need not be necessarily onto. But if f(x) is onto then as f(x) is one one also, (x+1) being something +ve,f−1(x) will exist where (x+1)2−1=y ⇒x+1=√y+1 (+ve square root as x+1≥0) ⇒x=−1+√y+1 ⇒f−1(x)=√x+1−1 Then f(x)=f−1(x) ⇒(x+1)2−1=√x+1−1 ⇒(x+1)2=√x+1 ⇒(x+1)4=(x+1) ⇒(x+1)[(x+1)3−1]=0 ⇒x=−1,0 ∴ The statement −1 is correct but statement- 2 is false.