EAMCET Engineering 2010 Solved Paper

Given, f(x)=−x2,f:A→B
where, A and B∈R
R→ Real No's x∈A
Here, the domain of the function is positive real number only
ie, Domain =A∈R+
Example {1,2,3,...} and B={1,4,9,...}
Both set have f:A→B unique image.
But, A={−1,1,2,...} and B={1,1,4,...}
In f:A→B have not unique image and for Range x2=y,x=±√y
ie, (Range ∈R+)
In onto function the (Range = Co-domain =B ) ie,(B∈R+)
So,
(A) f is one-one and onto, if A=B=R+
(B f is one-one but not onto, if A=R+,B=R
(C) f is onto but not one-one, if A=R,B=R+
(D) f is neither one-one nor onto, if A=B=R Hence, the answer is
(A) →4, (B) →1, (C) →3, (D) →2