Question

Let $R$ denote the set of all real numbers and $R^{+}$ denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$ define $f: A \rightarrow B$ by $f(x)=x^{2}$ for $x \in A$. Observe the two lists given below

Column I		Column II
A.	$f$ is one-one and onto, if	1.	$A=R^{+}, B=R$
B.	$f$ is one-one but not onto, if	2.	$A=B=R$
C.	$f$ is onto but not one-one, if	3.	$A=R, B=R^{+}$
D.	$f$ is neither one-one nor onto, if	4.	$A=B=R^{+}$

• A. (A)-(1), (B)-(2), (C)-(3), (D)-(4)
• B. (A)-(4), (B)-(2), (C)-(1), (D)-(3)
• C. (A)-(4), (B)-(1), (C)-(3), (D)-(2)
• D. (A)-(4), (B)-(2), (C)-(3), (D)-(1)

Answer 1

Answer: (C)

Given, $f(x)=-x^{2}, f: A \rightarrow B$
where, $A$ and $B \in R$
$R \rightarrow$ Real No's $x \in A$
Here, the domain of the function is positive real number only
ie, Domain $=A \in R^{+}$
Example $\{1,2,3, \ldots\}$ and $B=\{1,4,9, \ldots\}$
Both set have $f: A \rightarrow B$ unique image.
But, $A=\{-1,1,2, \ldots\}$ and $B=\{1,1,4, \ldots\}$
In $f: A \rightarrow B$ have not unique image and for Range
$x^{2}=y, x=\pm \sqrt{y}$
ie, $\left(\right.$ Range $\left.\in R^{+}\right)$
In onto function the (Range $=$ Co-domain $=B$ )
ie, $\left(B \in R^{+}\right)$
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
$(A) \rightarrow 4,(B) \rightarrow 1,(C) \rightarrow 3,(D) \rightarrow 2$