Let R denote the set of all real numbers and | vedclass

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(A) The function is $f(x) = x^2$.For $f$ to be one-one,$f(x_1) = f(x_2) \implies x_1^2 = x_2^2$. This implies $x_1 = x_2$ only if $x_1, x_2$ have the

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(A) The function is $f(x) = x^2$.For $f$ to be one-one,$f(x_1) = f(x_2) \implies x_1^2 = x_2^2$. This implies $x_1 = x_2$ only if $x_1, x_2$ have the same sign.For $f$ to be onto,the range of $f$ must equal the codomain $B$. The range of $f(x) = x^2$ for $x \in A$ is ${x^2 : x \in A}$.Analysis:$(A)$ $A = B = R^{+}$: $f(x) = x^2$ maps positive reals to positive reals. It is one-one (since $x_1^2 = x_2^2$ and $x_1, x_2 > 0 \implies x_1 = x_2$) and onto (since for any $y \in R^{+}$,$x = \sqrt{y} \in R^{+}$ exists). Thus,$A \rightarrow 4$.$(B)$ $A = R^{+}, B = R$: $f(x) = x^2$ is one-one (as $x > 0$),but not onto because negative values in $B$ are not covered by the range $(R^{+})$. Thus,$B \rightarrow 1$.$(C)$ $A = R, B = R^{+}$: $f(x) = x^2$ is not one-one (as $f(1) = f(-1) = 1$) but is onto (since every $y \in R^{+}$ has a pre-image $x = \pm \sqrt{y} \in R$). Thus,$C \rightarrow 3$.$(D)$ $A = B = R$: $f(x) = x^2$ is not one-one (as $f(1) = f(-1)$) and not onto (as negative values in $B$ are not covered). Thus,$D \rightarrow 2$.The correct matching is $A-4, B-1, C-3, D-2$.

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	$4$. $A = B = R^{+}$

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