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(D) function $f : A 	o B$ is bijective if it is both one-to-one (injective) and onto (surjective). Statement $1$ states that $f$ is an onto function,

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Statement $1$ is true,Statement $2$ is true; Statement $2$ is not the correct explanation for Statement $1$.

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(D) function $f : A 	o B$ is bijective if it is both one-to-one (injective) and onto (surjective). Statement $1$ states that $f$ is an onto function,which is true by the definition of a bijective function. Statement $2$ states that there exists a function $g : B 	o A$ such that $f \circ g = I_B$. Since $f$ is bijective,it is invertible,meaning there exists an inverse function $f^{-1} : B 	o A$ such that $f \circ f^{-1} = I_B$. Thus,$g = f^{-1}$ satisfies the condition. Therefore,Statement $2$ is also true. However,Statement $2$ describes the property of invertibility resulting from bijectivity,while Statement $1$ is simply the definition of one part of bijectivity. Statement $2$ is not the reason why $f$ is onto; rather,both are consequences of $f$ being bijective. Thus,Statement $2$ is not the correct explanation for Statement $1$.

$(A)$ $f: R \rightarrow R$ is such that $f(x)=px+q$ $(p \neq 0)$,$\forall x \in R$	$I.$ $f$ is neither one-one nor onto
$(B)$ $f: R \rightarrow R^{+} \cup\{0\}$ is such that $f(x)=x^2$,$\forall x \in R$	$II.$ $f$ is both one-one and onto
$(C)$ $f: N \rightarrow N$ is such that $f(n)=n^2+2n+3$,$\forall n \in N$	$III.$ $f$ is one-one but not onto
$(D)$ $f: R \rightarrow R$ is such that $f(x)=2(\cos ^2 5x+\sin ^2 5x)$ $\forall x \in R$	$IV.$ $f$ is onto but not one-one
	$V.$ $f$ is a constant function and also a bijection

Let $A$ and $B$ be non-empty sets in $\mathbb{R}$ and $f : A \to B$ be a bijective function.
Statement $1$ : $f$ is an onto function.
Statement $2$ : There exists a function $g : B \to A$ such that $f \circ g = I_B$.

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Let $A$ and $B$ be non-empty sets in $\mathbb{R}$ and $f : A \to B$ be a bijective function. Statement $1$ : $f$ is an onto function. Statement $2$ : There exists a function $g : B \to A$ such that $f \circ g = I_B$.