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(A) To determine the nature of the function $f: N 	imes N ightarrow N$ defined by $f(m, n) = mn$:$1$. Check for one-one:Consider $f(1, 4) = 1 	ime

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many-one and onto

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(A) To determine the nature of the function $f: N 	imes N ightarrow N$ defined by $f(m, n) = mn$:$1$. Check for one-one:Consider $f(1, 4) = 1 	imes 4 = 4$ and $f(2, 2) = 2 	imes 2 = 4$.Since $f(1, 4) = f(2, 2)$ but $(1, 4) 
eq (2, 2)$,the function is not one-one. Therefore,it is many-one.$2$. Check for onto:For $f$ to be onto,for every $n \in N$,there must exist $(m, k) \in N 	imes N$ such that $f(m, k) = mk = n$.However,consider the prime number $7 \in N$. The only factors of $7$ are $1$ and $7$. Thus,$f(1, 7) = 7$ and $f(7, 1) = 7$. This works for primes.Wait,let us check if every $n \in N$ has a pre-image. For any $n \in N$,we can always choose $(1, n) \in N 	imes N$ such that $f(1, n) = 1 	imes n = n$.Since every element $n$ in the codomain $N$ has at least one pre-image $(1, n)$ in the domain $N 	imes N$,the function is onto.Conclusion: The function is many-one and onto. The correct option is $A$.

$(A)$ $f: R \rightarrow R$ 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

If $f: N \times N \rightarrow N$ defined by $f(m, n) = mn$,then $f$ is . . . . . . .

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