Let A be the set of all 50 students of Class | vedclass

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(N/A) $1$. To check for one-one: Let $x_1$ and $x_2$ be two different students in set $A$. Since no two different students can have the same roll numb

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(N/A) $1$. To check for one-one: Let $x_1$ and $x_2$ be two different students in set $A$. Since no two different students can have the same roll number,$f(x_1) 
eq f(x_2)$. Thus,$f$ is one-one.$2$. To check for onto: The codomain of $f$ is the set of natural numbers $N = \{1, 2, 3, ...\}$. The range of $f$ is the set of roll numbers assigned to the $50$ students,which is $\{1, 2, 3, ..., 50\}$.$3$. Since the range $\{1, 2, 3, ..., 50\}$ is a proper subset of the codomain $N$ (e.g.,$51 \in N$ but $51$ is not in the range),there exists at least one element in $N$ that has no pre-image in $A$.$4$. Therefore,$f$ is not onto.

Let $A$ be the set of all $50$ students of Class $X$ in a school. Let $f: A \rightarrow N$ be a function defined by $f(x) = \text{roll number of the student } x$. Show that $f$ is one-one but not onto.

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