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Mathematics · 2025 · 3 marks
CBSE 2025 · Region 6 · Set 1 · Q28
(a) A student wants to pair up natural numbers in such a way that they satisfy the equation 2 x + y =41, x, y ∈ N. Find the domain and range of the…
A student wants to pair up natural numbers in such a way that they satisfy the equation $\displaystyle 2 \mathrm{x}+\mathrm{y}=41, \mathrm{x}, \mathrm{y} \in \mathrm{N}$. Find the domain and range of the relation. Check if the relation thus formed is reflexive, symmetric and transitive. Hence, state whether it is an equivalence relation or not.
Show that the function $\displaystyle \mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$, where N is a set of natural numbers, given by $\displaystyle \mathrm{f}(\mathrm{n})=\left\{\begin{array}{l}\mathrm{n}-1, \text { if } \mathrm{n} \text { is even } \\ \mathrm{n}+1, \text { if } \mathrm{n} \text { is odd }\end{array}\right.$ is a bijection.
Marking-scheme solution
$\displaystyle \mathrm{R}=\{(1,39),(2,37), \ldots,(20,1)\}$
Domain $\displaystyle =\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$
Range $\displaystyle =\{1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39\}$
$\displaystyle (1,1)$ does not belong to R hence not reflexive
$\displaystyle ( 1,39 )$ belongs to R but $\displaystyle ( 39,1 )$ does not belong to R hence not symmetric
$\displaystyle (11, 19)$ and $\displaystyle (19, 3)$ belong to R but $\displaystyle (11, 3)$ does not belong to R hence not transitive. Hence R is not an equivalence relation.
Let $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{y})$
Let x and y are both odd or both even
Then either $\displaystyle \mathrm{x}+1=\mathrm{y}+1$ or $\displaystyle \mathrm{x}-1=\mathrm{y}-1$ gives
\[\mathrm{x}=\mathrm{y}
\]
$\displaystyle \mathrm{x}$ odd and $\displaystyle \mathrm{y}$ even is rejected as
$\displaystyle \mathrm{x}+1=\mathrm{y}-1$ gives $\displaystyle \mathrm{x}-\mathrm{y}=-2$ not possible as odd number and even number cannot differ by $\displaystyle 2$
Hence f is one-one
For onto: Let $\displaystyle \mathrm{f}(\mathrm{x})=\mathrm{y}$ gives $\displaystyle \mathrm{x}=\mathrm{y}+1$ or $\displaystyle \mathrm{x}=\mathrm{y}-1$
If y is odd, x is even and if y is even, x is odd.
\[\text { Range }=\mathrm{N}=\text { co-domain, hence onto }
\]
As f is both one-one and onto hence bijective
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CBSE Class 12 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.