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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 2 · Set 1 · Q36
A school is organizing a debate competition with participants as speakers S ={ S_1, S_2, S_3, S_4} and these are judged by judges J ={ J_1, J_2,…
A school is organizing a debate competition with participants as speakers $\displaystyle \mathrm{S}=\left\{\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}, \mathrm{~S}_{4}\right\}$ and these are judged by judges $\displaystyle \mathrm{J}=\left\{\mathrm{J}_{1}, \mathrm{~J}_{2}, \mathrm{~J}_{3}\right\}$. Each speaker can be assigned one judge. Let R be a relation from set S to J defined as $\displaystyle \mathrm{R}=\{(x, \mathrm{y})$ : speaker $\displaystyle x$ is judged by judge $\displaystyle \mathrm{y}, x \in \mathrm{~S}, \mathrm{y} \in \mathrm{J}\}$.
Based on the above, answer the following :
(i)
How many relations can be there from S to J ?
(ii)
A student identifies a function from S to J as $\displaystyle \mathrm{f}=\left\{\left(\mathrm{S}_{1}, \mathrm{~J}_{1}\right),\left(\mathrm{S}_{2}, \mathrm{~J}_{2}\right)\right.$, $\displaystyle \left.\left(\mathrm{S}_{3}, \mathrm{~J}_{2}\right),\left(\mathrm{S}_{4}, \mathrm{~J}_{3}\right)\right\}$ Check if it is bijective.
(iii)
(a)
How many one-one functions can be there from set S to set J ?
OR
(b)
Another student considers a relation $\displaystyle \mathrm{R}_{1}=\left\{\left(\mathrm{S}_{1}, \mathrm{~S}_{2}\right),\left\{\mathrm{S}_{2}, \mathrm{~S}_{4}\right)\right\}$ in set S . Write minimum ordered pairs to be included in $\displaystyle \mathrm{R}_{1}$ so that $\displaystyle \mathrm{R}_{1}$ is reflexive but not symmetric.
Marking-scheme solution
(i)
The number of relations $\displaystyle =2^{4 \times 3}=2^{12}$
(ii)
Since, $\displaystyle \mathrm{S}_{2}$ and $\displaystyle \mathrm{S}_{3}$ have been assigned the same judge $\displaystyle J_{2}$, the function is not one-one.
Hence, it is not bijective.
(iii)
(a)
There cannot exist any one-one function from S to J as $\displaystyle n(\mathrm{S})>n(\mathrm{J})$. Hence, the number of one-one functions from S to J is 0.
OR
(b)
To make $\displaystyle R_{1}$ reflexive and not symmetric we need to add the following ordered pairs:
$\displaystyle \left(S_{1}, S_{1}\right),\left(S_{2}, S_{2}\right),\left(S_{3}, S_{3}\right),\left(S_{4}, S_{4}\right)$
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