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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q32
(a) A relation R is defined on Z, the set of integers, as R ={(x, y ):|x- y | is divisible by a prime number ' p ', x, y ∈ Z } check whether R is an…
(a)
A relation R is defined on Z , the set of integers, as $\displaystyle \mathrm{R}=\{(x, \mathrm{y}):|x-\mathrm{y}|$ is divisible by a prime number ' p ', $\displaystyle x, \mathrm{y} \in \mathrm{Z}\}$ check whether $\displaystyle \mathrm{R}$ is an equivalence relation or not.
OR
(b)
A function $\displaystyle \mathrm{f}: \mathrm{R}-\left\{\frac{3}{5}\right\} \longrightarrow \mathrm{R}-\left\{\frac{3}{5}\right\}$ is defined as $\displaystyle \mathrm{f}(x)=\frac{3 x+2}{5 x-3}$. Show that $\displaystyle \mathrm{f}$ is one-one and onto.
Official answer
Part (a): R is not an equivalence relation because it is not transitive. Part (b): f is bijective (one-one and onto).
Marking-scheme solution
Let $\displaystyle x\in \mathrm{Z}$. Then $\displaystyle |x-x|=0$, which is divisible by a prime number p.
Hence, $\displaystyle (x,x)\in \mathrm{R}$. Thus, R is reflexive.
Let $\displaystyle x,\mathrm{y}\in \mathrm{Z}$ such that $\displaystyle (x,\mathrm{y})\in \mathrm{R}$.
Then $\displaystyle |x-\mathrm{y}|$ is divisible by the prime number p.
$\displaystyle \Rightarrow |\mathrm{y}-x|$ is divisible by the prime number p as $\displaystyle |x-\mathrm{y}|=|\mathrm{y}-x|$
Hence, $\displaystyle (\mathrm{y},x)\in \mathrm{R}$. Thus, R is symmetric.
Let $\displaystyle x,\mathrm{y},z\in \mathrm{Z}$ such that $\displaystyle (x,\mathrm{y}),(\mathrm{y},z)\in \mathrm{R}$.
Then $\displaystyle |x-\mathrm{y}|$ is divisible by the prime number p and $\displaystyle |\mathrm{y}-z|$ is divisible by the prime number p.
$\displaystyle \Rightarrow x-\mathrm{y}$ is divisible by the prime number p and $\displaystyle \mathrm{y}-z$ is divisible by the prime number p.
$\displaystyle \Rightarrow x-\mathrm{y}+\mathrm{y}-z$ is divisible by the prime number p
$\displaystyle \Rightarrow x-z$ is divisible by the prime number p
$\displaystyle \Rightarrow |x-z|$ is divisible by the prime number p
Hence, $\displaystyle (x,z)\in \mathrm{R}$. Thus, R is transitive.
Since, R is reflexive, symmetric and transitive, therefore, R is an equivalence relation.
Let $\displaystyle x_1,x_2\in \mathrm{R}-\left\{\dfrac{3}{5}\right\}$ such that $\displaystyle \mathrm{f}(x_1)=\mathrm{f}(x_2)$
$\displaystyle \Rightarrow \dfrac{3x_1+2}{5x_1-3}=\dfrac{3x_2+2}{5x_2-3}$
$\displaystyle \Rightarrow 15x_1x_2-9x_1+10x_2-6=15x_1x_2+10x_1-9x_2-6$
$\displaystyle \Rightarrow x_1=x_2$
Hence, f is one-one.
Let $\displaystyle \mathrm{y}\in \mathrm{R}-\left\{\dfrac{3}{5}\right\}$ (Codomain).
Then $\displaystyle \mathrm{f}(x)=\mathrm{y}$
or, $\displaystyle \dfrac{3x+2}{5x-3}=\mathrm{y} \Rightarrow x=\dfrac{3\mathrm{y}+2}{5\mathrm{y}-3}\in$ Domain.
Hence, for every $\displaystyle \mathrm{y}\in \mathrm{R}-\left\{\dfrac{3}{5}\right\}$ (codomain),
there exists $\displaystyle x=\dfrac{3\mathrm{y}+2}{5\mathrm{y}-3}\in \mathrm{R}-\left\{\dfrac{3}{5}\right\}$ (domain) such that $\displaystyle \mathrm{f}\left(\dfrac{3\mathrm{y}+2}{5\mathrm{y}-3}\right)=\mathrm{y}$
Thus, f is onto.
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