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Mathematics · 2026 · 3 marks
CBSE 2026 · Region 4 · Set 1 · Q26
(a) Let A = R -{3} and B = R -{1}. A function f: A → B is defined by f ( x )=((x -2)/(x -3)). Find whether f is one-one and onto.
(a)
Let $\displaystyle \mathrm{A}=\mathbb{\mathrm{R}}-\{3\}$ and $\displaystyle \mathrm{B}=\mathbb{\mathrm{R}}-\{1\}$. A function $\displaystyle \mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ is defined by $\displaystyle \mathrm{f}(\mathrm{x})=\left(\frac{\mathrm{x}-2}{\mathrm{x}-3}\right)$. Find whether f is one-one and onto.
OR
(b)
Let n be a fixed positive integer. A relation R is defined in set Z such that $\displaystyle \mathrm{R}=\{(\mathrm{x}, \mathrm{y}):(\mathrm{x}-\mathrm{y})$ is divisible by $\displaystyle \mathrm{n}, \mathrm{x}, \mathrm{y} \in \mathrm{Z}\}$. Determine if $\displaystyle \mathrm{R}$ is an equivalence relation.
Marking-scheme solution
Let $\displaystyle f(a)=f(b)$ for some $\displaystyle a, b \in \mathbb{R}-\{3\}$
so, $\displaystyle \left(\dfrac{a-2}{a-3}\right)=\left(\dfrac{b-2}{b-3}\right) \Rightarrow a b-2 b-3 a+6=a b-3 b-2 a+6$
$\displaystyle \Rightarrow a=b$, Thus f is one-one function.
Now, let us assume for some $\displaystyle y \in \mathbb{R}-\{1\}, y=f(x)=\dfrac{x-2}{x-3}$
For getting, $\displaystyle x=\dfrac{3 y-2}{y-1}$
Range $\displaystyle =\mathbb{R}-\{1\}=$ co-domain, Thus f is onto function.
Let for some $\displaystyle a \in \mathbb{Z},(a, a) \in R$
$\displaystyle \Rightarrow 0$ is divisible by n, which is true
$\displaystyle \therefore(a, a) \in R$, so R is reflexive.
Let $\displaystyle (a, b) \in R \Rightarrow a-b=n p, p \in \mathbb{Z}$
$\displaystyle \Rightarrow b-a=n(-p)=n q, q \in \mathbb{Z} \Rightarrow b-a$ is also divisible by n.
$\displaystyle \therefore(b, a) \in R$, so R is symmetric.
Let $\displaystyle (a, b) \in R,(b, c) \in R$ for some $\displaystyle a, b, c \in \mathbb{Z}$
$\displaystyle \Rightarrow a-b=n p$ and $\displaystyle b-c=n r$, where $\displaystyle p, r \in \mathbb{Z}$
$\displaystyle \Rightarrow a-b+b-c=n(p+r) \Rightarrow(a-c)$ is divisible by n.
$\displaystyle \therefore(a, c) \in R$, so R is transitive.
Thus, R is an equivalence relation.
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.