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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 2 · Set 1 · Q21
(a) Check whether f: R -{3} → R defined as f (x)=(x-2)/(x-3) is onto or not.
(a)
Check whether $\displaystyle \mathrm{f}: \mathrm{R}-\{3\} \rightarrow \mathrm{R}$ defined as $\displaystyle \mathrm{f}(x)=\frac{x-2}{x-3}$ is onto or not.
OR
(b)
Check whether $\displaystyle \mathrm{f}: \mathrm{Z} \times \mathrm{Z} \rightarrow \mathrm{Z} \times \mathrm{Z}$ (where Z is the set of integers) defined as $\displaystyle \mathrm{f}(x, \mathrm{y})=(2 \mathrm{y}, 3 x)$ is injective or not.
Marking-scheme solution
Let $\displaystyle \mathrm{y}=\dfrac{x-2}{x-3}$ i.e. $\displaystyle x=\dfrac{3\mathrm{y}-2}{\mathrm{y}-1}$
Here $\displaystyle \mathrm{y} \neq 1 \therefore \mathrm{R}_{\mathrm{f}}=\mathrm{R}-\{1\} \neq$ Codomain
$\displaystyle \therefore \mathrm{f}$ is not onto.
Let $\displaystyle \left(x_{1}, \mathrm{y}_{1}\right),\left(x_{2}, \mathrm{y}_{2}\right) \in \mathrm{Z} \times \mathrm{Z}$ such that $\displaystyle \mathrm{f}\left(x_{1}, \mathrm{y}_{1}\right)=\mathrm{f}\left(x_{2}, \mathrm{y}_{2}\right)$
$\displaystyle \Rightarrow\left(2 \mathrm{y}_{1}, 3 x_{1}\right)=\left(2 \mathrm{y}_{2}, 3 x_{2}\right)$
$\displaystyle \Rightarrow \mathrm{y}_{1}=\mathrm{y}_{2}, x_{1}=x_{2}$
So, $\displaystyle \left(x_{1}, \mathrm{y}_{1}\right)=\left(x_{2}, \mathrm{y}_{2}\right)$
$\displaystyle \therefore \mathrm{f}$ is injective.
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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.