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Mathematics · 2025 · 2 marks
CBSE 2025 · Region 7 · Set 1 · Q21
Let f: A → B be defined by f ( x )=(x -2)/(x -3), where A = R -{3} and B = R -{1}. Discuss the bijectivity of the function.
Let $\displaystyle \mathrm{f}: \mathrm{A} \rightarrow \mathrm{B}$ be defined by $\displaystyle \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}-2}{\mathrm{x}-3}$, where $\displaystyle \mathrm{A}=\mathrm{R}-\{3\}$ and $\displaystyle \mathrm{B}=\mathrm{R}-\{1\}$. Discuss the bijectivity of the function.
Marking-scheme solution
Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{A}$ such that $\displaystyle \mathrm{f}\left(\mathrm{x}_{1}\right)=\mathrm{f}\left(\mathrm{x}_{2}\right) \Rightarrow \frac{\mathrm{x}_{1}-2}{\mathrm{x}_{1}-3}=\frac{\mathrm{x}_{2}-2}{\mathrm{x}_{2}-3} \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}, \therefore$ ' $\displaystyle \mathrm{f}$ ' is one-one.
For each $\displaystyle y \in \mathrm{B}$, there exists $\displaystyle \mathrm{x}=\frac{3 y-2}{y-1} \in \mathrm{R}-\{3\}$, such that $\displaystyle \mathrm{f}(\mathrm{x})=y, \therefore$ ' $\displaystyle \mathrm{f}$ ' is onto
⇒ ' $\displaystyle \mathbf{f}$ ' is one-one $\displaystyle \boldsymbol{\&}$ onto, or ' $\displaystyle \mathbf{f}$ ' is a bijective function.
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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.