CBSE 2025 · Region 4 · Set 1 · Q26 · 3 marks
If $\displaystyle f: \mathrm{R}^{+} \rightarrow \mathrm{R}$ is defined as $\displaystyle f(\mathrm{x})=\log _{a} \mathrm{x}(a>0$ and $\displaystyle a \neq 1)$, prove that $\displaystyle f$ is a bijection. ( $\displaystyle \mathrm{R}^{+}$is a set of all positive real numbers.)Let $\displaystyle \mathrm{A}=\{1,2,3\}$ and $\displaystyle \mathrm{B}=\{4,5,6\}$. A relation R from A to B is defined as $\displaystyle \mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}+\mathrm{y}=6, \mathrm{x} \in \mathrm{A}, \mathrm{y} \in \mathrm{B}\}$.(i)Write all elements of $\displaystyle \mathrm{R}$.(ii)Is R a function ? Justify.(iii)Determine domain and range of $\displaystyle \mathrm{R}$.
If $\displaystyle f: \mathrm{R}^{+} \rightarrow \mathrm{R}$ is defined as $\displaystyle f(\mathrm{x})=\log _{a} \mathrm{x}(a>0$ and $\displaystyle a \neq 1)$, prove that $\displaystyle f$ is a bijection. ( $\displaystyle \mathrm{R}^{+}$is a set of all positive real numbers.)
Let $\displaystyle \mathrm{A}=\{1,2,3\}$ and $\displaystyle \mathrm{B}=\{4,5,6\}$. A relation R from A to B is defined as $\displaystyle \mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x}+\mathrm{y}=6, \mathrm{x} \in \mathrm{A}, \mathrm{y} \in \mathrm{B}\}$.
(i)
Write all elements of $\displaystyle \mathrm{R}$.
(ii)
Is R a function ? Justify.
(iii)
Determine domain and range of $\displaystyle \mathrm{R}$.
Marking-scheme solution
\[f(\mathrm{x})=\log _{a} \mathrm{x} \quad(a>0, a \neq 1)
\]
Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}^{+}$such that $\displaystyle f\left(\mathrm{x}_{1}\right)=f\left(\mathrm{x}_{2}\right)$
$\displaystyle \Rightarrow \log _{a} \mathrm{x}_{1}=\log _{a} \mathrm{x}_{2}$
$\displaystyle \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2} \Rightarrow f$ is one-one.
Let $\displaystyle f(\mathrm{x})=\mathrm{y} \Rightarrow \log _{a} \mathrm{x}=\mathrm{y} \Rightarrow a^{\mathrm{y}}=\mathrm{x}$
∴ for every $\displaystyle \mathrm{y} \in \mathrm{R}$, there exists $\displaystyle \mathrm{x} \in \mathrm{R}^{+}$
$\displaystyle \therefore f$ is onto.
$\displaystyle f$ is a bijection.
(i)
$\displaystyle \mathrm{R}=\{(1,5),(2,4)\}$
(ii)
$\displaystyle \mathrm{R}$ is not a function as $\displaystyle 3 \in$ Ado not have an image in co-domain.
(iii)
Domain of $\displaystyle \mathrm{R}=\{1,2\}$, Range of $\displaystyle \mathrm{R}=\{4,5\}$
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CBSE Class 12 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.