CBSE 2025 · Region 7 · Set 1 · Q26 · 3 marks
Show that the function $\displaystyle f: \mathrm{R} \rightarrow \mathrm{R}$ defined by $\displaystyle f(\mathrm{x})=4 \mathrm{x}^{3}-5, \forall \mathrm{x} \in \mathrm{R}$ is one-one and onto.Let R be a relation defined on a set N of natural numbers such that $\displaystyle \mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{xy}$ is a square of a natural number, $\displaystyle \mathrm{x}, \mathrm{y} \in \mathrm{N}\}$. Determine if the relation $\displaystyle \mathrm{R}$ is an equivalence relation.
Show that the function $\displaystyle f: \mathrm{R} \rightarrow \mathrm{R}$ defined by $\displaystyle f(\mathrm{x})=4 \mathrm{x}^{3}-5, \forall \mathrm{x} \in \mathrm{R}$ is one-one and onto.
Let R be a relation defined on a set N of natural numbers such that $\displaystyle \mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{xy}$ is a square of a natural number, $\displaystyle \mathrm{x}, \mathrm{y} \in \mathrm{N}\}$. Determine if the relation $\displaystyle \mathrm{R}$ is an equivalence relation.
Marking-scheme solution
One-One: Let $\displaystyle \mathrm{x}_{1}, \mathrm{x}_{2} \in \mathrm{R}$ such that $\displaystyle \mathbf{f}\left(\mathrm{x}_{1}\right)=\mathbf{f}\left(\mathrm{x}_{2}\right) \Rightarrow 4 \mathrm{x}_{1}{ }^{3}-5=4 \mathrm{x}_{2}{ }^{3}-5 \Rightarrow \mathrm{x}_{1}{ }^{3}=\mathrm{x}_{2}{ }^{3} \Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}, \therefore$ ' $\displaystyle f$ ' is one-one Onto: $\displaystyle \mathrm{x} \in \mathrm{R}\left(D_{f}\right) \Rightarrow \mathrm{x}^{3} \in \mathrm{R} \Rightarrow 4 \mathrm{x}^{3}-5 \in \mathrm{R} \Rightarrow f(\mathrm{x}) \in \mathrm{R}, \quad \therefore \mathrm{R}_{f}=C o-\operatorname{domain}(f)$
∴ ' $\displaystyle \mathbf{f}$ ' is an onto function
⇒ ' $\displaystyle f$ ' is one-one & onto both
Reflexive: For any $\displaystyle \mathbf{x} \in \mathbf{N}, \mathbf{x} \cdot \mathbf{x}=\mathbf{x}^{\mathbf{2}}$, which is square of the natural number ' $\displaystyle \mathbf{x}$ '.
\[\Rightarrow(\mathrm{x}, \mathrm{x}) \in \mathrm{R}
\]
∴ ' $\displaystyle \mathbf{R}$ ' is a Reflexive relation.
Symmetric: Let $\displaystyle (\mathrm{x}, \mathrm{y}) \in \mathrm{R} \Rightarrow \mathrm{x} \mathrm{y}$ is a square of a natural number
$\displaystyle \Rightarrow \mathrm{yx}$ is a square of a natural number, $\displaystyle \because \mathrm{xy}=\mathrm{yx}$.
\[\Rightarrow(\mathrm{y}, \mathrm{x}) \in \mathbf{R}
\]
$\displaystyle \therefore{ }^{\prime}{ }^{\prime}{ }^{\prime}$ is a Symmetric relation.
Transitive: Let $\displaystyle (\mathrm{x}, \mathrm{y}),(\mathrm{y}, z) \in \mathrm{R} \Rightarrow \mathrm{x} \mathrm{y}=a^{2}, \mathrm{y} z=b^{2}$ for some $\displaystyle a, b \in \mathrm{N}$,
\[\therefore \frac{\mathbf{a}^{2}}{\mathbf{y}}=\mathbf{x}, \frac{\mathbf{b}^{2}}{\mathbf{y}}=\mathbf{z} \in \mathbf{N}
\]
∴ ' $\displaystyle \mathrm{R}$ ' is a Transitive relation.
Hence, $\displaystyle \mathrm{R}$ is an Equivalence relation
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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.