(a) Show that the function f: R → R defined by f( x )=4 x^3-5, ∀ x ∈ R is one-one and onto.

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