Chapter 1: Relations and Functions

Ans: Here, ${\text{f}}:{\mathbf{R}} \to {\mathbf{R}}$ is defined as ${\text{f}}(x) = {x^4}$.

Let $x,y \in R$ such that ${\text{f}}(x) = {\text{f}}(y)$.

$ \Rightarrow {x^4} = {y^4}$

$ \Rightarrow x =  \pm y$

${\text{f}}(x) = {\text{f}}(y)$ does not imply that $x = y$

For example, ${\text{f}}(1) = {\text{f}}( - 1) = 1$

Thus, ${\text{f}}$ is not one-one.

Consider an element \[2\] in co-domain ${\mathbf{R}}$. It is clear that there does not exist any $x$ in 

domain ${\text{R}}$ such that ${\text{f}}(x) = 2$

Thus, ${\text{f}}$ is not onto.

Hence, the function ${\text{f}}$ is neither one-one nor onto.

The correct answer is \[{\text{D}}.\]

Ans: Here, ${\text{f}}:{\text{R}} \to {\text{R}}$ is defined as ${\text{f}}(x) = 3x$.

Let $x,\,\,y \in {\text{R}}$ such that ${\text{f}}(x) = {\text{f}}(y)$.

$ \Rightarrow 3x = 3y$

$ \Rightarrow x = y$

Thus, ${\text{f}}$ is one-one.

Now, for any real number \[\left( y \right)\] in co-domain ${\text{R}}$, there exists \[\dfrac{y}{3}\] in ${\text{R}}$ such that ${\text{f}}\left( {\dfrac{y}{3}} \right) = 3\left( {\dfrac{y}{3}} \right) = y$

Thus, ${\text{f}}$ is onto.

Hence, the function ${\text{f}}$ is one-one and onto.

Therefore, the correct answer is \[{\text{A}}.\]