Let f: R → R be defined as f(x) = 10x. Then (Where R is the set of real numbers)

The correct option is: f is both one-one and onto.

To determine the nature of the function $f:R\to R$ defined as $f\left(x\right)=10x$, we check for both injectivity (one-one) and surjectivity (onto).

1. Checking if the function is one-one (injective):
A function $f$ is one-one if distinct elements in the domain have distinct images in the codomain. Mathematically, for any $x_1,x_2\in R$:
$f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$

Let us assume:
$f\left(x_1\right)=f\left(x_2\right)$

Substituting the function definition:
$10x_1=10x_2$

Dividing both sides by 10, we get:
$x_1=x_2$

Since $f\left(x_1\right)=f\left(x_2\right)$ directly implies $x_1=x_2$, the function $f$ is one-one.

2. Checking if the function is onto (surjective):
A function $f:R\to R$ is onto if every element $y$ in the codomain $R$ has a pre-image $x$ in the domain $R$ such that $f\left(x\right)=y$.

Let $y\in R$ (codomain). Let us set:
$y=f\left(x\right)$

This gives:
$y=10x$

Solving for $x$ in terms of $y$:
$x=\frac{y}{10}$

For any real number $y$, the value $\frac{y}{10}$ is also a real number. Therefore, for every $y\in R$ (codomain), there exists $x=\frac{y}{10}\in R$ (domain) such that:
$f\left(\frac{y}{10}\right)=10\left(\frac{y}{10}\right)=y$

Since every element in the codomain has a corresponding pre-image in the domain, the function $f$ is onto.

Conclusion:
Since $f$ is both one-one and onto, the correct option is f is both one-one and onto.