Step 1: Define the Problem:
This problem requires verifying if the function f(x) = 10x adheres to the criteria for being one-one (injective) and onto (surjective).
One-one (Injective): A function is one-one if distinct inputs produce distinct outputs. Formally, if f(x\textsubscript{1}) = f(x\textsubscript{2}), then it must follow that x\textsubscript{1} = x\textsubscript{2}.
Onto (Surjective): A function f: R $\to$ R is onto if every element y in the codomain R has at least one corresponding element x in the domain R such that f(x) = y. Equivalently, the range of the function must equal its codomain.
Step 3: Verification Process:
One-one Check:
Assume f(x\textsubscript{1}) = f(x\textsubscript{2}) for x\textsubscript{1}, x\textsubscript{2} $\in$ R.
\[ 10x_1 = 10x_2 \]
Dividing by 10 yields:
\[ x_1 = x_2 \]
The implication f(x\textsubscript{1}) = f(x\textsubscript{2}) $\implies$ x\textsubscript{1} = x\textsubscript{2} confirms the function is one-one.
Onto Check:
For any element y in the codomain R, we seek an x in the domain R such that f(x) = y.
\[ f(x) = y \]
\[ 10x = y \]
\[ x = \frac{y}{10} \]
For any real number y, x = \frac{y}{10} is also a real number. This demonstrates that for every y in the codomain, a pre-image x exists in the domain. Consequently, the function is onto.
Step 4: Conclusion:
As f(x) = 10x satisfies both one-one and onto conditions, option (1) is validated.