Step 1: Define Function Properties:
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 mapping from set R to set R is onto if every element in the codomain R has at least one corresponding element in the domain R that maps to it. This means the function's range equals its codomain.
Step 3: Verification Process:
One-one Check:
Assume f(x\textsubscript{1}) = f(x\textsubscript{2}), where x\textsubscript{1}, x\textsubscript{2} $\in$ R.
\[ 10x_1 = 10x_2 \]
Divide both sides by 10:
\[ x_1 = x_2 \]
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 arbitrary element y in the codomain R, we must find an x in the domain R such that f(x) = y.
\[ f(x) = y \]
\[ 10x = y \]
Solve for x:
\[ x = \frac{y}{10} \]
For any real number y, the derived value \(x = \frac{y}{10}\) is also a real number. This establishes that for every y in the codomain, a pre-image x exists in the domain. Consequently, the function is onto.
Step 4: Conclusion:
Given that f(x) = 10x is both one-one and onto, option (1) is the correct choice.