**One-One and Onto**

To determine whether the function f(x) = 10x from R to [0,∞] is one-one and onto, we need to analyze its properties.

**One-One (Injective)**

A function is said to be one-one or injective if every element in the domain maps to a unique element in the co-domain.

In the given function f(x) = 10x, let's assume two values x₁ and x₂ in the domain such that x₁ ≠ x₂. Now, we need to check if f(x₁) ≠ f(x₂) holds true.

f(x₁) = 10x₁
f(x₂) = 10x₂

If f(x₁) = f(x₂), then the function is not one-one. However, if f(x₁) ≠ f(x₂), then the function is one-one.

Let's assume f(x₁) = f(x₂):
10x₁ = 10x₂

Dividing both sides by 10, we get:
x₁ = x₂

This implies that x₁ = x₂, which contradicts our initial assumption that x₁ ≠ x₂. Therefore, f(x) = 10x is a one-one function.

**Onto (Surjective)**

A function is said to be onto or surjective if every element in the co-domain has a corresponding element in the domain.

In the given function f(x) = 10x, the co-domain is [0,∞]. To check if the function is onto, we need to ensure that for every y in the co-domain, there exists an x in the domain such that f(x) = y.

Let's consider an arbitrary y in [0,∞]. We need to find an x such that f(x) = y.

f(x) = 10x = y

Dividing both sides by 10, we get:
x = y/10

Since y can take any non-negative value, we can see that for every y in [0,∞], there exists an x such that f(x) = y. Therefore, f(x) = 10x is an onto function.

Hence, the function f(x) = 10x is both one-one and onto, making it a bijection. Therefore, the correct answer is option 'B'.