Question:

Step 1: Check if the function is one-one (injective) by finding its derivative. We need to determine if $f$ is strictly monotonic. Let us compute the derivative of $f(x) = \dfrac{2x}{1+2x}$ using the quotient rule: $$f'(x) = \frac{2(1+2x) - 2x \cdot 2}{(1+2x)^2} = \frac{2 + 4x - 4x}{(1+2x)^2} = \frac{2}{(1+2x)^2}$$ Since $(1+2x)^2 > 0$ for all $x \in [0,\infty)$, we have $f'(x) > 0$ for all $x$ in the domain. This means $f$ is strictly increasing on $[0,\infty)$. A strictly increasing function is injective, so **$f$ is one-one**. Step 2: Check if the function is onto (surjective) by analyzing its range. To determine the range of $f$, we examine its behavior at the boundary points and as $x \to \infty$. At $x = 0$: $$f(0) = \frac{2(0)}{1+2(0)} = 0$$ As $x \to \infty$: $$f(x) = \frac{2x}{1+2x} = \frac{2x}{2x(1/(2x)+1)} = \frac{1}{1/(2x)+1} \to \frac{1}{0+1} = 1$$ Since $f$ is continuous and strictly increasing on $[0,\infty)$, the range of $f$ is $[0, 1)$. However, the codomain is $[0,\infty)$. Since the range $[0,1)$ is a proper subset of the codomain $[0,\infty)$, the function does not map onto the entire codomain. For example, no value of $x$ satisfies $f(x) = 2$. Therefore, **$f$ is NOT onto**. Step 3: Conclude the classification. Since $f$ is one-one but not onto, the correct answer is **Option 2: one-one but not onto**.