1. **Problem statement:** We have a function $f: \mathbb{R}^+ \to \mathbb{R}$ defined by $f(x) = 4x + 3$. We need to determine if it is one-one (injective), onto (surjective), both, or neither. 2. **Definitions:** - A function is **one-one (injective)** if different inputs map to different outputs. - A function is **onto (surjective)** if every element in the codomain has a preimage in the domain. 3. **Check if $f$ is one-one:** Assume $f(x_1) = f(x_2)$. $$4x_1 + 3 = 4x_2 + 3$$ Subtract 3 from both sides: $$4x_1 = 4x_2$$ Divide both sides by 4: $$x_1 = x_2$$ Since equal outputs imply equal inputs, $f$ is one-one. 4. **Check if $f$ is onto:** The codomain is $\mathbb{R}$ (all real numbers), but the domain is $\mathbb{R}^+$ (non-negative real numbers). For $y$ in $\mathbb{R}$, solve for $x$: $$y = 4x + 3 \implies x = \frac{y - 3}{4}$$ Since $x$ must be in $\mathbb{R}^+$ (i.e., $x \geq 0$), we require: $$\frac{y - 3}{4} \geq 0 \implies y - 3 \geq 0 \implies y \geq 3$$ Thus, the range of $f$ is $[3, \infty)$, not all real numbers. Therefore, $f$ is **not onto** $\mathbb{R}$. 5. **Conclusion:** $f$ is one-one but not onto. **Final answer:** (A) one-one but not onto