1. **State the problem:** We want to find for which values of $k$ the function $f(x) = x^3 - 6x^2 + kx + 2$ is one-to-one. 2. **Recall the rule for one-to-one functions:** A function is one-to-one if it is strictly increasing or strictly decreasing everywhere. For differentiable functions, this means its derivative does not change sign (no local maxima or minima). 3. **Find the derivative:** $$f'(x) = 3x^2 - 12x + k$$ 4. **Analyze the derivative:** For $f$ to be one-to-one, $f'(x)$ must not have two distinct real roots (which would indicate local extrema). So the quadratic $3x^2 - 12x + k$ must have zero or one real root. 5. **Use the discriminant:** The discriminant of $f'(x)$ is $$\Delta = (-12)^2 - 4 \cdot 3 \cdot k = 144 - 12k$$ 6. **Condition for no two distinct roots:** $$\Delta \leq 0 \implies 144 - 12k \leq 0$$ 7. **Solve inequality:** $$144 \leq 12k \implies k \geq 12$$ 8. **Conclusion:** The function is one-to-one if and only if $k \geq 12$. **Final answer:** B) $k \geq 12$