Question:

Step 1: Analyze the monotonicity of each piece of the function. We need to verify that each piece of the function is monotonically increasing by computing its derivative. For $x \leq 0$: $f(x) = -e^{-x} + k$, so $f'(x) = e^{-x} > 0$ ✓ For $0 < x < 1$: $f(x) = e^x + 1$, so $f'(x) = e^x > 0$ ✓ For $x \geq 1$: $f(x) = ex^2 + \lambda$, so $f'(x) = 2ex > 0$ for $x \geq 1$ ✓ All pieces are individually monotonically increasing. Step 2: Apply the continuity condition at $x = 0$. For the function to be one-one and monotonically increasing across the entire domain, we need the function to be non-decreasing at the junction points. At $x = 0$, the left-hand limit is: $$\lim_{x \to 0^-} f(x) = -e^0 + k = -1 + k = k - 1$$ The right-hand limit is: $$\lim_{x \to 0^+} f(x) = e^0 + 1 = 2$$ For the function to be monotonically increasing (non-decreasing at the junction), we require: $$f(0^-) \leq f(0^+)$$ $$k - 1 \leq 2$$ $$k \leq 3$$ Therefore, the maximum value of $k$ is $\boxed{3}$. Step 3: Apply the continuity condition at $x = 1$. At $x = 1$, the left-hand limit is: $$\lim_{x \to 1^-} f(x) = e^1 + 1 = e + 1$$ The right-hand limit is: $$\lim_{x \to 1^+} f(x) = e(1)^2 + \lambda = e + \lambda$$ For the function to be monotonically increasing (non-decreasing at the junction), we require: $$f(1^-) \leq f(1^+)$$ $$e + 1 \leq e + \lambda$$ $$\lambda \geq 1$$ Therefore, the minimum value of $\lambda$ is $\boxed{1}$. Step 4: Calculate the difference between maximum $k$ and minimum $\lambda$. The difference is: $$\text{Maximum } k - \text{Minimum } \lambda = 3 - 1 = 2$$ With $k = 3$ and $\lambda = 1$, the function is continuous and strictly increasing everywhere, ensuring it is one-one. **Final Answer: The difference is $2$, which corresponds to Option 3.**