Question:

Step 1: Verify that each piece of the function is individually increasing. We need to check the derivative of each piece to ensure monotonicity within its domain. For $x \leq 0$: $f(x) = -e^{-x} + k$ $$f'(x) = e^{-x} > 0 \text{ for all } x \leq 0$$ This piece is always increasing. ✓ For $0 < x < 1$: $f(x) = e^x + 1$ $$f'(x) = e^x > 0 \text{ for all } 0 < x < 1$$ This piece is always increasing. ✓ For $x \geq 1$: $f(x) = ex^2 + \lambda$ $$f'(x) = 2ex > 0 \text{ for all } x \geq 1$$ This piece is always increasing. ✓ Step 2: Apply the continuity/monotonicity condition at $x = 0$. For the function to be monotonically increasing across $x = 0$, the left-hand limit must not exceed the right-hand limit. Left-hand limit as $x \to 0^-$: $$\lim_{x \to 0^-} f(x) = -e^{0} + k = -1 + k$$ Right-hand limit as $x \to 0^+$: $$\lim_{x \to 0^+} f(x) = e^{0} + 1 = 2$$ For monotonicity to be maintained: $$-1 + k \leq 2$$ $$k \leq 3$$ Therefore, the maximum value of $k$ is $\boxed{3}$. Step 3: Apply the continuity/monotonicity condition at $x = 1$. For the function to be monotonically increasing across $x = 1$, the left-hand limit must not exceed the right-hand limit. Left-hand limit as $x \to 1^-$: $$\lim_{x \to 1^-} f(x) = e^{1} + 1 = e + 1$$ Right-hand limit as $x \to 1^+$: $$\lim_{x \to 1^+} f(x) = e(1)^2 + \lambda = e + \lambda$$ For monotonicity to be maintained: $$e + 1 \leq e + \lambda$$ $$1 \leq \lambda$$ $$\lambda \geq 1$$ Therefore, the minimum value of $\lambda$ is $\boxed{1}$. Step 4: Calculate the difference between the maximum value of $k$ and minimum value of $\lambda$. $$\text{Difference} = k_{\max} - \lambda_{\min} = 3 - 1 = 2$$ The answer is **Option 3: 2**.