Question:

Step 1: Interpret the given condition. The condition states: for all $x_1 \in [0,1]$, there exists some $x_2 \in [1,2]$ such that $f(x_1) \geq g(x_2)$. This is equivalent to saying: for every $x_1 \in [0,1]$, we have $f(x_1) \geq \min_{x_2 \in [1,2]} g(x_2)$. For this to hold for all $x_1 \in [0,1]$, we need: $$\min_{x_1 \in [0,1]} f(x_1) \geq \min_{x_2 \in [1,2]} g(x_2)$$ Step 2: Find the minimum value of $f(x)$ on $[0,1]$. We have $f(x) = x - \dfrac{1}{x+1}$. Taking the derivative: $f'(x) = 1 + \dfrac{1}{(x+1)^2} > 0$ for all $x \in [0,1]$. Since $f'(x) > 0$, the function $f(x)$ is strictly increasing on $[0,1]$. Therefore, the minimum occurs at $x = 0$: $$\min_{x_1 \in [0,1]} f(x_1) = f(0) = 0 - \frac{1}{0+1} = -1$$ Step 3: Find the minimum value of $g(x)$ on $[1,2]$ in terms of parameter $a$. We have $g(x) = x^2 - 2ax + 4$. Completing the square: $$g(x) = (x-a)^2 + 4 - a^2$$ The vertex of this parabola is at $x = a$. The minimum on $[1,2]$ depends on where $a$ lies: **Case 1:** If $a < 1$, the minimum on $[1,2]$ occurs at $x = 1$: $$g(1) = 1 - 2a + 4 = 5 - 2a$$ **Case 2:** If $1 \leq a \leq 2$, the minimum occurs at $x = a$: $$g(a) = 4 - a^2$$ **Case 3:** If $a > 2$, the minimum on $[1,2]$ occurs at $x = 2$: $$g(2) = 4 - 4a + 4 = 8 - 4a$$ Step 4: Apply the constraint $\min g(x_2) \leq -1$. We need $\min_{x_2 \in [1,2]} g(x_2) \leq -1$. **Case 1:** $a < 1$: We need $5 - 2a \leq -1$, which gives $2a \geq 6$, so $a \geq 3$. This contradicts $a < 1$, so this case yields no solution. **Case 2:** $1 \leq a \leq 2$: We need $4 - a^2 \leq -1$, which gives $a^2 \geq 5$, so $a \geq \sqrt{5} \approx 2.236$. This contradicts $a \leq 2$, so this case yields no solution. **Case 3:** $a > 2$: We need $8 - 4a \leq -1$, which gives $4a \geq 9$, so $a \geq \dfrac{9}{4}$. Step 5: Verify and conclude. Since $\dfrac{9}{4} = 2.25 > 2$, the condition $a > 2$ is satisfied when $a \geq \dfrac{9}{4}$. Therefore, the minimum value of $a$ is $\boxed{\dfrac{9}{4}}$. The answer is **Option 4**.