Let `f(x) = x^2 + ax + b`. If the maximum and the minimum values of f(x) are 3 and 2 respectively for `0 <= x <= 2`, then the possible ordered pair(s) of (a,b) is/are

Options

A. `(-2,3)`
B. `(-3//2,2)`
C. `(-5//2, 3)`
D. `(-5//2, 2)`

Answer

To solve the problem, we need to find the ordered pairs \((a, b)\) such that the function \(f(x) = x^2 + ax + b\) has a maximum value of 3 and a minimum value of 2 for \(0 \leq x \leq 2\).

### Step 1: Determine the critical point
The first step is to find the derivative of \(f(x)\):
\[ f'(x) = 2x + a \]
Setting the derivative equal to zero to find the critical points:
\[ 2x + a = 0 \implies x = -\frac{a}{2} \]

### Step 2: Analyze the critical point
Since we are interested in the interval \(0 \leq x \leq 2\), we need to consider the position of the critical point:
1. If \(-\frac{a}{2} < 0\), then the critical point is outside the interval, and we only need to evaluate \(f(0)\) and \(f(2)\).
2. If \(0 \leq -\frac{a}{2} \leq 2\), then the critical point lies within the interval, and we evaluate \(f(0)\), \(f(2)\), and \(f\left(-\frac{a}{2}\right)\).

### Step 3: Evaluate \(f(0)\) and \(f(2)\)
Calculating \(f(0)\):
\[ f(0) = b \]
Calculating \(f(2)\):
\[ f(2) = 2^2 + 2a + b = 4 + 2a + b \]

### Step 4: Set up equations based on maximum and minimum values
From the problem statement:
- The maximum value is \(f(0) = b = 3\).
- The minimum value is \(f(2) = 4 + 2a + b = 2\).

Substituting \(b = 3\) into the equation for \(f(2)\):
\[ 4 + 2a + 3 = 2 \implies 2a + 7 = 2 \implies 2a = -5 \implies a = -\frac{5}{2} \]

Thus, one ordered pair is:
\[ (a, b) = \left(-\frac{5}{2}, 3\right) \]

### Step 5: Consider the case where the critical point is within the interval
Next, we consider the case where \(-\frac{a}{2}\) is within the interval \(0 \leq x \leq 2\). This means:
\[ 0 \leq -\frac{a}{2} \leq 2 \implies 0 \leq a \leq -4 \]
This is not possible since \(a\) cannot be both non-negative and negative.

### Step 6: Check the case where the maximum occurs at \(f(2)\)
Now, we check if the maximum occurs at \(f(2)\):
Setting \(f(2) = 3\):
\[ 4 + 2a + b = 3 \implies 2a + b = -1 \]
And the minimum at \(f(0) = 2\):
\[ b = 2 \]
Substituting \(b = 2\) into the equation:
\[ 2a + 2 = -1 \implies 2a = -3 \implies a = -\frac{3}{2} \]

Thus, another ordered pair is:
\[ (a, b) = \left(-\frac{3}{2}, 2\right) \]

### Conclusion
The possible ordered pairs \((a, b)\) are:
1. \(\left(-\frac{5}{2}, 3\right)\)
2. \(\left(-\frac{3}{2}, 2\right)\)