Let f(x) = `4x^(2)-4ax+a^(2)-2a+2` and the golbal minimum value of f(x) for x in [0,2] is equal to 3
The number of values of a for which the global minimum value equal to 3 for x in [0,2] occurs at the endpoint of interval[0,2] is

Options

A. 1
B. 2
C. 3
D. 0

Explanation

To solve the problem, we are given the function:

\[ f(x) = 4x^2 - 4ax + (a^2 - 2a + 2) \]

We need to find the number of values of \( a \) such that the global minimum of \( f(x) \) on the interval \([0, 2]\) is equal to 3, and this minimum occurs at one of the endpoints (either \( x = 0 \) or \( x = 2 \)).

### Step 1: Evaluate \( f(0) \) and \( f(2) \)

First, we calculate the function values at the endpoints:

1. **At \( x = 0 \)**:
\[ f(0) = 4(0)^2 - 4a(0) + (a^2 - 2a + 2) = a^2 - 2a + 2 \]

2. **At \( x = 2 \)**:
\[ f(2) = 4(2)^2 - 4a(2) + (a^2 - 2a + 2) = 16 - 8a + a^2 - 2a + 2 = a^2 - 10a + 18 \]

### Step 2: Set up equations for the global minimum

We need to find conditions under which either \( f(0) = 3 \) or \( f(2) = 3 \).

1. **Condition 1: \( f(0) = 3 \)**:
\[ a^2 - 2a + 2 = 3 \implies a^2 - 2a - 1 = 0 \]
Using the quadratic formula:
\[ a = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \]

2. **Condition 2: \( f(2) = 3 \)**:
\[ a^2 - 10a + 18 = 3 \implies a^2 - 10a + 15 = 0 \]
Using the quadratic formula:
\[ a = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 60}}{2} = \frac{10 \pm \sqrt{40}}{2} = \frac{10 \pm 2\sqrt{10}}{2} = 5 \pm \sqrt{10} \]

### Step 3: Count the valid values of \( a \)

From the calculations, we have:

- From \( f(0) = 3 \): \( a = 1 + \sqrt{2} \) and \( a = 1 - \sqrt{2} \)
- From \( f(2) = 3 \): \( a = 5 + \sqrt{10} \) and \( a = 5 - \sqrt{10} \)

### Step 4: Check if the minimum occurs at the endpoints

1. **For \( a = 1 + \sqrt{2} \)**:
- Check if \( a/2 \) lies in \([0, 2]\):
\[ \frac{1 + \sqrt{2}}{2} \approx 1.207 \quad (\text{valid}) \]

2. **For \( a = 1 - \sqrt{2} \)**:
- Check if \( a/2 \) lies in \([0, 2]\):
\[ \frac{1 - \sqrt{2}}{2} < 0 \quad (\text{not valid}) \]

3. **For \( a = 5 + \sqrt{10} \)**:
- Check if \( a/2 \) lies in \([0, 2]\):
\[ \frac{5 + \sqrt{10}}{2} > 2 \quad (\text{not valid}) \]

4. **For \( a = 5 - \sqrt{10} \)**:
- Check if \( a/2 \) lies in \([0, 2]\):
\[ \frac{5 - \sqrt{10}}{2} \approx 1.43 \quad (\text{valid}) \]

### Conclusion

The valid values of \( a \) for which the global minimum occurs at the endpoints are:

1. \( a = 1 + \sqrt{2} \)
2. \( a = 5 - \sqrt{10} \)

Thus, the total number of values of \( a \) is **2**.