Find all numbers a for each of which the least value of the quadratic trinomial `4x^2 - 4ax + a^2 - 2a + 2` on the interva `0 le x le 2` is equal to 3.

Answer

To solve the problem, we need to find all values of \( a \) for which the least value of the quadratic trinomial \( 4x^2 - 4ax + a^2 - 2a + 2 \) on the interval \( 0 \leq x \leq 2 \) is equal to 3.

### Step 1: Define the quadratic function
Let \( f(x) = 4x^2 - 4ax + (a^2 - 2a + 2) \).

### Step 2: Find the derivative
To find the minimum value of the quadratic function, we first find its derivative:
\[ f'(x) = 8x - 4a \]

### Step 3: Determine critical points
Set the derivative equal to zero to find critical points:
\[ 8x - 4a = 0 \implies x = \frac{a}{2} \]

### Step 4: Evaluate critical points and endpoints
We need to evaluate \( f(x) \) at the critical point \( x = \frac{a}{2} \) and at the endpoints \( x = 0 \) and \( x = 2 \).

#### Case 1: If \( 0 \leq \frac{a}{2} \leq 2 \)
- Evaluate \( f(0) \):
\[ f(0) = 4(0)^2 - 4a(0) + (a^2 - 2a + 2) = a^2 - 2a + 2 \]
- Evaluate \( f(2) \):
\[ f(2) = 4(2)^2 - 4a(2) + (a^2 - 2a + 2) = 16 - 8a + a^2 - 2a + 2 = a^2 - 10a + 18 \]
- Evaluate \( f\left(\frac{a}{2}\right) \):
\[ f\left(\frac{a}{2}\right) = 4\left(\frac{a}{2}\right)^2 - 4a\left(\frac{a}{2}\right) + (a^2 - 2a + 2) = a^2 - 2a + (a^2 - 2a + 2) = 2a^2 - 4a + 2 \]

### Step 5: Find the minimum value
The minimum value of \( f(x) \) on the interval \( [0, 2] \) is:
\[ \min\{f(0), f(2), f\left(\frac{a}{2}\right)\} \]

### Step 6: Set the minimum value equal to 3
We need to find \( a \) such that:
\[ \min\{a^2 - 2a + 2, a^2 - 10a + 18, 2a^2 - 4a + 2\} = 3 \]

#### Solve for \( f(0) = 3 \):
\[ a^2 - 2a + 2 = 3 \implies a^2 - 2a - 1 = 0 \]
Using the quadratic formula:
\[ a = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \]

#### Solve for \( f(2) = 3 \):
\[ a^2 - 10a + 18 = 3 \implies a^2 - 10a + 15 = 0 \]
Using the quadratic formula:
\[ a = \frac{10 \pm \sqrt{100 - 60}}{2} = 5 \pm \sqrt{10} \]

### Step 7: Combine results
The values of \( a \) that satisfy the condition are:
1. \( a = 1 + \sqrt{2} \)
2. \( a = 1 - \sqrt{2} \)
3. \( a = 5 + \sqrt{10} \)
4. \( a = 5 - \sqrt{10} \)

### Final Answer
The values of \( a \) for which the least value of the quadratic trinomial on the interval \( [0, 2] \) is equal to 3 are:
\[ a = 1 + \sqrt{2}, \quad a = 1 - \sqrt{2}, \quad a = 5 + \sqrt{10}, \quad a = 5 - \sqrt{10} \]