Let `f(x) =4x^2-4ax+a^2-2a+2` be a quadratic polynomial in `x`,`a` be any real number. If x-coordinate of vertex of parabola y =f(x) is less than 0 and f(x) has minimum value 3 for `x in [0,2]` then value of a is (a) `1+sqrt2` (b) `1-sqrt2` (c) `1-sqrt3` (d) `1+sqrt3`

Options

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

Explanation

To solve the problem, we need to analyze the quadratic polynomial given by:

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

### Step 1: Determine the x-coordinate of the vertex
The x-coordinate of the vertex of a quadratic function in the form \( ax^2 + bx + c \) is given by:

\[ x = -\frac{b}{2a} \]

For our polynomial, we identify:
- \( a = 4 \)
- \( b = -4a \)

Substituting these values into the vertex formula gives:

\[ x = -\frac{-4a}{2 \cdot 4} = \frac{4a}{8} = \frac{a}{2} \]

### Step 2: Apply the condition that the x-coordinate of the vertex is less than 0
We are given that the x-coordinate of the vertex is less than 0:

\[ \frac{a}{2} < 0 \]

This implies:

\[ a < 0 \]

### Step 3: Find the minimum value of \( f(x) \) in the interval [0, 2]
To find the minimum value of \( f(x) \) on the interval [0, 2], we need to evaluate \( f(x) \) at the endpoints and at the vertex (if it lies within the interval).

1. **Evaluate at \( x = 0 \)**:

\[ f(0) = 4(0)^2 - 4a(0) + (a^2 - 2a + 2) = a^2 - 2a + 2 \]

2. **Evaluate at \( x = 2 \)**:

\[ f(2) = 4(2)^2 - 4a(2) + (a^2 - 2a + 2) = 16 - 8a + (a^2 - 2a + 2) = a^2 - 10a + 18 \]

3. **Evaluate at the vertex \( x = \frac{a}{2} \)** (only if \( 0 \leq \frac{a}{2} \leq 2 \)):
Since \( a < 0 \), \( \frac{a}{2} < 0 \) is not in the interval [0, 2]. Therefore, we only need to consider \( f(0) \) and \( f(2) \).

### Step 4: Set the minimum value to 3
We know that the minimum value of \( f(x) \) in the interval [0, 2] is 3. Therefore, we set up the equations:

1. From \( f(0) \):

\[ a^2 - 2a + 2 = 3 \]

This simplifies to:

\[ a^2 - 2a - 1 = 0 \]

2. From \( f(2) \):

\[ a^2 - 10a + 18 = 3 \]

This simplifies to:

\[ a^2 - 10a + 15 = 0 \]

### Step 5: Solve the quadratic equations
1. **For \( a^2 - 2a - 1 = 0 \)**:

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. **For \( a^2 - 10a + 15 = 0 \)**:

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 6: Determine valid values of \( a \)
From the first equation, we have two potential solutions: \( a = 1 + \sqrt{2} \) and \( a = 1 - \sqrt{2} \). Since \( \sqrt{2} \) is approximately 1.414, \( 1 + \sqrt{2} \) is positive, and \( 1 - \sqrt{2} \) is negative.

From the second equation, both \( 5 + \sqrt{10} \) and \( 5 - \sqrt{10} \) are positive.

Since we need \( a < 0 \), the only valid solution is:

\[ a = 1 - \sqrt{2} \]

### Conclusion
Thus, the value of \( a \) is:

\[ \boxed{1 - \sqrt{2}} \]