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 the vertex of parabola y =f(x) is less thna 0 and f(x) has minimum value 3 for `x in [0,2]` then value of a is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write down the quadratic polynomial
The quadratic polynomial is given as:
\[ f(x) = 4x^2 - 4ax + (a^2 - 2a + 2) \]

### Step 2: Find the x-coordinate of the vertex
The x-coordinate of the vertex of a quadratic polynomial \( f(x) = Ax^2 + Bx + C \) is given by:
\[ x_v = -\frac{B}{2A} \]
In our case, \( A = 4 \) and \( B = -4a \), so:
\[ x_v = -\frac{-4a}{2 \cdot 4} = \frac{4a}{8} = \frac{a}{2} \]

### Step 3: Set the condition for the vertex
We are given that the x-coordinate of the vertex is less than 0:
\[ \frac{a}{2} < 0 \]
This implies:
\[ a < 0 \]

### Step 4: Determine the minimum value of f(x)
Since \( f(x) \) is a quadratic polynomial that opens upwards (as the coefficient of \( x^2 \) is positive), it has a minimum value at the vertex. We need to find \( f(0) \) because the function is increasing in the interval [0, 2].

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

### Step 5: Set the minimum value condition
We are given that the minimum value of \( f(x) \) in the interval [0, 2] is 3:
\[ f(0) = a^2 - 2a + 2 = 3 \]

### Step 6: Solve the equation
Now, we simplify the equation:
\[ a^2 - 2a + 2 - 3 = 0 \]
\[ a^2 - 2a - 1 = 0 \]

### Step 7: Use the quadratic formula to find a
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -2, c = -1 \):
\[ a = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ a = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ a = \frac{2 \pm \sqrt{8}}{2} \]
\[ a = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ a = 1 \pm \sqrt{2} \]

### Step 8: Determine the valid value of a
We have two potential solutions:
1. \( a = 1 + \sqrt{2} \)
2. \( a = 1 - \sqrt{2} \)

Since we established earlier that \( a < 0 \), we reject \( a = 1 + \sqrt{2} \) (which is positive) and accept:
\[ a = 1 - \sqrt{2} \]

### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{1 - \sqrt{2}} \]