For the quadratic polynomial `f (x) =4x ^(2)-8ax+a.` the statements (s) which hold good is/are:

Options

A. There is only one integral 'a' for which f (x) is non- negative `AA x in R`
B. For `a lt 0,` the number zero lies between the zeroes of the polynomial
C. `f (x) =0` has two distinct solutions in `(0,1) ` for `a in ((1)/(7), (4)/(7))`
D. The minimum value of `f (x)` for minimum value of a for which f (x) is non-negative `AA x in R` is 0

Answer

To solve the problem, we need to analyze the quadratic polynomial \( f(x) = 4x^2 - 8ax + a \) and determine which statements hold true regarding the values of \( a \) for which the function is non-negative.

### Step 1: Understanding Non-negativity of the Quadratic Polynomial
For the quadratic polynomial \( f(x) \) to be non-negative for all \( x \in \mathbb{R} \), the discriminant must be less than or equal to zero. The discriminant \( D \) for a quadratic polynomial \( ax^2 + bx + c \) is given by:

\[ D = b^2 - 4ac \]

In our case, \( a = 4 \), \( b = -8a \), and \( c = a \). Thus, the discriminant becomes:

\[ D = (-8a)^2 - 4 \cdot 4 \cdot a = 64a^2 - 16a \]

### Step 2: Setting the Discriminant Less Than or Equal to Zero
To ensure \( f(x) \) is non-negative, we set the discriminant less than or equal to zero:

\[ 64a^2 - 16a \leq 0 \]

### Step 3: Factoring the Discriminant
Factoring out \( 16a \):

\[ 16a(4a - 1) \leq 0 \]

### Step 4: Finding Critical Points
The critical points are found by setting each factor to zero:

1. \( 16a = 0 \) gives \( a = 0 \)
2. \( 4a - 1 = 0 \) gives \( a = \frac{1}{4} \)

### Step 5: Analyzing the Sign of the Expression
We analyze the intervals determined by the critical points \( a = 0 \) and \( a = \frac{1}{4} \):

- For \( a < 0 \): \( 16a < 0 \) and \( 4a - 1 < 0 \) → product is positive.
- For \( 0 < a < \frac{1}{4} \): \( 16a > 0 \) and \( 4a - 1 < 0 \) → product is negative.
- For \( a = 0 \): \( 16a(4a - 1) = 0 \).
- For \( a = \frac{1}{4} \): \( 16a(4a - 1) = 0 \).
- For \( a > \frac{1}{4} \): \( 16a > 0 \) and \( 4a - 1 > 0 \) → product is positive.

Thus, the expression \( 16a(4a - 1) \leq 0 \) holds for:

\[ a \in [0, \frac{1}{4}] \]

### Step 6: Identifying Integral Values
The integral values of \( a \) in the interval \( [0, \frac{1}{4}] \) are only \( 0 \). Thus, there is only one integral value for which \( f(x) \) is non-negative.

### Step 7: Analyzing the Statements
1. **Statement 1**: There is only one integral value \( a \) for which \( f(x) \) is non-negative. **(True)**
2. **Statement 2**: If \( a < 0 \), then \( 0 \) lies between the roots of the polynomial. **(True)**
3. **Statement 3**: If \( f(x) = 0 \), then \( a \) is between \( \frac{1}{7} \) and \( \frac{4}{7} \). **(False)**
4. **Statement 4**: The minimum value of \( a \) for which \( f(x) \) is non-negative is \( 0 \). **(True)**

### Conclusion
The correct statements are 1, 2, and 4.