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 regarding the quadratic polynomial \( f(x) = 4x^2 - 8ax + a \), we need to analyze the conditions under which the polynomial is non-negative and the nature of its roots.

### Step-by-Step Solution:

1. **Identify the Form of the Quadratic Polynomial**:
The given polynomial is \( f(x) = 4x^2 - 8ax + a \).

2. **Determine the Discriminant**:
The discriminant \( D \) of a quadratic polynomial \( ax^2 + bx + c \) is given by \( D = b^2 - 4ac \).
Here, \( a = 4 \), \( b = -8a \), and \( c = a \).
\[ D = (-8a)^2 - 4 \cdot 4 \cdot a = 64a^2 - 16a \]

3. **Set the Discriminant Greater Than or Equal to Zero**:
For \( f(x) \) to be non-negative for all \( x \), the discriminant must be non-positive (i.e., \( D \leq 0 \)).
\[ 64a^2 - 16a \leq 0 \]
Factor the expression:
\[ 16a(4a - 1) \leq 0 \]

4. **Find the Roots of the Inequality**:
The roots of the equation \( 16a(4a - 1) = 0 \) are:
\[ a = 0 \quad \text{and} \quad a = \frac{1}{4} \]

5. **Analyze the Sign of the Expression**:
We will test the intervals determined by the roots \( a = 0 \) and \( a = \frac{1}{4} \):
- For \( a < 0 \): \( 16a(4a - 1) > 0 \) (positive)
- For \( 0 < a < \frac{1}{4} \): \( 16a(4a - 1) < 0 \) (negative)
- For \( a > \frac{1}{4} \): \( 16a(4a - 1) > 0 \) (positive)

Thus, \( f(x) \geq 0 \) when \( a \in [0, \frac{1}{4}] \).

6. **Check for Integral Values**:
The only integral value of \( a \) in the interval \( [0, \frac{1}{4}] \) is \( a = 0 \).

7. **Evaluate the Second Statement**:
The second statement claims that if \( a < 0 \), then 0 lies between the zeros of the polynomial.
- Since \( a < 0 \) implies that the sum of the roots \( \alpha + \beta = 2a < 0 \) and the product \( \alpha \beta < 0 \), it indicates that one root is positive and the other is negative, thus 0 lies between the roots. This statement is true.

8. **Evaluate the Third Statement**:
The third statement claims that \( f(x) = 0 \) has two distinct solutions in the interval \( (0, 1) \).
- For distinct solutions, \( D > 0 \) implies \( 64a^2 - 16a > 0 \), which gives \( a < 0 \) or \( a > \frac{1}{4} \). However, we need to check if the roots lie in \( (0, 1) \). This requires further analysis and is not guaranteed for \( a \) in the given range.

9. **Evaluate the Fourth Statement**:
The fourth statement asks for the minimum value of \( a \) for which \( f(x) \) is non-negative. As determined earlier, the minimum value of \( a \) is \( 0 \).

### Conclusion:
The statements that hold true are:
- There is only one integral \( a \) for which \( f(x) \) is non-negative (True).
- For \( a < 0 \), the number 0 lies between the zeros of the polynomial (True).
- The third statement regarding distinct solutions needs further verification.
- The minimum value of \( a \) for which \( f(x) \) is non-negative is \( 0 \) (True).