If `ax^(2)-bx+c=0` have two distinct roots lying in the interval `(0,1),a,b,c Epsilon N` The least value of `log_(5)abc` is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to find the least value of \( \log_5(abc) \) given the quadratic equation \( ax^2 - bx + c = 0 \) has two distinct roots lying in the interval \( (0, 1) \) where \( a, b, c \in \mathbb{N} \).

### Step-by-step Solution:

1. **Understanding the Roots Condition**:
The quadratic equation \( ax^2 - bx + c = 0 \) has two distinct roots if its discriminant is positive. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac > 0 \]

2. **Roots in the Interval (0, 1)**:
For the roots \( \alpha \) and \( \beta \) to lie in the interval \( (0, 1) \), we need:
- \( f(0) = c > 0 \)
- \( f(1) = a - b + c > 0 \)

This implies:
\[ c > 0 \quad \text{and} \quad a - b + c > 0 \]

3. **Using the Roots**:
The roots can be expressed using Vieta's formulas:
\[ \alpha + \beta = \frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]
Since \( \alpha, \beta \in (0, 1) \), we have:
\[ 0 < \alpha + \beta < 2 \quad \text{and} \quad 0 < \alpha \beta < 1 \]

4. **Setting Up Inequalities**:
From \( \alpha + \beta < 2 \):
\[ \frac{b}{a} < 2 \implies b < 2a \]
From \( \alpha \beta < 1 \):
\[ \frac{c}{a} < 1 \implies c < a \]

5. **Finding Minimum Values**:
To minimize \( abc \), we need to find the smallest integers \( a, b, c \) that satisfy the inequalities:
- \( c < a \)
- \( b < 2a \)

Let's start with \( a = 5 \) (the smallest integer greater than 4):
- Then \( c \) can be \( 1, 2, 3, \) or \( 4 \).
- For \( c = 1 \), \( b < 10 \), so the smallest \( b \) can be is \( 5 \) (since \( b \) must also be a natural number).

6. **Calculating \( abc \)**:
Now we have \( a = 5, b = 5, c = 1 \):
\[ abc = 5 \times 5 \times 1 = 25 \]

7. **Finding \( \log_5(abc) \)**:
\[ \log_5(abc) = \log_5(25) = 2 \]

### Conclusion:
The least value of \( \log_5(abc) \) is \( 2 \).