Consider the quadration`ax^(2) - bx + c =0,a,b,c in N` which has two distinct real roots belonging to the interval (1,2).
The least value of b is

Options

A. 10
B. 11
C. 13
D. 15

Explanation

To solve the problem, we need to find the least value of \( b \) for the quadratic equation \( ax^2 - bx + c = 0 \) such that it has two distinct real roots in the interval \( (1, 2) \), where \( a, b, c \) are natural numbers.

### Step-by-Step Solution:

1. **Understanding the Roots Condition**:
- For the quadratic equation to have two distinct real roots, the discriminant must be positive:
\[ D = b^2 - 4ac > 0 \]

2. **Roots in the Interval**:
- Let the roots be \( \alpha \) and \( \beta \) such that \( 1 < \alpha < 2 \) and \( 1 < \beta < 2 \).
- By Vieta's formulas, we know:
\[ \alpha + \beta = \frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]

3. **Finding Bounds for \( \alpha + \beta \)**:
- Since both roots are between 1 and 2:
\[ 2 < \alpha + \beta < 4 \]
- This implies:
\[ 2a < b < 4a \]

4. **Finding Bounds for \( \alpha \beta \)**:
- The product of the roots gives:
\[ 1 < \alpha \beta < 4 \]
- This implies:
\[ a < c < 4a \]

5. **Discriminant Condition**:
- We need to ensure the discriminant is positive:
\[ b^2 - 4ac > 0 \]
- Using the bounds for \( b \) and \( c \):
- Substitute \( c \) with its upper bound \( c = 4a - 1 \) (to maximize \( b \)):
\[ b^2 - 4a(4a - 1) > 0 \]
- Rearranging gives:
\[ b^2 > 16a^2 - 4a \]

6. **Minimizing \( b \)**:
- We want the smallest \( b \) that satisfies \( 2a < b < 4a \) and \( b^2 > 16a^2 - 4a \).
- The smallest integer \( b \) should be \( 2a + 1 \) (to ensure \( b \) is greater than \( 2a \)):
\[ (2a + 1)^2 > 16a^2 - 4a \]
- Expanding:
\[ 4a^2 + 4a + 1 > 16a^2 - 4a \]
- Rearranging gives:
\[ 0 > 12a^2 - 8a - 1 \]

7. **Finding the Roots of the Quadratic**:
- Solve \( 12a^2 - 8a - 1 = 0 \) using the quadratic formula:
\[ a = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 12 \cdot (-1)}}{2 \cdot 12} \]
\[ a = \frac{8 \pm \sqrt{64 + 48}}{24} = \frac{8 \pm \sqrt{112}}{24} = \frac{8 \pm 4\sqrt{7}}{24} = \frac{2 \pm \sqrt{7}}{6} \]

8. **Finding the Least Value of \( b \)**:
- The smallest integer \( a \) satisfying \( a \in \mathbb{N} \) is \( a = 1 \):
- For \( a = 1 \):
\[ 2(1) < b < 4(1) \Rightarrow 2 < b < 4 \]
- The least integer \( b \) satisfying this is \( b = 3 \).

### Conclusion:
The least value of \( b \) such that the quadratic equation \( ax^2 - bx + c = 0 \) has two distinct real roots in the interval \( (1, 2) \) is:
\[ \boxed{3} \]