Statement -1 : There is just one quadratic equation with real coefficient one of whose roots is ` 1/(sqrt2 +1)`
and
Statement -2 : In a quadratic equation with rational coefficients the irrational roots are in conjugate pairs.

Options

A. Statement -1 is True, Statement -2 is True, Statement -2 is a correct explanation for statement -18
B. Statement -1 is True, Statement -2 is True , Statement -2 is NOT a correct explanation for Statement -18
C. Statement -1 is True, Statement -2 is False
D. Statement -1 is Flase, Statement -2 is True

Answer

To analyze the given statements, we will break down each statement and evaluate its validity step by step.

### Step 1: Evaluate Statement 1
**Statement 1:** There is just one quadratic equation with real coefficients one of whose roots is \( \frac{1}{\sqrt{2} + 1} \).

1. **Identify the root:** The given root is \( \frac{1}{\sqrt{2} + 1} \).
2. **Rationalize the root:** To work with this root more easily, we can rationalize it:
\[ \frac{1}{\sqrt{2} + 1} \cdot \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{\sqrt{2} - 1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1 \]
So, one root is \( \sqrt{2} - 1 \).

3. **Form the quadratic equation:** If we denote the other root as \( \beta \), the sum of the roots \( \alpha + \beta \) must be equal to \( -\frac{b}{a} \) (where \( a \) and \( b \) are coefficients of the quadratic equation). Since \( \beta \) can be any real number, we can create multiple quadratic equations with this root.

4. **Conclusion for Statement 1:** Since \( \beta \) can take an infinite number of values, there are infinitely many quadratic equations that can be formed with \( \frac{1}{\sqrt{2} + 1} \) as one of the roots. Therefore, Statement 1 is **false**.

### Step 2: Evaluate Statement 2
**Statement 2:** In a quadratic equation with rational coefficients, the irrational roots are in conjugate pairs.

1. **Understanding conjugate pairs:** If a quadratic equation has rational coefficients and one of its roots is irrational, say \( \alpha = a + b\sqrt{c} \) (where \( a, b, c \) are rational numbers), the other root must be \( \beta = a - b\sqrt{c} \) to ensure that the coefficients remain rational.

2. **Sum and product of roots:** The sum of the roots \( \alpha + \beta = (a + b\sqrt{c}) + (a - b\sqrt{c}) = 2a \) is rational. The product of the roots \( \alpha \beta = (a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - (b\sqrt{c})^2 = a^2 - b^2c \) is also rational.

3. **Conclusion for Statement 2:** Thus, if one root is irrational, the other must be its conjugate, confirming that irrational roots in a quadratic equation with rational coefficients are indeed in conjugate pairs. Therefore, Statement 2 is **true**.

### Final Conclusion
- **Statement 1:** False
- **Statement 2:** True