Let `x=(a+2b)/(a+b)` and `y=(a)/(b)`, where a and b are positive integers. If `y^(2) gt 2`, then

Options

A. `x^(2) le 2`
B. `x^(2) lt 2`
C. `x^(2) gt 2`
D. `x^(2) ge 2`

Answer

To solve the problem step by step, we start with the given equations for \( x \) and \( y \):

1. **Define the equations**:
\[ x = \frac{a + 2b}{a + b} \]
\[ y = \frac{a}{b} \]

2. **Rearranging \( x \)**:
We can rewrite \( x \) as:
\[ x = \frac{a + 2b}{a + b} = \frac{a + b + b}{a + b} = 1 + \frac{b}{a + b} \]

3. **Expressing \( x \) in terms of \( y \)**:
Since \( y = \frac{a}{b} \), we can express \( a \) in terms of \( y \) and \( b \):
\[ a = by \]
Substituting \( a \) into the equation for \( x \):
\[ x = \frac{by + 2b}{by + b} = \frac{b(y + 2)}{b(y + 1)} = \frac{y + 2}{y + 1} \]

4. **Finding the relationship between \( x \) and \( y \)**:
We can rearrange this to express \( y \) in terms of \( x \):
\[ x(y + 1) = y + 2 \]
Expanding and rearranging gives:
\[ xy + x = y + 2 \implies xy - y = 2 - x \implies y(x - 1) = 2 - x \]
Thus,
\[ y = \frac{2 - x}{x - 1} \]

5. **Using the condition \( y^2 > 2 \)**:
We substitute \( y \) into the inequality:
\[ \left(\frac{2 - x}{x - 1}\right)^2 > 2 \]

6. **Clearing the fraction**:
Multiply both sides by \( (x - 1)^2 \) (which is positive since \( a \) and \( b \) are positive integers):
\[ (2 - x)^2 > 2(x - 1)^2 \]

7. **Expanding both sides**:
Expanding gives:
\[ 4 - 4x + x^2 > 2(x^2 - 2x + 1) \]
Simplifying the right side:
\[ 4 - 4x + x^2 > 2x^2 - 4x + 2 \]
Rearranging leads to:
\[ 4 - 2 > x^2 - x^2 \implies 2 > x^2 - 2 \]
Thus,
\[ x^2 < 2 \]

8. **Conclusion**:
The final result is:
\[ x^2 < 2 \]