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Thank you for the image. Let's solve the problem step by step for each case.

### Show that \( A \subseteq B \) or \( B \subseteq A \):

### **Case (a):**
- **Given Sets**:
  - \( A = \{1, 2, B\} \)
  - \( B = \{x \mid x^2 - 11x + 6 = 0\} \)

#### **Solution**:
1. **Solve the quadratic equation** \( x^2 - 11x + 6 = 0 \):
   - Discriminant \( \Delta = (-11)^2 - 4(1)(6) = 121 - 24 = 97 \).
   - Since the discriminant is positive, there are two distinct real roots.
   - The roots are \( x = \frac{11 \pm \sqrt{97}}{2} \).
   - These roots are not integers and hence do not include 1, 2, or B (assuming B is a placeholder).

2. **Conclusion**:
   - The elements in \( A \) (which are \( 1 \), \( 2 \), and \( B \)) do not match the elements in \( B \).
   - Therefore, neither \( A \subseteq B \) nor \( B \subseteq A \) can be concluded from this setup.

### **Case (b):**
- **Given Sets**:
  - \( A = \{1\} \times \{1, 2, B\} \)
  - \( B = (1, 2) \)

#### **Solution**:
1. **Cartesian Product \( A \)**:
   - \( A = \{(1, 1), (1, 2), (1, B)\} \).
2. **Compare with Set \( B \)**:
   - Set \( B \) contains the tuple \( (1, 2) \) only.
   - Since \( (1, 2) \) is in both sets, \( (1, 2) \subseteq (1, 2) \).

3. **Conclusion**:
   - \( A \) is not a subset of \( B \), but \( B \subseteq A \) because all elements of \( B \) are in \( A \).

### **Case (c):**
- **Given Sets**:
  - \( A = \{x \mid x^2 - 2x + 2 = 0\} \)
  - \( B = \{1, 2\} \)

#### **Solution**:
1. **Solve the quadratic equation** \( x^2 - 2x + 2 = 0 \):
   - Discriminant \( \Delta = 2^2 - 4(1)(2) = 4 - 8 = -4 \).
   - Since the discriminant is negative, there are no real solutions (the roots are complex).
   - Therefore, \( A \) is an empty set \( \emptyset \).

2. **Conclusion**:
   - Since \( A \) is empty, \( A \subseteq B \) is true because the empty set is a subset of any set.
   - \( B \subseteq A \) is false because \( B \) contains elements.

### **Case (d):**
- **Given Sets**:
  - \( A = \{1, 2, 3, 4\} \)
  - \( C = \{5, 6, 7, 8\} \)
  - \( B = \{n \mid n \in A \text{ and } n + m = B \text{ for some } m \in C\} \)

#### **Solution**:
1. **Interpretation**:
   - \( B \) consists of elements \( n \) from \( A \) such that \( n + m \) equals some value in \( B \) for some \( m \) in \( C \).

2. **Possible values**:
   - Example: If \( n = 1 \) and \( m = 5 \), then \( 1 + 5 = 6 \) (which could be in another \( B \)).

3. **Conclusion**:
   - This problem is challenging without a precise definition of \( B \), but generally, check the elements of \( B \) as defined and see if they match or subset \( A \).

Would you like further details or explanations on any specific case?

### Related Practice Questions:
1. How does the quadratic discriminant determine the nature of roots?
2. What does it mean when a set is defined by a Cartesian product?
3. How do you determine whether an empty set is a subset of another set?
4. Can you give an example where \( A \subseteq B \) but \( B \subseteq A \) is not true?
5. How would you approach proving that one set is a subset of another?

**Tip:** When dealing with quadratic equations in sets, always consider the nature of the roots (real or complex) as it can drastically change the outcome of set relationships.

Explore detailed solutions to set theory problems involving quadratic equations and subset relationships. This analysis covers cases where sets A and B are analyzed for subset relationships and solutions to quadratic equations. Suitable for advanced high school students and those interested in set theory and algebraic concepts.

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