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.