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Tips to identify the kind of prob In set theory

Mathematics

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d

c) (A∩B)∪(A∩C)\text{c) } (A \cap B) \cup (A \cap C)

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Here are tips for identifying set theory problems and the solutions to the questions:

Tips to Identify Set Theory Problems

Set theory problems deal with collections of objects, called sets, and the relationships and operations between them. Here are some common indicators:

  • Symbols: Look for specific set notation symbols:
    • ∪\cup (Union): "or", "at least one", "all elements in either set".
    • ∩\cap (Intersection): "and", "both", "common elements".
    • ′' or c^c (Complement): "not in", "elements outside the set".
    • ∖\setminus or −- (Difference): "in A but not in B".
    • ⊆\subseteq (Subset): "is contained in".
    • ∈\in (Element of): "is a member of".
    • ∅\emptyset (Empty set): A set with no elements.
    • UU (Universal set): The set of all possible elements under consideration.
  • Keywords: Phrases like "equivalence of sets," "simplify the set expression," "prove the identity," "number of elements," "survey results," "groups of people," "at least one," "neither," "only."
  • Context: Problems involving grouping, categorization, or relationships between different collections of items (e.g., students taking different subjects, people with different hobbies).
  • Venn Diagrams: If the problem mentions or can be easily visualized with overlapping circles, it's likely a set theory problem.

Solutions to the Questions

4. The equivalence of A∩(B∪C)A \cap (B \cup C) is:

Part a

(A∪B′)∪(A∪C)(A \cup B') \cup (A \cup C)

Part b

(A∩B)∪C(A \cap B) \cup C

Part c

(A∩B)∪(A∩C)(A \cap B) \cup (A \cap C)

Part d

c) (A∩B)∪(A∩C)\text{c) } (A \cap B) \cup (A \cap C)

(A∪B)∩(A∪C)(A \cup B) \cap (A \cup C)

This question asks for an equivalent expression using the Distributive Law of set operations.
The Distributive Law states that for any sets A, B, and C:

A∩(B∪C)=(A∩B)∪(A∩C)A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

This law is analogous to distributing multiplication over addition in algebra, e.g., a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c).

Comparing the given expression with the options, option (c) directly matches the Distributive Law.

The correct option is c) (A∩B)∪(A∩C)\boxed{\text{c) } (A \cap B) \cup (A \cap C)}.

5. For an AP, a=25,d=−3a=25, d=-3 find the value of nn if Sn=112S_n = 112.

  1. Identify the given values for the Arithmetic Progression (AP).

    First term: a=25a = 25
    Common difference: d=−3d = -3
    Sum of the first nn terms: Sn=112S_n = 112

  2. Use the formula for the sum of the first nn terms of an AP.

    The formula is:

    Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1)d]

  3. Substitute the given values into the formula.

    112=n2[2(25)+(n−1)(−3)]112 = \frac{n}{2} [2(25) + (n-1)(-3)]

  4. Simplify the equation.

    112=n2[50−3(n−1)]112 = \frac{n}{2} [50 - 3(n-1)] 112=n2[50−3n+3]112 = \frac{n}{2} [50 - 3n + 3] 112=n2[53−3n]112 = \frac{n}{2} [53 - 3n]

    Multiply both sides by 2 to eliminate the fraction:

    2×112=n(53−3n)2 \times 112 = n(53 - 3n) 224=53n−3n2224 = 53n - 3n^2

  5. Rearrange the equation into a standard quadratic form (Ax2+Bx+C=0Ax^2 + Bx + C = 0).

    3n2−53n+224=03n^2 - 53n + 224 = 0

  6. Solve the quadratic equation for nn using the quadratic formula n=−B±B2−4AC2An = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}.

    Here, A=3A=3, B=−53B=-53, C=224C=224.

    n=−(−53)±(−53)2−4(3)(224)2(3)n = \frac{-(-53) \pm \sqrt{(-53)^2 - 4(3)(224)}}{2(3)} n=53±2809−26886n = \frac{53 \pm \sqrt{2809 - 2688}}{6} n=53±1216n = \frac{53 \pm \sqrt{121}}{6} n=53±116n = \frac{53 \pm 11}{6}

    This gives two possible values for nn:

    n1=53+116=646=323n_1 = \frac{53 + 11}{6} = \frac{64}{6} = \frac{32}{3} n2=53−116=426=7n_2 = \frac{53 - 11}{6} = \frac{42}{6} = 7

    Since the number of terms (nn) must be a positive integer, n=323n = \frac{32}{3} is not a valid solution.
    Therefore, the value of nn is 7.

    The value of nn is 7\boxed{7}.

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