RBSE Class 11 Maths: Chapter 1 (Sets) – Exercise 1.5 Solutions

RBSE (Rajasthan Board of Secondary Education) Class 11 Maths Chapter 1, “Sets,” covers fundamental concepts. Exercise 1.5 specifically focuses on the Complement of a Set and related properties, including De Morgan’s Laws.

Key Concepts to Remember:

• Universal Set (U): The overarching set that contains all elements relevant to a particular discussion.
• Complement of a Set (A’): The complement of a set A (denoted as A’, Ac, or A) is the set of all elements in the universal set U that are not in A. A′=U−A={x:x∈U and x∈/A}
• Properties of Complement Sets:
  • Complement Laws:
    • A∪A′=U
    • A∩A′=Φ
  • De Morgan’s Laws:
    • (A∪B)′=A′∩B′ (The complement of the union is the intersection of the complements)
    • (A∩B)′=A′∪B′ (The complement of the intersection is the union of the complements)
  • Law of Double Complementation: (A′)′=A
  • Laws of Empty Set and Universal Set:
    • Φ′=U (The complement of the empty set is the universal set)
    • U′=Φ (The complement of the universal set is the empty set)
• Venn Diagrams: Useful for visualizing set operations and verifying De Morgan’s Laws.

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Here are the typical types of questions found in Exercise 1.5, along with their solutions:

Question 1: Let U={1,2,3,4,5,6,7,8,9}, A={1,2,3,4}, B={2,4,6,8} and C={3,4,5,6}. Find:

(i) A′ * Solution: A′=U−A={1,2,3,4,5,6,7,8,9}−{1,2,3,4}={5,6,7,8,9}

(ii) B′ * Solution: B′=U−B={1,2,3,4,5,6,7,8,9}−{2,4,6,8}={1,3,5,7,9}

(iii) (A∪C)′ * Solution: First, find A∪C={1,2,3,4}∪{3,4,5,6}={1,2,3,4,5,6} Then, (A∪C)′=U−(A∪C)={1,2,3,4,5,6,7,8,9}−{1,2,3,4,5,6}={7,8,9}

(iv) (A∪B)′ * Solution: First, find A∪B={1,2,3,4}∪{2,4,6,8}={1,2,3,4,6,8} Then, (A∪B)′=U−(A∪B)={1,2,3,4,5,6,7,8,9}−{1,2,3,4,6,8}={5,7,9}

(v) (A′)′ * Solution: We know that A′={5,6,7,8,9} from (i). (A′)′=U−A′={1,2,3,4,5,6,7,8,9}−{5,6,7,8,9}={1,2,3,4} Alternatively, by the Law of Double Complementation, (A′)′=A. So, (A′)′={1,2,3,4}.

(vi) (B−C)′ * Solution: First, find B−C={2,4,6,8}−{3,4,5,6}={2,8} (Elements in B but not in C). Then, (B−C)′=U−(B−C)={1,2,3,4,5,6,7,8,9}−{2,8}={1,3,4,5,6,7,9}

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Question 2: If U={a,b,c,d,e,f,g,h}, find the complements of the following sets:

(i) A={a,b,c} * Solution: A′=U−A={d,e,f,g,h}

(ii) B={d,e,f,g} * Solution: B′=U−B={a,b,c,h}

(iii) C={a,c,e,g} * Solution: C′=U−C={b,d,f,h}

(iv) D={f,g,h,a} * Solution: D′=U−D={b,c,d,e}

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Question 3: Taking the set of natural numbers as the universal set, U=N={1,2,3,…}, write down the complements of the following sets:

(i) {x:x is an even natural number} * Solution: {x:x is an odd natural number}

(ii) {x:x is an odd natural number} * Solution: {x:x is an even natural number}

(iii) {x:x is a positive multiple of 3} * Solution: {x:x∈N and x is not a multiple of 3}

(iv) {x:x is a prime number} * Solution: {x:x∈N and x is a composite number or x=1} (Recall that 1 is neither prime nor composite).

(v) {x:x is a natural number divisible by 3 and 5} * Solution: {x:x∈N and x is not divisible by 3 or x is not divisible by 5}

(vi) {x:x is a perfect square} * Solution: {x:x∈N and x is not a perfect square}

(vii) {x:x is a perfect cube} * Solution: {x:x∈N and x is not a perfect cube}

(viii) {x:x+5=8} * Solution: The set is {3}. Its complement is {x:x∈N and x=3}

(ix) {x:2x+5=9} * Solution: Solving 2x+5=9⟹2x=4⟹x=2. The set is {2}. Its complement is {x:x∈N and x=2}

(x) {x:x≥7} * Solution: {x:x∈N and x<7}={1,2,3,4,5,6}

(xi) {x:x∈N and 2x+1>10} * Solution: Solving 2x+1>10⟹2x>9⟹x>4.5. Since x is a natural number, this means x≥5. So the set is {5,6,7,…}. Its complement is {x:x∈N and x≤4}={1,2,3,4}

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Question 4: If U={1,2,3,4,5,6,7,8,9}, A={2,4,6,8} and B={2,3,5,7}. Verify that:

(i) (A∪B)′=A′∩B′ (De Morgan’s First Law)

• Solution:
  • LHS: (A∪B)′ A∪B={2,4,6,8}∪{2,3,5,7}={2,3,4,5,6,7,8} (A∪B)′=U−(A∪B)={1,2,3,4,5,6,7,8,9}−{2,3,4,5,6,7,8}={1,9}
  • RHS: A′∩B′ A′=U−A={1,2,3,4,5,6,7,8,9}−{2,4,6,8}={1,3,5,7,9} B′=U−B={1,2,3,4,5,6,7,8,9}−{2,3,5,7}={1,4,6,8,9} A′∩B′={1,3,5,7,9}∩{1,4,6,8,9}={1,9}
  • Since LHS = RHS ($ {1, 9} = {1, 9}$), the statement (A∪B)′=A′∩B′ is verified.

(ii) (A∩B)′=A′∪B′ (De Morgan’s Second Law)

• Solution:
  • LHS: (A∩B)′ A∩B={2,4,6,8}∩{2,3,5,7}={2} (A∩B)′=U−(A∩B)={1,2,3,4,5,6,7,8,9}−{2}={1,3,4,5,6,7,8,9}
  • RHS: A′∪B′ A′={1,3,5,7,9} (from part i) B′={1,4,6,8,9} (from part i) A′∪B′={1,3,5,7,9}∪{1,4,6,8,9}={1,3,4,5,6,7,8,9}
  • Since LHS = RHS ($ {1, 3, 4, 5, 6, 7, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}$), the statement (A∩B)′=A′∪B′ is verified.

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Question 5: Draw appropriate Venn diagram for each of the following:

(i) (A∪B)′ * Venn Diagram: Draw a universal rectangle (U). Inside it, draw two intersecting circles, A and B. Shade the area outside the union of A and B (A∪B). This shaded region represents (A∪B)′.

(ii) A′∩B′ * Venn Diagram: Draw a universal rectangle (U). Inside it, draw two intersecting circles, A and B. * Shade the area outside A (A’). * Shade the area outside B (B’). * The region where both shadings overlap is A′∩B′. This will be the same as the shaded region in (i), demonstrating De Morgan’s Law.

(iii) (A∩B)′ * Venn Diagram: Draw a universal rectangle (U). Inside it, draw two intersecting circles, A and B. Identify the intersection region A∩B. Shade the entire area outside this intersection.

(iv) A′∪B′ * Venn Diagram: Draw a universal rectangle (U). Inside it, draw two intersecting circles, A and B. * Shade the area outside A (A’). * Shade the area outside B (B’). * The entire shaded region (where either A’ or B’ is shaded) represents A′∪B′. This will be the same as the shaded region in (iii), demonstrating De Morgan’s Law.

(Please note: I cannot directly draw images here, but the descriptions guide you on how to draw them.)

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Question 6: Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60°, what is A’?

• Solution:
  • U = Set of all triangles in a plane.
  • A = Set of all triangles with at least one angle different from 60°.
  • This means A includes all triangles except those where all angles are 60°.
  • A triangle with all angles equal to 60° is an equilateral triangle.
  • Therefore, the complement A’ would be the set of triangles not having at least one angle different from 60°, which means all angles are 60°.
  • Solution: A′ is the set of all equilateral triangles.

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Question 7: Fill in the blanks to make each of the following a true statement:

(i) A∪A′=U​ (Complement Law)

(ii) Φ′∩A=A​ (Since Φ′=U, then U∩A=A)

(iii) A∩A′=Φ​ (Complement Law)

(iv) U′∩A=Φ​ (Since U′=Φ, then Φ∩A=Φ)