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Exercise 3.3 · Q2

Q.Let U={1,2,3,4,5,6,7,8}U = \{1, 2, 3, 4, 5, 6, 7, 8\}; A={1,2,3,4}A = \{1, 2, 3, 4\}; B={3,4,6}B = \{3, 4, 6\}; C={5,6,7,8}C = \{5, 6, 7, 8\}, find:

(i) A−(B∪C)A - (B \cup C)
(ii) A∩C′A \cap C'
(iii) B′∩C′B' \cap C'
(iv) B′∪A′B' \cup A'
(v) A−(B∪C)′A - (B \cup C)'
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✓ Free question

Using U={1,…,8}U=\{1,\ldots,8\} with A={1,2,3,4}A=\{1,2,3,4\}, B={3,4,6}B=\{3,4,6\}, C={5,6,7,8}C=\{5,6,7,8\}, all five combinations are computed via the complements A′={5,6,7,8}A'=\{5,6,7,8\}, B′={1,2,5,7,8}B'=\{1,2,5,7,8\}, C′={1,2,3,4}C'=\{1,2,3,4\}.

[!FORMULA] Complement: S′=U−S={x∈U:x∉S}S'=U-S=\{x\in U : x\notin S\}. Difference: P−Q={x∈P:x∉Q}P-Q=\{x\in P : x\notin Q\}. Union/intersection as usual.

  1. First find the complements: A′=U−A={5,6,7,8}A'=U-A=\{5,6,7,8\}; B′=U−B={1,2,5,7,8}B'=U-B=\{1,2,5,7,8\}; C′=U−C={1,2,3,4}C'=U-C=\{1,2,3,4\}.

  2. (i) A−(B∪C)A-(B\cup C): B∪C={3,4,6}∪{5,6,7,8}={3,4,5,6,7,8}B\cup C=\{3,4,6\}\cup\{5,6,7,8\}=\{3,4,5,6,7,8\}. Removing these from A={1,2,3,4}A=\{1,2,3,4\} leaves {1,2}\{1,2\}. A−(B∪C)={1,2}A-(B\cup C)=\{1,2\}.

  3. (ii) A∩C′A\cap C': C′={1,2,3,4}C'=\{1,2,3,4\}. A∩C′={1,2,3,4}∩{1,2,3,4}={1,2,3,4}A\cap C'=\{1,2,3,4\}\cap\{1,2,3,4\}=\{1,2,3,4\} (since AA and CC are disjoint, A∩C′=AA\cap C'=A).

  4. (iii) B′∩C′B'\cap C': B′={1,2,5,7,8}B'=\{1,2,5,7,8\}, C′={1,2,3,4}C'=\{1,2,3,4\}. Common elements: 1,21,2. B′∩C′={1,2}B'\cap C'=\{1,2\}.

  5. (iv) B′∪A′B'\cup A': B′={1,2,5,7,8}B'=\{1,2,5,7,8\}, A′={5,6,7,8}A'=\{5,6,7,8\}. Union: {1,2,5,6,7,8}\{1,2,5,6,7,8\}. B′∪A′={1,2,5,6,7,8}B'\cup A'=\{1,2,5,6,7,8\}.

  6. (v) A−(B∪C)′A-(B\cup C)': from step 2, B∪C={3,4,5,6,7,8}B\cup C=\{3,4,5,6,7,8\}, so (B∪C)′=U−(B∪C)={1,2}(B\cup C)'=U-(B\cup C)=\{1,2\}. Then A−(B∪C)′={1,2,3,4}−{1,2}={3,4}A-(B\cup C)'=\{1,2,3,4\}-\{1,2\}=\{3,4\}.

✓Final answer

(i) A−(B∪C)={1,2}A-(B\cup C)=\{1,2\};

(ii) A∩C′={1,2,3,4}A\cap C'=\{1,2,3,4\};

(iii) B′∩C′={1,2}B'\cap C'=\{1,2\};

(iv) B′∪A′={1,2,5,6,7,8}B'\cup A'=\{1,2,5,6,7,8\};

(v) A−(B∪C)′={3,4}A-(B\cup C)'=\{3,4\}.

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