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NCERT Exemplar · Q2

Q.If P={x:x<3, x∈N}P = \{x : x < 3,\ x \in \mathbf{N}\}, Q={x:x≤2, x∈W}Q = \{x : x \le 2,\ x \in \mathbf{W}\}. Find (P∪Q)×(P∩Q)(P \cup Q) \times (P \cap Q), where W\mathbf{W} is the set of whole numbers.

Yanam BieapShort· 2mImportance★★★★★est
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✓ Free question

(P∪Q)×(P∩Q)={0,1,2}×{1,2}={(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}(P \cup Q) \times (P \cap Q) = \{0,1,2\} \times \{1,2\} = \{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\} — 6 ordered pairs.

List the sets. Using the Indian convention N={1,2,3,… }\mathbf{N} = \{1,2,3,\dots\} and W={0,1,2,3,… }\mathbf{W} = \{0,1,2,3,\dots\}:

  • P={x:x<3, x∈N}={1,2}P = \{x : x < 3,\ x \in \mathbf{N}\} = \{1, 2\}
  • Q={x:x≤2, x∈W}={0,1,2}Q = \{x : x \le 2,\ x \in \mathbf{W}\} = \{0, 1, 2\}

Union and intersection:

P∪Q={0,1,2},P∩Q={1,2}P \cup Q = \{0, 1, 2\}, \qquad P \cap Q = \{1, 2\}

Cartesian product — first element from {0,1,2}\{0,1,2\}, second from {1,2}\{1,2\}:

(P∪Q)×(P∩Q)={(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}(P \cup Q) \times (P \cap Q) = \{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}

This has ∣P∪Q∣×∣P∩Q∣=3×2=6|P \cup Q| \times |P \cap Q| = 3 \times 2 = 6 pairs, as expected.

✓Final answer

(P∪Q)×(P∩Q)={(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)}(P \cup Q) \times (P \cap Q) = \{(0,1),(0,2),(1,1),(1,2),(2,1),(2,2)\}.

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