(i) Set A has 3 elements, and set B has...

MathematicsTypes of Sets and Set OperationsJEE Advanced 1980Easy

(i) Set $A$ has 3 elements, and set $B$ has 6 elements. What can be the minimum number of elements in the set $A \cup B$ ? (ii) $P, Q, R$ are subsets of a set $A$ . Is the following equality true? $R \times (P^{c} \cup Q^{c})^{c} = (R \times P) \cap (R \times Q)$ ?

Answer:(i) 6 (ii) Yes

Visualized Solution (English)

Visualizing the Sets .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

Given: $n (A) = 3$ and $n (B) = 6$ .
We need to find the minimum value of $n (A \cup B)$ .
Recall that $A \cup B$ represents the set of elements in either $A$ , $B$ , or both.

The Principle of Inclusion-Exclusion

Using the formula: $n (A \cup B) = n (A) + n (B) - n (A \cap B)$
Substituting the known values: $n (A \cup B) = 3 + 6 - n (A \cap B)$
$n (A \cup B) = 9 - n (A \cap B)$

Maximizing the Intersection .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n (A \cap B)

To minimize $n (A \cup B)$ , we must maximize $n (A \cap B)$ .
The maximum possible number of common elements is $min (n (A), n (B))$ .
$max (n (A \cap B)) = min (3, 6) = 3$ .

Calculating Minimum .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n (A \cup B)

$Min n (A \cup B) = 9 - 3 = 6$ .
This occurs when $A \subset B$ .

Analyzing Part (ii): Set Equality

Verify: $R \times (P^{c} \cup Q^{c})^{c} = (R \times P) \cap (R \times Q)$
Let's simplify the Left Hand Side (LHS) first.

Applying De Morgan's Law

LHS: $R \times (P^{c} \cup Q^{c})^{c}$
By De Morgan's Law: $(P^{c} \cup Q^{c})^{c} = (P^{c})^{c} \cap (Q^{c})^{c}$
Since $(A^{c})^{c} = A$ , we get: $P \cap Q$
So, LHS = $R \times (P \cap Q)$

Distributive Property of Cartesian Product

RHS: $(R \times P) \cap (R \times Q)$
Using the distributive property: $A \times (B \cap C) = (A \times B) \cap (A \times C)$
Therefore, RHS = $R \times (P \cap Q)$

Conclusion and Summary

Since LHS = RHS = $R \times (P \cap Q)$ , the equality is True.
Key Takeaway: De Morgan's Laws and Distributive Properties are powerful tools for simplifying set expressions.

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