If $X$ and $Y$ are two sets and $X'$ denotes the complement of $X$, then $X \cap (X \cup Y)'$ is equal to (A) $X$ (B) $Y$ (C) $\phi$ (D) $X \cap Y$

If X and Y are two sets and X' denotes the complement of X, then X ∩ (X ∪ Y)' is equal to (A) X (B) Y (C) φ (D

NCERT Exemplar · Q43

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Concept understanding — Set Operations

The idea in plain words

Think of your two favourite groups of friends — the ones who play cricket and the ones who play football. Some friends are in both groups, some in only one, and some in neither. Set operations are simply the mathematical ways to answer questions like: "Who's in at least one team?" or "Who's only in the cricket team?".

The whole secret? Each operation is just a different way of combining or comparing two collections — like sorting your friends into different buckets.

Why this works

Sets are just labelled buckets that hold distinct items. The universal set $U$ is the "whole world" of things we're talking about — say, all your friends. Then each operation picks out a specific bucket:

Operation	What it asks	Bucket contains
Union ( $A \cup B$ )	In either?	Everything from $A$ or $B$ (or both)
Intersection ( $A \cap B$ )	In both?	Only the overlap
Difference ( $A \setminus B$ )	In $A$ but not $B$ ?	Just the part of $A$ that doesn't touch $B$
Complement ( $A^c$ )	Not in $A$ ?	Everything outside $A$ (inside $U$ )
Symmetric Difference ( $A \triangle B$ )	In exactly one?	The two crescent-shaped parts, excluding the overlap

Step by step

Let's take two concrete sets so you can see each operation in action:

A = \{1, 2, 3\}, \quad B = \{3, 4, 5\}

Step 1: Union — gather everything from both, but don't repeat anything.

A \cup B = \{1, 2, 3, 4, 5\}

Step 2: Intersection — only what's common to both.

A \cap B = \{3\}

Step 3: Difference ( $A$ minus $B$ ) — start with $A$ , remove anything that's also in $B$ .

A \setminus B = \{1, 2\}

Step 4: Complement — needs a universal set. Let $U = \{1, 2, 3, 4, 5\}$ . Then:

A^c = \{4, 5\}

Step 5: Symmetric Difference — combine the two differences:

A \triangle B = (A \setminus B) \cup (B \setminus A) = \{1, 2\} \cup \{4, 5\} = \{1, 2, 4, 5\}

A common slip …

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We simplify the given set expression by applying De Morgan's Law and properties of set intersection, ultimately finding that the expression simplifies to the empty set. The final result is $\boxed{\phi}$ .

When working with sets, expressions involving unions, intersections, and complements can often look complex. The goal is usually to simplify these expressions using fundamental laws of set theory, much like simplifying algebraic expressions. This process helps us understand the underlying relationship between the sets involved.

The expression we need to simplify is $X \cap (X \cup Y)'$ . Here, $X$ and $Y$ are arbitrary sets, and $X'$ denotes the complement of set $X$ . The complement $X'$ contains all elements not in $X$ (within a universal set). The union $X \cup Y$ contains all elements that are in $X$ or in $Y$ (or both). The intersection $X \cap Y$ contains all elements that are in $X$ and in $Y$ .

Let's break down the simplification step by step.

Identify the innermost operation and apply De Morgan's Law.

The expression contains $(X \cup Y)'$ . This is the complement of a union. De Morgan's Laws provide a way to express the complement of a union (or intersection) in terms of complements of individual sets.
$(A \cup B)' = A' \cap B'$
Applying this to $(X \cup Y)'$ , we get:

$(X \cup Y)' = X' \cap Y'$
Substitute the simplified term back into the original expression.

Now, replace $(X \cup Y)'$ with $X' \cap Y'$ in the original expression:

$X \cap (X \cup Y)' = X \cap (X' \cap Y')$
Apply the Associative Law of Intersection.

The intersection operation is associative, meaning the grouping of sets does not affect the result when only intersections are involved.

$A \cap (B \cap C) = (A \cap B) \cap C$

Applying this law, we can rearrange the parentheses:

$X \cap (X' \cap Y') = (X \cap X') \cap Y'$
Evaluate the intersection of a set and its complement. …

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Q.If $X$ and $Y$ are two sets and $X'$ denotes the complement of $X$ , then $X \cap (X \cup Y)'$ is equal to
(A) $X$
(B) $Y$
(C) $\phi$
(D) $X \cap Y$

Concept understanding — Set Operations

The idea in plain words

Why this works

Step by step

A common slip …

Step 1 — Compute the union $B\cup C$

Step 2 — Compute the difference $A-(B\cup C)$

Q.If XXX and YYY are two sets and X′X'X′ denotes the complement of XXX, then X∩(X∪Y)′X \cap (X \cup Y)'X∩(X∪Y)′ is equal to (A) XXX (B) YYY (C) ϕ\phiϕ (D) X∩YX \cap YX∩Y

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Q.If $X$ and $Y$ are two sets and $X'$ denotes the complement of $X$ , then $X \cap (X \cup Y)'$ is equal to
(A) $X$
(B) $Y$
(C) $\phi$
(D) $X \cap Y$

Unlock everything free for 14 days