Let $S$ = set of points inside the square, $T$ = the set of points inside the triangle and $C$ = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then (A) $S \cap T \cap C = \phi$ (B) $S \cup T \cup C = C$ (C) $S \cup T \cup C = S$ (D) $S \cup T = S \cap C$

> [!TLDR] > The triangle and circle are inside the square, so every point of the triangle and circle is also a point of the square. Therefore the union of all three sets is just the square itself, making option **(C)** correct. We have three sets: $S$ (square), $T$ (triangle), $C$ (circle). The key condition is that both the triangle and the circle are **contained in** the square, and they also intersect each other. That means every point that belongs to $T$ or $C$ automatically belongs to $S$. Let’s see what each option is really saying. 1. **Option (A):** $S \cap T \cap C = \phi$ This says there is no point common to all three — the square, triangle, and circle have no triple intersection. But the triangle and circle intersect (given), and that intersection lies inside the square (since both are inside the square). So any point in $T \cap C$ is also in $S$, meaning $S \cap T \cap C = T \cap C$, which is non-empty. So (A) is false. 2. **Option (B):** $S \cup T \cup C = C$ This claims the union of all three is just the circle. But the square contains points that are not in the circle (e.g., the corners of the square). So the union is larger than $C$. False. 3. **Option (C):** $S \cup T \cup C = S$ Since $T \subseteq S$ and $C \subseteq S$, every element of $T$ and $C$ is already in $S$. So the union of $S$, $T$, and $C$ is just $S$ itself. This is true. 4. **Option (D):** $S \cup T = S \cap C$ …

Let S = set of points inside the square, T = the set of points inside the triangle and C = the set of points i

NCERT Exemplar · Q33

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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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The triangle and circle are inside the square, so every point of the triangle and circle is also a point of the square. Therefore the union of all three sets is just the square itself, making option (C) correct.

We have three sets: $S$ (square), $T$ (triangle), $C$ (circle). The key condition is that both the triangle and the circle are contained in the square, and they also intersect each other. That means every point that belongs to $T$ or $C$ automatically belongs to $S$ . Let’s see what each option is really saying.

Option (A): $S \cap T \cap C = \phi$

This says there is no point common to all three — the square, triangle, and circle have no triple intersection. But the triangle and circle intersect (given), and that intersection lies inside the square (since both are inside the square). So any point in $T \cap C$ is also in $S$ , meaning $S \cap T \cap C = T \cap C$ , which is non-empty. So (A) is false.
Option (B): $S \cup T \cup C = C$

This claims the union of all three is just the circle. But the square contains points that are not in the circle (e.g., the corners of the square). So the union is larger than $C$ . False.
Option (C): $S \cup T \cup C = S$

Since $T \subseteq S$ and $C \subseteq S$ , every element of $T$ and $C$ is already in $S$ . So the union of $S$ , $T$ , and $C$ is just $S$ itself. This is true.
Option (D): $S \cup T = S \cap C$ …

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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 $A-(B\cup C)$

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