If $A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}$, $B = \{2, 4, \ldots, 18\}$ and $\mathbb{N}$ the set of natural numbers is the universal set, then $A' \cup ((A \cup B) \cap B')$ is (A) $\phi$ (B) $\mathbb{N}$ (C) $A$ (D) $B$

If A = \1, 3, 5, 7, 9, 11, 13, 15, 17\, B = \2, 4, …, 18\ and mathbbN the set of natural numbers is the univer

NCERT Exemplar · Q41

A = \{1, 3, 5, 7, 9, 11, 13, 15, 17\}

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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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Simplifying $A' \cup ((A \cup B) \cap B')$ gives $A' \cup A = \mathbb{N}$ , the universal set — option (B).

Here $A = \{1,3,5,\dots,17\}$ (odd numbers) and $B = \{2,4,\dots,18\}$ (even numbers), so $A$ and $B$ are disjoint, and the universal set is $\mathbb{N}$ .

Simplify the inner bracket using the distributive law:

$(A \cup B) \cap B' = (A \cap B') \cup (B \cap B') = (A \cap B') \cup \phi = A \cap B'$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$A = \{x \in \mathbb{R} \mid \operatorname{Sin}^{-1}(\sqrt{x^2+x+1}) \in [-\frac{\pi}{2}, \frac{\pi}{2}]\}$
›Reveal solutionSolution
Working out $A=[-1,0]$ and $B=[\pi/3,\pi/2]$ shows these two sets are disjoint (one is a set of $x$ -values near 0, the other a set of angle-values near $\pi/2$ ), which makes $A^C\cap B$ simply equal to $B$ itself, matching option (C).

Concept and Intuition

The trick is recognising that $Sin^{-1}(u) \in [-\pi/2,\pi/2]$ is automatically true for every $u$ in the domain of $Sin^{-1}$ (that's the very definition of the principal value range), so set $A$ 's defining condition reduces to just requiring $\sqrt{x^2+x+1}$ to be a valid arcsine input, i.e. lying in $[-1,1]$ . Set $B$ is then the actual set of angle outputs produced as $x$ ranges over $A$ — an entirely different kind of set (radians, not $x$ -values), which is why it ends up disjoint from $A$ .

Step-by-Step Solution

Find $A$ : Since $x^2+x+1 = (x+\tfrac12)^2+\tfrac34 > 0$ always, $\sqrt{x^2+x+1}$ is real and non-negative for all real $x$ . The condition $Sin^{-1}(\sqrt{x^2+x+1}) \in [-\pi/2,\pi/2]$ holds automatically whenever $Sin^{-1}$ is defined, i.e. whenever $\sqrt{x^2+x+1}\in[-1,1]$ . Since the square root is $\geq 0$ , this reduces to $\sqrt{x^2+x+1}\leq 1 \Leftrightarrow x^2+x+1\leq 1 \Leftrightarrow x^2+x\leq 0 \Leftrightarrow x(x+1)\leq 0 \Leftrightarrow x\in[-1,0]$ . So $A = [-1, 0]$ .

Find the range of $x^2+x+1$ on $A$ : this is a upward parabola with vertex at $x=-1/2$ , value $\tfrac34$ ; at the endpoints $x=-1$ and $x=0$ , the value is $1$ . So on $[-1,0]$ , $x^2+x+1$ ranges continuously over $[\tfrac34, 1]$ .

Find $B$ : $\sqrt{x^2+x+1}$ then ranges over $[\sqrt3/2, 1]$ . Since $Sin^{-1}$ is increasing, $B = Sin^{-1}([\sqrt3/2,1]) = [Sin^{-1}(\sqrt3/2), Sin^{-1}(1)] = [\pi/3, \pi/2]$ .

Compare $A$ and $B$ : $A=[-1,0]$ is a set of real numbers near the origin; $B=[\pi/3,\pi/2]\approx[1.047, 1.571]$ is a set of positive numbers greater than 1. These intervals do not overlap, so $A\cap B=\phi$ .

Check each option: …
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CA Foundation 2025Set may-20251 markMCQ
$A = \{1, 2, 3, 4\}$
›Reveal solutionSolution

$B\cup C=\{2,3,4,5,6,8\}$ ; removing these from $A$ leaves $\{1\}$ .

Step 1 — Compute the union $B\cup C$
$\{2,4,6,8\}\cup\{3,4,5,6\}=\{2,3,4,5,6,8\}$
Step 2 — Compute the difference $A-(B\cup C)$

Keep elements of $A=\{1,2,3,4\}$ NOT in the union. Elements $2,3,4$ are all present in the union; only $1$ survives.
$A-(B\cup C)=\{1\}$
Why the other options are wrong: (A) {1,2,3} and (B) {2,3,4,5} keep elements that ARE in the union; (D) {0} introduces $0$ , which is in no set. …
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AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
$E^c$
›Reveal solutionSolution
"Exactly one occurs" is the symmetric-difference event: (A but not B) union (B but not A).

Concept and Intuition

"Exactly one" excludes both the case where neither occurs and the case where both occur. It is the union of the two mutually exclusive possibilities: only $A$ happens, or only $B$ happens.

Step-by-Step Solution

" $A$ occurs, $B$ does not" $=A\cap B^c$ .

" $B$ occurs, $A$ does not" $=A^c\cap B$ .

These two cases are disjoint and together cover "exactly one occurs", so the event is $(A\cap B^c)\cup(A^c\cap B)$ .

Common Mistakes …
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