Q.Find the intersection of each pair of sets of question 1 above.
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🔒 Start your 14-day free trial to unlock the full solution →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∪B) | In either? | Everything from A or B (or both) |
| Intersection (A∩B) | In both? | Only the overlap |
| Difference (A∖B) | In A but not B? | Just the part of A that doesn't touch B |
| Complement (Ac) | Not in A? | Everything outside A (inside U) |
| Symmetric Difference (A△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},B={3,4,5}
Step 1: Union — gather everything from both, but don't repeat anything.
A∪B={1,2,3,4,5}
Step 2: Intersection — only what's common to both.
A∩B={3}
Step 3: Difference (A minus B) — start with A, remove anything that's also in B.
A∖B={1,2}
Step 4: Complement — needs a universal set. Let U={1,2,3,4,5}. Then:
Ac={4,5}
Step 5: Symmetric Difference — combine the two differences:
A△B=(A∖B)∪(B∖A)={1,2}∪{4,5}={1,2,4,5}
A common slip …
Intersection means the elements common to both sets. Using the five pairs of sets from Question 1 of this exercise:
- {1,3,5}∩{1,2,3}={1,3}
- {a,e,i,o,u}∩{a,b,c}={a}
- (multiples of 3) ∩ (naturals less than 6) ={3} …
Using the same five pairs of sets as Question 1 of this exercise, their intersections are: (i) {1,3},
(ii) {a},
(iii) {3},
(iv) ∅,
(v) ∅.
This question reuses the five pairs of sets from Question 1 of the same exercise and asks for the intersection (∩) of each pair -- the elements common to both sets.
(i) X={1,3,5}, Y={1,2,3}. The elements appearing in both are 1 and 3.
X∩Y={1,3}
(ii) A={a,e,i,o,u}, B={a,b,c}. Only a appears in both.
A∩B={a}
(iii) A={x:x is a natural number and a multiple of 3}={3,6,9,12,…}, B={x:x is a natural number less than 6}={1,2,3,4,5}. The only multiple of 3 that is also less than 6 is 3.
A∩B={3}
(iv) A={x:x is a natural number,1<x≤6}={2,3,4,5,6}, B={x:x is a natural number,6<x<10}={7,8,9}. No number can be both ≤6 and >6 at once.
A∩B=∅ …
Method: Direct Element Comparison (Set Intersection)
Concept: The intersection of two sets A and B, written A∩B, is the set of all elements that belong to both A and B.
Steps
- List all elements of the first set clearly.
- List all elements of the second set clearly.
- Compare each element of the first set with the elements of the second set.
- Pick only those elements that appear in both sets.
- Write the result as a set (curly braces, elements separated by commas).
Example (from a typical Question 1)
Let’s assume the pairs from Question 1 were:
- Pair (i): A={1,2,3,4}, B={3,4,5,6}
- Pair (ii): C={a,b,c}, D={b,c,d,e}
For Pair (i):
- Elements of A: 1,2,3,4
- Elements of B: 3,4,5,6
- Common elements: 3 and 4
- Answer: A∩B={3,4}
For Pair (ii): …
Here’s a breakdown of the common mistakes students make when finding the intersection of sets (based on a typical "Question 1" style from NCERT or state board textbooks), along with how to avoid each.
✗ Mistake 1: Confusing Intersection with Union
What students do:
They list all elements from both sets instead of only the common ones.
Example:
A={1,2,3}, B={2,3,4}
Wrong answer: A∩B={1,2,3,4} (this is actually A∪B).
Why it happens:
Students mix up the symbols ∩ (intersection) and ∪ (union).
✓ How to avoid:
- Memorise the shape: ∩ looks like an arch — think “Arch for And” (both sets must have it).
- Always ask: “Is this element in both sets?” If no, leave it out.
✗ Mistake 2: Forgetting to Check All Elements in Both Sets
What students do:
They only check elements from the first set against the second, but miss elements that appear only in the second set.
Example:
X={a,b,c}, Y={b,c,d}
Wrong: X∩Y={b} (missed c).
Why it happens:
Rushing — they stop after scanning the first set.
✓ How to avoid:
- Systematic method:
- List all elements of set 1.
- For each, check if it’s in set 2.
- Then repeat by checking elements of set 2 against set 1 (or just verify your list).
- Use a Venn diagram to visualise overlap.
✗ Mistake 3: Including Duplicate Elements in the Answer
What students do:
They write the same element twice in the intersection set.
Example:
A={1,2,2,3}, B={2,3,4}
Wrong: A∩B={2,2,3}
Why it happens:
They copy duplicates from the original sets without realising sets don’t have repetitions.
✓ How to avoid:
- Remember: Sets never contain duplicates.
- Write the intersection as a set of distinct elements only.
- After listing, check: “Did I write any number twice?”
✗ Mistake 4: Writing the Answer in Wrong Notation
What students do:
They write the answer as a list without curly braces, or use round brackets.
Example:
Wrong: A∩B=2,3 or (2,3)
Why it happens:
Carelessness with set notation.
✓ How to avoid:
- Always use curly braces {} for sets.
- Double-check your final answer format before moving to the next question.
✗ Mistake 5: Assuming Intersection Always Has Elements …
- AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQQ.If A={x∈R∣Sin−1(x2+x+1)∈[−2π,2π]} and B={y∈R∣y=Sin−1(x2+x+1),x∈A} then (A) A∩B=ϕ (B) A∩BC=[0,1] (C) AC∩B=[3π,2π] (D) A∪B=R−{[−1,0]∪[3π,2π]}
›Reveal solutionSolution
Working out A=[−1,0] and B=[π/3,π/2] shows these two sets are disjoint (one is a set of x-values near 0, the other a set of angle-values near π/2), which makes AC∩B simply equal to B itself, matching option (C).
Concept and Intuition
The trick is recognising that Sin−1(u)∈[−π/2,π/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 x2+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 x2+x+1=(x+21)2+43>0 always, x2+x+1 is real and non-negative for all real x. The condition Sin−1(x2+x+1)∈[−π/2,π/2] holds automatically whenever Sin−1 is defined, i.e. whenever x2+x+1∈[−1,1]. Since the square root is ≥0, this reduces to x2+x+1≤1⇔x2+x+1≤1⇔x2+x≤0⇔x(x+1)≤0⇔x∈[−1,0]. So A=[−1,0].
- Find the range of x2+x+1 on A: this is a upward parabola with vertex at x=−1/2, value 43; at the endpoints x=−1 and x=0, the value is 1. So on [−1,0], x2+x+1 ranges continuously over [43,1].
- Find B: x2+x+1 then ranges over [3/2,1]. Since Sin−1 is increasing, B=Sin−1([3/2,1])=[Sin−1(3/2),Sin−1(1)]=[π/3,π/2].
- Compare A and B: A=[−1,0] is a set of real numbers near the origin; B=[π/3,π/2]≈[1.047,1.571] is a set of positive numbers greater than 1. These intervals do not overlap, so A∩B=ϕ.
- Check each option: …
- CA Foundation 2025Set may-20251 markMCQQ.If A={1,2,3,4}, B={2,4,6,8} and C={3,4,5,6}, the value of A−{B∪C} is (A) {1, 2, 3} (B) {2, 3, 4, 5} (C) {1} (D) {0}
›Reveal solutionSolution
B∪C={2,3,4,5,6,8}; removing these from A leaves {1}.
Step 1 — Compute the union B∪C
{2,4,6,8}∪{3,4,5,6}={2,3,4,5,6,8}
Step 2 — Compute the difference A−(B∪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∪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. …
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.If A and B are any two events of a sample space, then set-theoretic description for the event: "Exactly one of the events A, B to occur" is (Here Ec denotes the compliment of the event E) (A) A∩Bc (B) (A−B)∪(A∪B) (C) (A∩Bc)∪(Ac∩B) (D) (A∩B)c∪(Ac∩Bc)
›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∩Bc.
- "B occurs, A does not" =Ac∩B.
- These two cases are disjoint and together cover "exactly one occurs", so the event is (A∩Bc)∪(Ac∩B).
Common Mistakes …
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