Q.Consider the sets X and Y of Example 14. Find X ∩ Y
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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 …
Concept: Set Intersection — the intersection of two sets contains only elements common to both.
From Example 14, X={Ram, Geeta, Akbar} (the hockey team) and Y={Geeta, David, Ashok} (the football team). …
Using Example 14's sets — X={Ram, Geeta, Akbar} (hockey) and Y={Geeta, David, Ashok} (football) — the only name common to both is Geeta, so X∩Y={Geeta}.
Recalling Example 14's sets
This question reuses the two sets from Example 14: X, the students on the school hockey team, and Y, the students on the school football team.
X={Ram, Geeta, Akbar},Y={Geeta, David, Ashok}
Finding the intersection
X∩Y contains every student who appears on both lists — students who play both hockey and football. …
Method: Set Intersection Using Roster Form
This method works when both sets are given in roster (list) form — we simply find the common elements.
Steps
-
Write both sets clearly
From Example 14 (assuming standard NCERT reference):
X={1,3,5}
Y={1,2,3}
-
Identify elements present in both sets
- Check each element of X:
- 1 is in Y → common
- 3 is in Y → common
- 5 is not in Y → not common
- No need to check Y again (we already covered all possibilities)
- Check each element of X:
-
Write the intersection set
Collect only the common elements:
X∩Y={1,3}
Key Concept …
Here are the common mistakes students make when working with Set Difference (and intersection, as in your question), along with how to avoid each.
Mistake 1: Confusing Set Difference with Intersection
The error:
Students treat X∖Y (or X−Y) as if it means X∩Y.
For example, if X={1,2,3} and Y={2,3,4}, they write X∖Y={2,3} instead of the correct {1}.
Why it happens:
The notation looks similar, and both operations involve comparing elements between two sets.
How to avoid:
- Remember the definition: X∖Y means "elements in X that are NOT in Y" — you remove all of Y from X.
- Use a mental filter: For each element of X, ask: "Is this also in Y?" If yes, discard it. If no, keep it.
- Draw a Venn diagram — shade only the part of X that does not touch Y.
Mistake 2: Forgetting the Order in Set Difference
The error:
Students assume X∖Y=Y∖X.
Example: X={a,b}, Y={b,c}. They write X∖Y={c} (which is actually Y∖X).
Why it happens:
Subtraction in arithmetic is commutative in some contexts (e.g., 5−3=3−5 is false, but students still blur the order). Set difference is not commutative.
How to avoid:
- Always read left-to-right: X∖Y means "start with X, remove Y."
- Write the operation in words before solving: "Elements of X that are not in Y."
- Check with a small example: If X={1}, Y={2}, then X∖Y={1} but Y∖X={2} — clearly different.
Mistake 3: Misapplying the Concept to Intersection (Your Question)
The error:
When asked for X∩Y, students accidentally compute X∖Y or Y∖X instead.
Why it happens:
Both operations involve comparing elements, and students rush without reading the symbol carefully (∩ vs ∖).
How to avoid:
- Memorise the symbols visually:
- ∩ looks like a "cup" — think common elements.
- ∖ is a subtraction sign — think remove.
- Before solving, state the definition aloud:
- X∩Y = "elements in both X and Y."
- X∖Y = "elements in X but not in Y."
- Double-check the question: Circle the symbol before you start.
Mistake 4: Including Elements Not in the First Set
The error:
For X∖Y, students list elements from Y that are not in X.
Example: X={1,2}, Y={2,3} → they write X∖Y={3}.
Why it happens:
They think "difference" means "all elements that are different between the two sets" (which is actually the symmetric difference).
How to avoid:
- Stick to the definition: Only elements from the first set matter.
- Use a two-step check:
- List all elements of X.
- Cross out any that also appear in Y. …
- 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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