Module 2 · JEE Mathematics
Sets, Relations & Functions — Organize, Connect, Transform
No formulas. No memorization. Just stories, animations, and "AHA!" moments.
🧸 Sets
🗺️ Relations
🥤 Functions
📈 Real Funcs
🎯 Practice
🧸 Part 1: Sets — The Toy Boxes
🗺️ Part 2: Relations — The Friendship Map
🥤 Part 3: Functions — The Vending Machine
📈 Part 4: Real Functions & Their Graphs
⚠️ Common Mistakes (Don't Fall For These!)
🧠 Sets, Relations & Functions Problem-Solving Toolkit
🔥 Advanced Worked Examples (JEE Level)
🎯 Practice Zone: Test Your Understanding (No Formulas!)
Sets Are Just Labeled Boxes Full of Things
Imagine you have two labeled boxes: Box A contains toys {1, 2, 3}. Box B contains {3, 4, 5}. That's all a set is — a collection of distinct objects with a label.
🧸 The Core Idea
Sets = Labeled Toy Boxes
Every set is like a box with a label. Inside the box are distinct items — no duplicates allowed. You can have a box of numbers, a box of letters, a box of colors, or even a box of other boxes!
Set notation: A = {1, 2, 3} means "Box A contains items 1, 2, and 3."
∈ means "is in the box" (3 ∈ A). ∉ means "is not in the box" (5 ∉ A).
Key rule: Order doesn't matter. {1, 2, 3} = {3, 1, 2}. And no duplicates: {1, 1, 2} is just {1, 2}.
Set notation: A = {1, 2, 3} means "Box A contains items 1, 2, and 3."
∈ means "is in the box" (3 ∈ A). ∉ means "is not in the box" (5 ∉ A).
Key rule: Order doesn't matter. {1, 2, 3} = {3, 1, 2}. And no duplicates: {1, 1, 2} is just {1, 2}.
💡 A set is a box. ∈ means "inside the box." ∉ means "outside the box." That's the whole language.
🧸 Sets = Toy Boxes
Union (∪): Pour both boxes into one big box. A ∪ B = {1, 2, 3, 4, 5}. The 3 only appears once — sets don't have duplicates!
Types of Sets
Sets come in different flavors. Here are the ones you'll see in JEE:
📦 Set Types at a Glance
Empty, Finite, Infinite, Equal, Subset, Superset, Universal
Empty set ∅: A box with nothing in it. {} or ∅.
Finite set: A box with a countable number of items. {1, 2, 3}.
Infinite set: A box that never ends. ℕ = {1, 2, 3, ...}.
Equal sets: Two boxes with exactly the same items. {1, 2} = {2, 1}.
Subset A ⊆ B: Every item in box A is also in box B. {1, 2} ⊆ {1, 2, 3}.
Superset A ⊇ B: Box A contains everything box B has and possibly more.
Universal set U: The biggest box — contains ALL items under consideration.
Finite set: A box with a countable number of items. {1, 2, 3}.
Infinite set: A box that never ends. ℕ = {1, 2, 3, ...}.
Equal sets: Two boxes with exactly the same items. {1, 2} = {2, 1}.
Subset A ⊆ B: Every item in box A is also in box B. {1, 2} ⊆ {1, 2, 3}.
Superset A ⊇ B: Box A contains everything box B has and possibly more.
Universal set U: The biggest box — contains ALL items under consideration.
🧠 Think of ⊆ as "fits inside." {1, 2} ⊆ {1, 2, 3} means the smaller box fits entirely inside the bigger one.
🔄 Intersection, Difference, Complement — Venn Diagram View
Intersection (∩): Items in BOTH boxes. Difference (−): Items in A but NOT in B. Complement ('): Everything in the universe U that's NOT in A. Power set: All possible subsets, including ∅ and A itself.
🔄 De Morgan's Law — The Complement Dance
De Morgan's laws are like logical double negatives. "Not (A or B)" means the same as "Not A and Not B." The complement of the union is the intersection of the complements.
🌟 Set Cardinality Formula
n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
Count everything in A, add everything in B, then subtract what you double-counted (the intersection). Like counting students in two classrooms — those in both get counted twice!
Worked Example: Venn Diagram with Three Sets
In a survey of 100 students: 50 like Math (M), 40 like Physics (P), 30 like Chemistry (C), 20 like both M and P, 15 like both P and C, 10 like both M and C, 5 like all three. How many like none?
1
Fill the Venn diagram from the center out
▶
Start with the triple intersection: M∩P∩C = 5.
M∩P only = 20 − 5 = 15 (total who like both minus those who like all three)
P∩C only = 15 − 5 = 10
M∩C only = 10 − 5 = 5
M only = 50 − 15 − 5 − 5 = 25
P only = 40 − 15 − 5 − 10 = 10
C only = 30 − 5 − 5 − 10 = 10
M∩P only = 20 − 5 = 15 (total who like both minus those who like all three)
P∩C only = 15 − 5 = 10
M∩C only = 10 − 5 = 5
M only = 50 − 15 − 5 − 5 = 25
P only = 40 − 15 − 5 − 10 = 10
C only = 30 − 5 − 5 − 10 = 10
2
Sum all the regions
▶
Total in at least one = 25 + 10 + 10 + 15 + 10 + 5 + 5 = 80
Students who like none = 100 − 80 = 20.
Students who like none = 100 − 80 = 20.
Worked Example: Power Set
Find the power set of A = {a, b, c}. How many subsets does it have?
1
Count the subsets
▶
For each of the 3 elements, you can either include it or exclude it. So 2 × 2 × 2 = 2³ = 8 subsets.
2
List them all
▶
∅ (empty set), {a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}
These 8 subsets form the power set P(A). The empty set and the original set A are always included!
These 8 subsets form the power set P(A). The empty set and the original set A are always included!
Worked Example: De Morgan's Law in Action
Given U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 4}, B = {3, 4, 5, 6}. Verify that (A ∪ B)' = A' ∩ B'.
1
Find A ∪ B and its complement
▶
A ∪ B = {1, 2, 3, 4, 5, 6}
(A ∪ B)' = U − (A ∪ B) = {7, 8}
(A ∪ B)' = U − (A ∪ B) = {7, 8}
2
Find A', B' and their intersection
▶
A' = U − A = {5, 6, 7, 8}
B' = U − B = {1, 2, 7, 8}
A' ∩ B' = {5, 6, 7, 8} ∩ {1, 2, 7, 8} = {7, 8}
(A ∪ B)' = {7, 8} = A' ∩ B' ✅
De Morgan's Law is verified!
B' = U − B = {1, 2, 7, 8}
A' ∩ B' = {5, 6, 7, 8} ∩ {1, 2, 7, 8} = {7, 8}
(A ∪ B)' = {7, 8} = A' ∩ B' ✅
De Morgan's Law is verified!
Relations Are Just Connections Between Things
A relation connects elements of one set to elements of another (or the same) set. Think of it like a friendship map: "Alice knows Bob," "Bob knows Charlie." Each arrow is an ordered pair.
🗺️ The Core Idea
Relations = Social Media Connections
Imagine a social network. People (set A) can follow other people (set B). Each follow is an ordered pair (follower, followee).
Ordered pair (a, b): The first element comes from set A (domain), the second from set B (codomain). Unlike sets, order matters here. (Alice, Bob) means Alice follows Bob — it's different from (Bob, Alice)!
Cartesian product A × B: The set of ALL possible ordered pairs. If A has 3 people and B has 3 people, A × B has 9 possible connections. A relation is just a SUBSET of these possible connections.
Ordered pair (a, b): The first element comes from set A (domain), the second from set B (codomain). Unlike sets, order matters here. (Alice, Bob) means Alice follows Bob — it's different from (Bob, Alice)!
Cartesian product A × B: The set of ALL possible ordered pairs. If A has 3 people and B has 3 people, A × B has 9 possible connections. A relation is just a SUBSET of these possible connections.
💡 A relation = a set of arrows. Domain = where arrows start. Codomain = where arrows could go. Range = where arrows actually go.
🗺️ Relations = Friendship Arrows
Domain = {Alice, Bob, Charlie}. Codomain = {Diana, Eve, Frank}. Range = {Diana, Eve, Frank} (all three have arrows pointing to them). The relation R = {(Alice, Diana), (Alice, Eve), (Bob, Diana), (Charlie, Frank)}.
Types of Relations on the Same Set
When domain and codomain are the SAME set (A to A), we get special types:
🔄 Reflexive, Symmetric, Transitive — The Three Musketeers
Reflexive: Every element relates to itself (self-loop). Symmetric: If A relates to B, B relates to A (bidirectional). Transitive: If A→B and B→C, then A→C (chain). An equivalence relation has ALL three properties — it groups elements into classes.
📐 Cartesian Product — All Possible Ordered Pairs
Cartesian product A × B: The set of ALL possible ordered pairs combining every element of A with every element of B. If |A| = m and |B| = n, then |A × B| = m × n. A relation is just a subset — you pick which pairs to include.
Worked Example: Identifying Relation Types
Let A = {1, 2, 3} and R = {(1,1), (1,2), (2,1), (2,2), (3,3)}. Is R reflexive? Symmetric? Transitive? An equivalence relation?
1
Check reflexive
▶
Reflexive means EVERY element relates to itself. Need (1,1), (2,2), (3,3). All three are present. ✅ Reflexive.
2
Check symmetric
▶
Symmetric means if (a,b) is in R, then (b,a) must also be in R.
(1,2) is in R → (2,1) is in R ✅
All other pairs are self-pairs which are trivially symmetric. ✅ Symmetric.
(1,2) is in R → (2,1) is in R ✅
All other pairs are self-pairs which are trivially symmetric. ✅ Symmetric.
3
Check transitive and equivalence
▶
Transitive: if (a,b) and (b,c) in R then (a,c) in R.
(1,2) and (2,1) → (1,1) ✅ (present)
(2,1) and (1,2) → (2,2) ✅ (present)
No other pair combinations apply. ✅ Transitive.
Since R is reflexive + symmetric + transitive = Equivalence relation ✅. It partitions A into {1, 2} and {3}.
(1,2) and (2,1) → (1,1) ✅ (present)
(2,1) and (1,2) → (2,2) ✅ (present)
No other pair combinations apply. ✅ Transitive.
Since R is reflexive + symmetric + transitive = Equivalence relation ✅. It partitions A into {1, 2} and {3}.
🌟 Equivalence Relations = Sorting Machine
An equivalence relation partitions a set into disjoint groups (equivalence classes). Elements in the same group are "equivalent" to each other.
Example: "Has the same birthday" is an equivalence relation. It groups people by their birth date. Everyone in a group shares the same birthday! "Congruence modulo n" is another classic example.
Worked Example: Number of Relations
If |A| = m and |B| = n, how many possible relations can exist from A to B?
1
Think of the grid
▶
|A × B| = m × n. A relation is any subset of A × B. The number of subsets of a set with k elements is 2k.
Therefore, number of relations from A to B = 2m×n.
Example: |A| = 2, |B| = 3 → mn = 6 → 2⁶ = 64 possible relations!
Therefore, number of relations from A to B = 2m×n.
Example: |A| = 2, |B| = 3 → mn = 6 → 2⁶ = 64 possible relations!
A Function Is a Vending Machine: One Button → One Snack
A function is a special kind of relation. A vending machine: each button (input) gives EXACTLY ONE snack (output). Not zero, not two — exactly one.
🥤 The Vending Machine Test
Is It a Function? The "One Button → One Snack" Rule
Function: Button 1 gives chips, Button 2 gives cookies. Every button gives exactly one snack. ✅
Not a function: Button 1 gives chips AND cookies. One input → two outputs. ❌
Still a function: Button 1 and Button 2 both give chips. Different inputs → same output is fine!
Domain: The set of buttons you can press.
Codomain: All snacks that could come out.
Range: Snacks that actually come out for some button.
Not a function: Button 1 gives chips AND cookies. One input → two outputs. ❌
Still a function: Button 1 and Button 2 both give chips. Different inputs → same output is fine!
Domain: The set of buttons you can press.
Codomain: All snacks that could come out.
Range: Snacks that actually come out for some button.
🧠 Functions are the strict parent of relations: "One input, exactly one output. No exceptions!"
🥤 Functions = Vending Machines
Domain = buttons you can press (inputs). Codomain = all snacks in the machine (possible outputs). Range = snacks that actually drop for some button. Each button → exactly one snack. That's a function!
🔄 One-One (Injective) vs Many-One — Mapping Diagrams
One-one (injective): Each input maps to a unique output (distinct arrows, no sharing). Many-one: Multiple inputs share the same output. Onto (surjective): Every possible output is used. Bijective: One-one AND onto = perfect pairing!
Composition of Functions
Think of composition as a conveyor belt. f(g(x)) means: first apply g, then apply f to the result.
⛓️ Composition f(g(x)) — The Conveyor Belt
Watch the ball move: x enters g → g(x) comes out → that goes into f → final result f(g(x)). Like a factory assembly line where each machine transforms the product further.
Worked Example: Domain and Range
Find the domain and range of f(x) = √(x − 3).
1
Domain: what x values are allowed?
▶
The square root of a negative number is not real. So we need x − 3 ≥ 0 → x ≥ 3.
Domain: [3, ∞)
Domain: [3, ∞)
2
Range: what y values come out?
▶
√(anything) is always ≥ 0. For x = 3, f(3) = 0. As x → ∞, f(x) → ∞.
Range: [0, ∞)
Range: [0, ∞)
Worked Example: Composite Function
If f(x) = 2x + 1 and g(x) = x², find f(g(3)) and g(f(3)). Are they equal?
1
Compute f(g(3)) — apply g first, then f
▶
g(3) = 3² = 9
f(g(3)) = f(9) = 2(9) + 1 = 18 + 1 = 19
f(g(3)) = f(9) = 2(9) + 1 = 18 + 1 = 19
2
Compute g(f(3)) — apply f first, then g
▶
f(3) = 2(3) + 1 = 7
g(f(3)) = g(7) = 7² = 49
f(g(3)) = 19 ≠ 49 = g(f(3))
Composition is NOT commutative: f(g(x)) ≠ g(f(x)) in general. Order matters!
g(f(3)) = g(7) = 7² = 49
f(g(3)) = 19 ≠ 49 = g(f(3))
Composition is NOT commutative: f(g(x)) ≠ g(f(x)) in general. Order matters!
Worked Example: Checking Bijectivity
Is the function f: ℝ → ℝ defined by f(x) = 2x + 3 bijective? Find its inverse if it exists.
1
Check one-one (injective)
▶
If f(a) = f(b), then 2a + 3 = 2b + 3 → 2a = 2b → a = b.
Different inputs always give different outputs. ✅ One-one.
Different inputs always give different outputs. ✅ One-one.
2
Check onto (surjective)
▶
For any y in ℝ, can we find x such that f(x) = y?
y = 2x + 3 → x = (y − 3)/2. For any real y, x is real. ✅ Onto.
Since f is both one-one AND onto, it is bijective.
y = 2x + 3 → x = (y − 3)/2. For any real y, x is real. ✅ Onto.
Since f is both one-one AND onto, it is bijective.
3
Find the inverse
▶
Since f is bijective, its inverse exists. Swap x and y: x = 2y + 3 → y = (x − 3)/2.
f(f⁻¹(y)) = f((y − 3)/2) = 2(y − 3)/2 + 3 = y ✅
f⁻¹(x) = (x − 3)/2
Verify: f⁻¹(f(x)) = f⁻¹(2x + 3) = (2x + 3 − 3)/2 = x ✅f(f⁻¹(y)) = f((y − 3)/2) = 2(y − 3)/2 + 3 = y ✅
🌟 Invertible Functions
A function is invertible if you can "reverse the vending machine" — press a button that exactly undoes what the original did. This is only possible if the function is bijective (one-one AND onto).
f⁻¹(f(x)) = x and f(f⁻¹(y)) = y. Like a perfect undo button! Only bijective functions are invertible.
🧩 One-Liner Summary
Functions at a Glance
One-one (injective): Different inputs → different outputs. No two buttons give the same snack.
Onto (surjective): Every snack in the machine is reachable by some button.
Bijective: Both one-one AND onto. Perfect matching. The only kind that's invertible!
Many-one: Not one-one. Two different inputs give the same output.
Into: Not onto. Some snacks in the machine never come out.
Onto (surjective): Every snack in the machine is reachable by some button.
Bijective: Both one-one AND onto. Perfect matching. The only kind that's invertible!
Many-one: Not one-one. Two different inputs give the same output.
Into: Not onto. Some snacks in the machine never come out.
🧠 Injective = no collisions. Surjective = full coverage. Bijective = perfect 1-to-1 mapping.
Seeing Functions: The V-Shape, the Staircase, the Step
Some functions have such distinctive graphs that they get special names. These appear constantly in JEE — learn their shapes, not their formulas!
Modulus Function: f(x) = |x|
The absolute value (modulus) measures distance from zero. |x| = x if x ≥ 0, |x| = -x if x < 0. The graph is a perfect V-shape.
📐 The V-Shape — f(x) = |x|
The modulus function creates a V-shape. For x ≥ 0, it's the line y = x (45° slope). For x < 0, it's y = -x (reflects across the y-axis). The vertex at (0, 0) is the minimum — the closest you can get to 0.
Signum Function: sgn(x)
The signum function tells you the SIGN of a number: -1 if negative, 0 if zero, 1 if positive.
⏫ The Step Function — sgn(x)
Signum: A step function with three possible values. At x = 0, sgn(0) = 0 (the only zero output). For any positive x, sgn(x) = 1. For any negative x, sgn(x) = -1. The open circles at the jumps show discontinuity.
Greatest Integer Function: f(x) = [x] (Floor)
The greatest integer function [x] returns the largest integer ≤ x. Also called the floor function. Its graph is a staircase!
🏗️ The Staircase — f(x) = [x] (Greatest Integer)
[x] creates a staircase ascending to the right. Each step is flat (constant value) from integer n to just before n+1. At the integer n itself, the value jumps up by 1. Note: [-2.3] = -3 (rounds DOWN, not towards zero!).
Fractional Part Function: f(x) = {x}
The fractional part {x} = x - [x]. It extracts the decimal part of x. The graph is a sawtooth wave!
🦷 The Sawtooth — {x} = x - [x] (Fractional Part)
Fractional part {x}: A sawtooth wave that resets to 0 at every integer. It rises linearly from 0 to almost 1, then drops back to 0. It's always in the range [0, 1). The function is periodic with period 1.
🌟 Real Functions Cheat Sheet
|x| = distance from 0 (V-shape) |
sgn(x) = sign of x (step) |
[x] = floor (staircase) |
{x} = x - [x] (sawtooth)
Learn the SHAPES, not the formulas. In JEE, visual recognition of these graphs saves precious minutes.
Worked Example: Modulus Equation
Solve |2x − 5| = 9.
1
Set up two cases
▶
|2x − 5| = 9 means the distance from 2x − 5 to 0 is 9.
So either 2x − 5 = 9 or 2x − 5 = −9.
So either 2x − 5 = 9 or 2x − 5 = −9.
2
Solve both cases
▶
Case 1: 2x − 5 = 9 → 2x = 14 → x = 7
Case 2: 2x − 5 = −9 → 2x = −4 → x = −2
Solution set: {−2, 7}
Case 2: 2x − 5 = −9 → 2x = −4 → x = −2
Solution set: {−2, 7}
These Traps Trip Up Even Top Students
⚠️ Mistake 1
Confusing ∈ and ⊆
Wrong: If A = {{1, 2}, 3}, then 1 ∈ A ❌
Right: 1 ∉ A because A contains {1, 2} (a set) and 3, not the element 1 directly.
∈ is for elements (items inside the box). ⊆ is for subsets (boxes inside boxes).
{1, 2} ∈ A ✅ (the set {1,2} is an element of A)
1 ∉ A ❌ (1 is inside {1,2}, which is in A, but not directly in A)
Right: 1 ∉ A because A contains {1, 2} (a set) and 3, not the element 1 directly.
∈ is for elements (items inside the box). ⊆ is for subsets (boxes inside boxes).
{1, 2} ∈ A ✅ (the set {1,2} is an element of A)
1 ∉ A ❌ (1 is inside {1,2}, which is in A, but not directly in A)
⚠️ ∈ = "is an element of" (direct member). ⊆ = "is a subset of" (fits inside). Don't mix them!
⚠️ Mistake 2
Forgetting That Order Matters in Relations
Wrong: (a, b) = (b, a) in all cases ❌
Right: (a, b) ≠ (b, a) unless a = b.
Ordered pairs are the foundation of relations. (Alice, Bob) means Alice follows Bob. (Bob, Alice) means Bob follows Alice. Completely different relationships!
In a Cartesian product, (1, 2) and (2, 1) are TWO DIFFERENT pairs.
Right: (a, b) ≠ (b, a) unless a = b.
Ordered pairs are the foundation of relations. (Alice, Bob) means Alice follows Bob. (Bob, Alice) means Bob follows Alice. Completely different relationships!
In a Cartesian product, (1, 2) and (2, 1) are TWO DIFFERENT pairs.
⚠️ (a, b) ≠ (b, a). Order always matters in ordered pairs!
⚠️ Mistake 3
Thinking "Onto" Means "Every Input Has an Output"
Wrong: "A function is onto because every input maps to something." ❌
Right: Every function has this property — that's what makes it a function!
Onto (surjective) means every element of the CODOMAIN is used, not every element of the domain. It's about covering the target set, not about having defined outputs.
Think: "Every possible snack actually comes out for some button" — that's onto.
Right: Every function has this property — that's what makes it a function!
Onto (surjective) means every element of the CODOMAIN is used, not every element of the domain. It's about covering the target set, not about having defined outputs.
Think: "Every possible snack actually comes out for some button" — that's onto.
⚠️ Into = some codomain elements unused. Onto = ALL codomain elements used. Don't confuse "function" with "onto"!
⚠️ Mistake 4
Greatest Integer of Negative Numbers
Wrong: [-2.3] = -2 ❌
Right: [-2.3] = -3 ✅
[x] is the greatest integer ≤ x. For -2.3, the integers ≤ -2.3 are ... -5, -4, -3. The greatest among these is -3.
Think: [-2.3] = -3 because -3 ≤ -2.3 < -2. The floor function always rounds DOWN, not towards zero!
Right: [-2.3] = -3 ✅
[x] is the greatest integer ≤ x. For -2.3, the integers ≤ -2.3 are ... -5, -4, -3. The greatest among these is -3.
Think: [-2.3] = -3 because -3 ≤ -2.3 < -2. The floor function always rounds DOWN, not towards zero!
⚠️ [x] rounds DOWN. For negatives, [-2.3] = -3, not -2. "Floor" means lower, not closer to zero.
4 Strategies That Crack Almost Any Problem
🧠 Strategy 1
"Draw the Venn Diagram"
For any set problem with multiple sets and overlaps, draw a Venn diagram. Start from the innermost intersection and work outward. Label each region. Then add up the regions.
Example: In a class of 100, 60 play cricket, 50 play football, 30 play both. How many play neither?
Draw two overlapping circles. The intersection is 30. Cricket only = 60−30 = 30. Football only = 50−30 = 20. Total who play = 30+30+20 = 80. Neither = 100−80 = 20.
Example: In a class of 100, 60 play cricket, 50 play football, 30 play both. How many play neither?
Draw two overlapping circles. The intersection is 30. Cricket only = 60−30 = 30. Football only = 50−30 = 20. Total who play = 30+30+20 = 80. Neither = 100−80 = 20.
💡 Venn diagrams turn abstract set problems into simple addition problems. Always start from the center!
🧠 Strategy 2
"The Vertical and Horizontal Line Tests"
These two tests are all you need for function classification:
Vertical line test: Is it a function? Draw a vertical line across the graph. If it hits the graph more than once, it's NOT a function (one input → multiple outputs).
Horizontal line test: Is it one-one? Draw a horizontal line across the graph. If it hits the graph more than once, it's NOT one-one (two inputs share the same output).
Every function passes the vertical line test. One-one functions also pass the horizontal line test.
Vertical line test: Is it a function? Draw a vertical line across the graph. If it hits the graph more than once, it's NOT a function (one input → multiple outputs).
Horizontal line test: Is it one-one? Draw a horizontal line across the graph. If it hits the graph more than once, it's NOT one-one (two inputs share the same output).
Every function passes the vertical line test. One-one functions also pass the horizontal line test.
💡 Vertical line = "is it a function?" Horizontal line = "is it one-one?" Two tests, zero calculations!
🧠 Strategy 3
"Check Reflexive First"
When identifying relation types, always check reflexive first. It's the easiest — just check if EVERY element has a self-loop (a,a).
If it's not reflexive, it can't be an equivalence relation. Save time by checking this first!
Then check symmetric (if a→b then b→a). Then transitive (if a→b and b→c then a→c). Only if all three hold is it an equivalence relation.
If it's not reflexive, it can't be an equivalence relation. Save time by checking this first!
Then check symmetric (if a→b then b→a). Then transitive (if a→b and b→c then a→c). Only if all three hold is it an equivalence relation.
💡 Reflexive is the quickest check. No self-loops? Not an equivalence relation. Move on!
🧠 Strategy 4
"f(g(x)) = Apply Inside-Out"
Composition f(g(x)) looks confusing, but it's just "apply g to x, then apply f to the result." Read from the inside out.
Example: f(x) = √x, g(x) = x + 1. f(g(3)) = f(3+1) = f(4) = √4 = 2.
The notation f∘g means "f after g" — first g, then f. Think of it as a pipeline: x → g → f → result.
Example: f(x) = √x, g(x) = x + 1. f(g(3)) = f(3+1) = f(4) = √4 = 2.
The notation f∘g means "f after g" — first g, then f. Think of it as a pipeline: x → g → f → result.
💡 f(g(x)) = "inside-out." Start with x, apply g, then apply f. Conveyor belt!
Multi-Step Problems That Combine Multiple Concepts
Example 1: Sets + Cardinality + Venn Diagram
In a group of 200 students: 120 take Physics (P), 90 take Chemistry (C), 80 take Math (M). 50 take P and C, 40 take C and M, 30 take P and M, 20 take all three. How many take exactly two subjects? How many take none?
1
Find the "exactly two" regions
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P∩C only = 50 − 20 = 30 (total P∩C minus those who also take M)
C∩M only = 40 − 20 = 20
P∩M only = 30 − 20 = 10
Exactly two subjects = 30 + 20 + 10 = 60
C∩M only = 40 − 20 = 20
P∩M only = 30 − 20 = 10
Exactly two subjects = 30 + 20 + 10 = 60
2
Find "exactly one" and "none"
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P only = 120 − 30 − 10 − 20 = 60
C only = 90 − 30 − 20 − 20 = 20
M only = 80 − 10 − 20 − 20 = 30
Total in at least one = 60 + 20 + 30 + 30 + 20 + 10 + 20 = 190
None = 200 − 190 = 10
C only = 90 − 30 − 20 − 20 = 20
M only = 80 − 10 − 20 − 20 = 30
Total in at least one = 60 + 20 + 30 + 30 + 20 + 10 + 20 = 190
None = 200 − 190 = 10
Example 2: Relations + Equivalence Classes
Let R be a relation on ℤ (integers) defined by aRb if a² ≡ b² (mod 5). Show that R is an equivalence relation and find the equivalence classes.
1
Show reflexive
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For any a ∈ ℤ, a² ≡ a² (mod 5) trivially. So aRa holds for all a. ✅ Reflexive.
2
Show symmetric and transitive
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Symmetric: If a² ≡ b² (mod 5), then b² ≡ a² (mod 5). So aRb → bRa. ✅
Transitive: If a² ≡ b² (mod 5) and b² ≡ c² (mod 5), then a² ≡ c² (mod 5). So aRb and bRc → aRc. ✅
Since reflexive + symmetric + transitive, R is an equivalence relation.
Transitive: If a² ≡ b² (mod 5) and b² ≡ c² (mod 5), then a² ≡ c² (mod 5). So aRb and bRc → aRc. ✅
Since reflexive + symmetric + transitive, R is an equivalence relation.
3
Find equivalence classes
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Squares modulo 5 are: 0²≡0, 1²≡1, 2²≡4, 3²≡9≡4, 4²≡16≡1.
So the squares are only 0, 1, or 4 mod 5. Equivalence classes:
• [0] = {a : a² ≡ 0 mod 5} = {..., -5, 0, 5, 10, ...} (multiples of 5)
• [1] = {a : a² ≡ 1 mod 5} = numbers ≡ ±1 mod 5 = {..., -4, -1, 1, 4, 6, 9, ...}
• [2] = {a : a² ≡ 4 mod 5} = numbers ≡ ±2 mod 5 = {..., -3, -2, 2, 3, 7, 8, ...}
Three equivalence classes partition ℤ!
So the squares are only 0, 1, or 4 mod 5. Equivalence classes:
• [0] = {a : a² ≡ 0 mod 5} = {..., -5, 0, 5, 10, ...} (multiples of 5)
• [1] = {a : a² ≡ 1 mod 5} = numbers ≡ ±1 mod 5 = {..., -4, -1, 1, 4, 6, 9, ...}
• [2] = {a : a² ≡ 4 mod 5} = numbers ≡ ±2 mod 5 = {..., -3, -2, 2, 3, 7, 8, ...}
Three equivalence classes partition ℤ!
Example 3: Function + Domain + Composite
If f(x) = 1/(x − 2) and g(x) = √(x + 3), find the domain of f(g(x)) and g(f(x)).
1
Domain of f(g(x))
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f(g(x)) = f(√(x + 3)) = 1/(√(x + 3) − 2)
Restrictions:
1. Inside √: x + 3 ≥ 0 → x ≥ −3
2. Denominator ≠ 0: √(x + 3) ≠ 2 → x + 3 ≠ 4 → x ≠ 1
Domain: [−3, 1) ∪ (1, ∞)
Restrictions:
1. Inside √: x + 3 ≥ 0 → x ≥ −3
2. Denominator ≠ 0: √(x + 3) ≠ 2 → x + 3 ≠ 4 → x ≠ 1
Domain: [−3, 1) ∪ (1, ∞)
2
Domain of g(f(x))
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g(f(x)) = g(1/(x − 2)) = √(1/(x − 2) + 3) = √((1 + 3(x − 2))/(x − 2)) = √((3x − 5)/(x − 2))
Restrictions:
1. Denominator of f: x ≠ 2
2. Inside √: (3x − 5)/(x − 2) ≥ 0
Solve inequality: critical points at x = 5/3 and x = 2.
Sign analysis: x < 5/3 → (+)/(−) = − ❌ | 5/3 ≤ x < 2 → (+)/(−) = − ❌ | x > 2 → (+)/(+) = + ✅
Domain: (−∞, 5/3] ∪ (2, ∞)
Restrictions:
1. Denominator of f: x ≠ 2
2. Inside √: (3x − 5)/(x − 2) ≥ 0
Solve inequality: critical points at x = 5/3 and x = 2.
Sign analysis: x < 5/3 → (+)/(−) = − ❌ | 5/3 ≤ x < 2 → (+)/(−) = − ❌ | x > 2 → (+)/(+) = + ✅
Domain: (−∞, 5/3] ∪ (2, ∞)
Practice Zone: Think, Don't Compute
These questions test if you truly understand the ideas. No calculation needed — just reasoning!