Sets, Relations and Functions: JEE Main Maths formulas and notes

Q: How do I solve “Number of elements, subsets and power set” questions in JEE Main?

A set is given in roster form or by a rule. Find the number of its elements, subsets, proper subsets or sets between two given sets. 1. List the elements that satisfy the rule and count them. 2. A set of n elements has 2^n subsets, 2^n − 1 proper subsets. 3. For A ⊆ X ⊆ B the number is 2^(n(B) − n(A)). Example: {x ∈ ℤ : x² < 10} has 7 elements and 128 subsets.

Q: How do I solve “Set operations and De Morgan’s laws” questions in JEE Main?

Two or three sets are given, or a set identity. Find A − B, A Δ B, a complement, or decide which identity is true. 1. Write A − B as A ∩ B′. 2. Use De Morgan to take a complement. 3. Check an identity with a small example or a Venn diagram. Example: A = {1, 2, 3, 4}, B = {3, 4, 5}: A Δ B = {1, 2, 5}.

Q: How do I solve “Two-set counting (tea, coffee, cricket, football)” questions in JEE Main?

A group likes two things. Some numbers are given. Find those who like both, only one, or neither. 1. Write n(A ∪ B) = n(A) + n(B) − n(A ∩ B). 2. Find the unknown from the formula. 3. Neither = total − union. Example: 40, 30 and 55 in the union: 15 like both.

Q: How do I solve “Three-set counting with exactly one, exactly two” questions in JEE Main?

Three sets with pair and triple intersections are given. Find the union, the number in exactly one or exactly two sets, or the number in none. 1. Union = Σ1 − Σ2 + T. 2. Exactly two = Σ2 − 3T. 3. Exactly one = Σ1 − 2Σ2 + 3T. Example: 30, 25, 20; 10, 8, 6; 3: union 54, exactly one 36.

Q: How do I solve “Greatest and least value of an overlap” questions in JEE Main?

Sizes of two or three sets in a population are given. Find the least or the greatest number of elements in the intersection. 1. Least n(A ∩ B) = max(0, a + b − N). 2. Greatest = min(a, b). 3. Three sets: least triple = max(0, a + b + c − 2N). Example: N = 100, a = 70, b = 60: between 30 and 60.

Q: How do I solve “Pairs of subsets and special subsets” questions in JEE Main?

S has n elements. Count ordered pairs (A, B) with A ⊆ B, A ∩ B = ∅ or A ∪ B = S, or subsets with a stated property. 1. Give each element its possible places. 2. Multiply over the n elements. 3. Fixed elements are not counted again. Example: n = 3, A ⊆ B: 3³ = 27 pairs.

Q: How do I solve “Cartesian product and number of relations” questions in JEE Main?

Sizes of A and B are given. Find n(A × B), the number of relations, or a statement on products. 1. n(A × B) = mn. 2. Relations from A to B: 2^mn. 3. Use A × (B ∩ C) = (A × B) ∩ (A × C). Example: n(A) = 3, n(B) = 2: 6 pairs and 64 relations.

Q: How do I solve “Domain and range of a relation” questions in JEE Main?

A relation is given by a rule on ℕ or on a small set. Write it as pairs, find the domain and range, or the inverse. 1. List the pairs. 2. Domain = first coordinates, range = second. 3. Inverse: swap each pair. Example: x + 2y = 8 on ℕ: domain {2, 4, 6}, range {1, 2, 3}.

Q: How do I solve “Type of a given relation” questions in JEE Main?

A relation is given on a set of numbers or lines. Decide which of reflexive, symmetric, transitive, antisymmetric it is. 1. Test reflexive: is every (a, a) present? 2. Test symmetric: has every pair its reverse? 3. Test transitive: chain two pairs. Example: |x − y| ≤ 1 on ℤ: reflexive, symmetric, not transitive.

Q: How do I solve “Proving an equivalence relation and finding the classes” questions in JEE Main?

A rule such as x − y divisible by m, or "same remainder", is given. Prove it is an equivalence relation or find the class of an element. 1. Show reflexive, symmetric and transitive. 2. The class of a is the set of all elements related to a. 3. The classes are disjoint and cover the set. Example: x − y divisible by 3 on ℤ: 3 classes.

Sets, Relations and Functions give 2 to 3 questions in JEE Mains in most years. Sets and counting give a short numerical question (a Venn diagram or a power set); relations ask for the type of a relation or the count of one type; functions ask for one-one and onto, a composition, an inverse, or the domain and range. About 9 hours in 5 days is enough. The chapter is also the base for limits, calculus and probability, so the effort pays again.

Practise this chapter free in Rank mode

Open Physica, choose Rank mode, JEE Mains, then Maths and this chapter for the guide, formulas, most-asked questions, past-year questions and a mock test.

Quick revision

Subsets 2^n; proper 2^n − 1; A ⊆ X ⊆ B gives 2^b−a; n(A × B) = mn; relations A → B: 2^mn
A − B = A ∩ B′; De Morgan: (A ∪ B)′ = A′ ∩ B′, (A ∩ B)′ = A′ ∪ B′; n(A Δ B) = n(A) + n(B) − 2n(A ∩ B)
n(A ∪ B) = n(A) + n(B) − n(A ∩ B); three sets: Σ1 − Σ2 + T; exactly two = Σ2 − 3T; exactly one = Σ1 − 2Σ2 + 3T
Reflexive: all (a, a); symmetric: every pair has its reverse; transitive: no missing shortcut; equivalence = all three
Relations on n elements: 2^n² all, 2^n²−n reflexive, 2^(n(n+1)/2) symmetric, 2^(n(n−1)/2) reflexive and symmetric, 3^(n(n−1)/2) asymmetric; equivalence 1, 2, 5, 15, 52
m → n functions: n^m all, n!/(n−m)! one-one, 2^m − 2 onto a 2-set, 3^m − 3×2^m + 3 onto a 3-set, ⁿC_m strictly increasing
One-one: horizontal line cuts once at most; onto: range = codomain; bijection: inverse exists, graph is a mirror image in y = x
(g∘f)(x) = g(f(x)), f first; not commutative; (g∘f)⁻¹ = f⁻¹∘g⁻¹; f self-inverse ⇔ f(f(x)) = x
Domain: denominator ≠ 0, root ≥ 0, log > 0; range: solve for x, complete the square, use bounds; |sin x|: period π; [x]: staircase, {x} = x − [x] with period 1

Formulas with solved examples

Sets and their representation

Representation, types and equality

roster form {1, 2, 3}
set-builder form {x : x ∈ ℕ, x < 4}
∅ has no element and is a subset of every set
singleton has one element
A = B ⇔ A ⊆ B and B ⊆ A
ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ
A ⊆ A for every A
{∅} is not empty: it has one element

How to apply. Convert set-builder form to roster form by listing the values that satisfy the condition. Check every condition (integer, positive, inequality) before you count.

Example. {x : x ∈ ℤ, x² < 10} = {−3, −2, −1, 0, 1, 2, 3}, so it has 7 elements. The set {(x, y) : x² + y² = 25, x, y ∈ ℤ} has 12 elements: (±5, 0), (0, ±5), (±3, ±4), (±4, ±3).

Subsets and power set

n(A) = n ⇒ number of subsets = 2^n
proper subsets = 2^n − 1
non-empty proper subsets = 2^n − 2
n(P(A)) = 2^n
subsets with exactly k elements = ⁿC_k
subsets containing a fixed element = 2^n−1
if A ⊆ X ⊆ B, n(A) = a, n(B) = b: number of X = 2^b−a

How to apply. Each element is either in a subset or out of it: two choices per element, so 2^n. Fixed elements remove their choice.

Example. A = {1, 2}, B = {1, 2, 3, 4, 5, 6}: the sets X with A ⊆ X ⊆ B number 2^6−2 = 16. A set of 4 elements has 16 subsets, 15 proper ones and 14 non-empty proper ones.

Cartesian product and ordered pairs

n(A × B) = n(A) × n(B)
(a, b) = (c, d) ⇔ a = c and b = d
A × B ≠ B × A unless A = B (or one is empty)
A × (B ∩ C) = (A × B) ∩ (A × C)
A × (B ∪ C) = (A × B) ∪ (A × C)
n(A × A) = n²
number of subsets of A × B = 2^mn

How to apply. The order inside a pair matters. A subset of A × B is a relation from A to B, so the number of relations is the number of subsets.

Example. n(A) = 3, n(B) = 2: A × B has 6 ordered pairs and 2⁶ = 64 relations from A to B.

Intervals and modulus inequalities

(a, b) = {x : a < x < b}
[a, b] includes the ends
|x| < a ⇔ −a < x < a
|x| > a ⇔ x < −a or x > a (a > 0)
|x − c| ≤ r ⇔ c − r ≤ x ≤ c + r
ℝ − ℚ = irrational numbers

How to apply. Write the condition as a statement about x, then as an interval. Use a round bracket for < and a square bracket for ≤.

Example. |x − 2| ≤ 3 ⇔ −3 ≤ x − 2 ≤ 3 ⇔ x ∈ [−1, 5].

Operations, laws and Venn diagrams

Union, intersection, difference, complement

A ∪ B = {x : x ∈ A or x ∈ B}
A ∩ B = {x : x ∈ A and x ∈ B}
A − B = A ∩ B′
A′ = U − A
A Δ B = (A − B) ∪ (B − A)
n(A Δ B) = n(A) + n(B) − 2n(A ∩ B)
A ⊆ B ⇔ A ∪ B = B ⇔ A ∩ B = A ⇔ B′ ⊆ A′
A, B are disjoint if A ∩ B = ∅

How to apply. Rewrite every difference as A ∩ B′ and every set condition as "and", "or", "not" before you work.

Example. A = {1, 2, 3, 4}, B = {3, 4, 5}: A − B = {1, 2}, B − A = {5}, A Δ B = {1, 2, 5}, and n(A Δ B) = 4 + 3 − 2 × 2 = 3.

De Morgan’s laws

(A ∪ B)′ = A′ ∩ B′
(A ∩ B)′ = A′ ∪ B′
(A ∪ B ∪ C)′ = A′ ∩ B′ ∩ C′
A − (B ∪ C) = (A − B) ∩ (A − C)
A − (B ∩ C) = (A − B) ∪ (A − C)
A ∩ (B − C) = (A ∩ B) − (A ∩ C)

How to apply. To take a complement, change ∪ into ∩, ∩ into ∪, and complement each set. This is the usual way to count "neither".

Example. U = {1, …, 10}, A = {1, 2, 3, 4}, B = {3, 4, 5, 6}: (A ∪ B)′ = {7, 8, 9, 10} = A′ ∩ B′, and (A ∩ B)′ has 10 − 2 = 8 elements = n(A′ ∪ B′).

Laws of sets

A ∪ B = B ∪ A, A ∩ B = B ∩ A
associative
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∪ A = A, A ∩ A = A
A ∪ ∅ = A, A ∩ U = A, A ∪ U = U, A ∩ ∅ = ∅
A ∪ A′ = U, A ∩ A′ = ∅, (A′)′ = A
absorption: A ∪ (A ∩ B) = A, A ∩ (A ∪ B) = A
A ∪ (A′ ∩ B) = A ∪ B

How to apply. Use these to simplify a long expression: expand with the distributive law, then cancel with A ∪ A′ = U and A ∩ A′ = ∅.

Example. A = {1, 2}, B = {2, 3}, U = {1, 2, 3, 4}: A′ ∩ B = {3}, so A ∪ (A′ ∩ B) = {1, 2, 3} = A ∪ B.

Region names in a Venn diagram

only A = A − B = A ∩ B′
only B = B − A
both = A ∩ B
exactly one = A Δ B
neither = A′ ∩ B′ = (A ∪ B)′
inside the rectangle U every element is in exactly one region
for A ⊆ B the circle of A lies inside the circle of B

How to apply. Draw the diagram, mark each region with a letter, and write the sentence as a list of regions. Then add the numbers.

Example. "Students who play cricket but not football" is the region A − B. "Students who play neither" is the region outside both circles.

Counting pairs of subsets

S has n elements. Ordered pairs (A, B) of subsets: all pairs 4^n
A ⊆ B: 3^n
A ∩ B = ∅: 3^n
A ∪ B = S: 3^n
each element has 3 choices in each of the last three cases

How to apply. Decide how many places an element may occupy: for A ⊆ B an element is in neither, in B only, or in both (3 places).

Example. n = 3: there are 4³ = 64 pairs in all and 3³ = 27 pairs with A ⊆ B, 27 with A ∩ B = ∅ and 27 with A ∪ B = S.

Counting with sets

Two sets

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
only A = n(A) − n(A ∩ B)
exactly one of the two = n(A) + n(B) − 2n(A ∩ B)
neither = N − n(A ∪ B), N the total
n(A ∩ B) ≤ min(n(A), n(B))
n(A ∩ B) ≥ max(0, n(A) + n(B) − N)

How to apply. Write the formula first, put in the known numbers, and solve for the unknown. Never add n(A) and n(B) alone: the overlap is counted twice.

Example. In a class 40 like tea and 30 like coffee, 55 like at least one. Both: 40 + 30 − 55 = 15. Exactly one: 55 − 15 = 40.

Three sets

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C)
n(A′ ∩ B′ ∩ C′) = N − n(A ∪ B ∪ C)
seven regions in the diagram: a, b, c (one set only), x, y, z (two sets only), t (all three)

How to apply. Subtract the three pairs, then add back the triple, because it was added 3 times and subtracted 3 times.

Example. n(A) = 30, n(B) = 25, n(C) = 20, n(A ∩ B) = 10, n(B ∩ C) = 8, n(A ∩ C) = 6, n(A ∩ B ∩ C) = 3: union = 75 − 24 + 3 = 54. The regions are a = 17, b = 10, c = 9, x = 7, y = 5, z = 3, t = 3, and they add up to 54.

Exactly one, exactly two, at least two

Σ1 = n(A) + n(B) + n(C), Σ2 = sum of the three pair intersections, T = n(A ∩ B ∩ C)
exactly one = Σ1 − 2Σ2 + 3T
exactly two = Σ2 − 3T
exactly three = T
at least two = Σ2 − 2T
at least one = Σ1 − Σ2 + T

How to apply. Pair intersections include the centre region three times, so take 3T off to get "exactly two". Check your answer by adding the cases: one + two + three = union.

Example. Use the numbers above: Σ1 = 75, Σ2 = 24, T = 3. Exactly one: 75 − 48 + 9 = 36. Exactly two: 24 − 9 = 15. Exactly three: 3. At least two: 24 − 6 = 18. Check: 36 + 15 + 3 = 54.

Percentages and "neither" problems

take N = 100 when percentages are given
n(A′ ∩ B′) = N − n(A ∪ B)
n(A′ ∩ B) = n(B) − n(A ∩ B)
n(A′ ∪ B′) = N − n(A ∩ B)
"at least one" = union, "none" = complement of the union

How to apply. Translate each phrase into a region or a complement, and work with the total N.

Example. 60 % read paper A, 50 % read paper B and 20 % read both: 90 % read at least one, so 10 % read neither and 40 % read exactly one.

Greatest and least overlap

n(A) = a, n(B) = b, total N: least n(A ∩ B) = max(0, a + b − N)
greatest n(A ∩ B) = min(a, b)
three sets: least n(A ∩ B ∩ C) = max(0, a + b + c − 2N)
greatest = min(a, b, c)

How to apply. For the least overlap push the sets apart until the union is N. Use the complements: the sets that miss the sets cannot exceed N altogether.

Example. N = 100, a = 70, b = 60: n(A ∩ B) lies between 30 and 60. N = 10, sets of 7, 6, 7 elements: the triple intersection can be anything from 0 to 6.

Relations: types and counting

Relation, domain, range, inverse

a relation R from A to B is any subset of A × B
domain = first coordinates
range = second coordinates (a subset of B, the codomain)
R⁻¹ = {(b, a) : (a, b) ∈ R}
relations from A to B: 2^mn
relations on A: 2^n²

How to apply. Make the list of pairs from the rule, then read the domain and range off the list.

Example. R = {(x, y) : x + 2y = 8, x, y ∈ ℕ} = {(2, 3), (4, 2), (6, 1)}. Domain {2, 4, 6}, range {1, 2, 3}, and R⁻¹ = {(3, 2), (2, 4), (1, 6)}.

Reflexive, symmetric, transitive, antisymmetric

reflexive: (a, a) ∈ R for every a
symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R
transitive: (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R
antisymmetric: (a, b) and (b, a) ∈ R ⇒ a = b
to disprove a property, one counter-example is enough
R is symmetric ⇔ R = R⁻¹

How to apply. Test reflexive first (look for a missing (a, a)), then symmetric (find a pair without its reverse), then transitive (chain two pairs and look for the shortcut).

Example. On {1, 2, 3} let R = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}. R is reflexive and symmetric, but not transitive: (1, 2) and (2, 3) are in R and (1, 3) is not.

Equivalence relation and classes

equivalence = reflexive + symmetric + transitive
the classes of an equivalence relation split A into disjoint parts that together make A
a ∼ b ⇔ a and b are in the same class
smallest equivalence relation on A: the identity {(a, a)}
largest: A × A
x R y ⇔ x − y divisible by m on ℤ gives m classes (the residues 0, …, m − 1)

How to apply. To show a relation is an equivalence prove all three properties. To count equivalence relations count the ways to split A into groups.

Example. x R y ⇔ 3 divides x − y on ℤ has the 3 classes {…, −3, 0, 3, …}, {…, −2, 1, 4, …}, {…, −1, 2, 5, …}.

Counting relations on a set of n elements

all relations: 2^n²
reflexive: 2^n²−n
symmetric: 2^(n(n+1)/2)
reflexive and symmetric: 2^(n(n−1)/2)
antisymmetric: 2^n × 3^(n(n−1)/2)
asymmetric: 3^(n(n−1)/2)
equivalence relations = number of partitions: 1, 2, 5, 15, 52 for n = 1 … 5
transitive: no simple formula

How to apply. The n² pairs split into n diagonal pairs and n(n − 1)/2 pairs of mirror images. Reflexive fixes the diagonal; symmetric makes the mirror pairs go together.

Example. For n = 3: 512 relations, 64 reflexive, 64 symmetric, 8 reflexive and symmetric, 216 antisymmetric, 27 asymmetric and 5 equivalence relations. For n = 4 there are 15 equivalence relations, and 5 of them contain (1, 2).

Standard relations to remember

"is parallel to" (lines): equivalence
"is perpendicular to": symmetric only
"x = y": equivalence
"x ≤ y" on ℝ: reflexive, antisymmetric, transitive (not symmetric)
"x < y": transitive and antisymmetric only
"x divides y" on ℕ: reflexive, antisymmetric, transitive
"|x − y| ≤ 1" on ℤ: reflexive and symmetric, not transitive
"is similar (congruent) to" for triangles: equivalence

How to apply. Remember these ten and most "find the type" questions are done. Check the one that is new with a counter-example.

Example. |x − y| ≤ 1 on ℤ: (0, 1) and (1, 2) are in the relation, but |0 − 2| = 2 > 1, so (0, 2) is not: not transitive.

Combining relations

if R and S are equivalence relations on A, R ∩ S is an equivalence relation
R ∪ S need not be transitive
R⁻¹ of an equivalence relation is the same relation
symmetric relations on n = 3 containing a given pair (a, b) with a ≠ b: 2^5 = 32
equivalence relations on 3 elements containing (1, 2): 2

How to apply. Check the three properties of the intersection pair by pair. For the union look for a chain that leaves A.

Example. On {1, 2, 3}: R has classes {1, 2}, {3} and S has classes {1}, {2, 3}. R ∩ S is the identity, an equivalence relation. R ∪ S contains (1, 2) and (2, 3) but not (1, 3), so it is not transitive.

Functions: one-one, onto and counting

Function, domain, codomain, range

a function assigns to each x of the domain exactly one y of the codomain
range = set of all values f(x) ⊆ codomain
vertical line test: a vertical line meets the graph at most once
f = g needs the same domain, codomain and rule
number of functions from A to B = n(B)^(n(A))

How to apply. Check the two conditions: every input has an output, and no input has two. Find the range by working out the values, not by looking at the codomain.

Example. f(x) = √(9 − x²): the domain is [−3, 3] and the range is [0, 3]. From a set of 2 elements to a set of 3 elements there are 3² = 9 functions.

One-one (injective) functions

f(a) = f(b) ⇒ a = b
equivalent: a ≠ b ⇒ f(a) ≠ f(b)
a strictly increasing or strictly decreasing function is one-one
an even function (not on one point) is many-one
polynomial of even degree on ℝ is many-one
n(A) = m, n(B) = n: one-one functions = n!/(n − m)! if m ≤ n, and 0 if m > n

How to apply. Try f(a) = f(b) and solve. If a second solution a ≠ b exists, it is many-one. For a counting question place the images one after the other: n, then n − 1, …

Example. f(x) = x² on ℝ: f(2) = f(−2), many-one. From 3 elements to 5 elements: 5 × 4 × 3 = 60 one-one functions; to 4 elements: 24.

Onto (surjective) functions

onto ⇔ range = codomain
onto functions from an m-set to a 2-set: 2^m − 2
to a 3-set: 3^m − 3×2^m + 3
general: Σ(−1)^k ⁿC_k (n − k)^m
for two finite sets of equal size: one-one ⇔ onto ⇔ bijective
bijections from an n-set to itself: n!

How to apply. Onto means every element of the codomain is hit. For counting, subtract the functions that miss some element (inclusion–exclusion).

Example. From 5 elements to 2: 2⁵ − 2 = 30. From 4 elements to 3: 3⁴ − 3 × 2⁴ + 3 = 36. From 5 elements to 3: 3⁵ − 3 × 2⁵ + 3 = 150.

Counting functions with a condition

f: A → B, n(A) = m, n(B) = n
all: n^m
constant: n
strictly increasing (order kept): ⁿC_m
non-decreasing: ⁿ⁺ᵐ⁻¹C_m
functions of A to itself with f(a) = a for one fixed a: n^n−1
idempotent f(f(x)) = f(x) on 3 elements: 10

How to apply. A strictly increasing function is fixed once you choose which m values of B are used, so count the choices of m values.

Example. {1, 2, 3} → {1, …, 5}: 125 functions in all, 60 one-one, 10 strictly increasing, 35 non-decreasing. {1, 2, 3, 4} → {1, 2, 3, 4} with f(1) = 1: 4³ = 64.

The four kinds of function

one-one and onto: bijection (inverse exists)
one-one and into
many-one and onto
many-one and into
x³ on ℝ: bijection
eˣ: ℝ → ℝ is one-one and into (range (0, ∞))
x²: ℝ → ℝ is many-one and into
sin x: ℝ → ℝ is many-one and into
to make a function onto, make the codomain equal to the range

How to apply. Decide one-one by the graph (horizontal line) and onto by comparing the range with the stated codomain. A question may change the codomain only to test onto.

Example. f(x) = x²: ℝ → ℝ is neither one-one nor onto. f: [0, ∞) → [0, ∞) with the same rule is a bijection.

Horizontal line test

one-one ⇔ every line y = c meets the graph at most once
onto ⇔ every line y = c with c in the codomain meets the graph at least once
a continuous one-one function on an interval is strictly monotonic
the derivative f′(x) keeps one sign ⇒ one-one

How to apply. Draw a rough graph, then sweep a horizontal line. Count the hits for each height.

Example. f(x) = x³ − 3x meets y = 0 at x = 0, √3 and −√3, so it is many-one; it takes every real value, so it is onto ℝ.

Composition and inverse

Composition of functions

(g∘f)(x) = g(f(x))
f acts first
domain of g∘f: x in the domain of f with f(x) in the domain of g
g∘f ≠ f∘g in general
h∘(g∘f) = (h∘g)∘f
f, g one-one ⇒ g∘f one-one
f, g onto ⇒ g∘f onto
g∘f one-one ⇒ f one-one
g∘f onto ⇒ g onto

How to apply. Substitute the whole of f(x) in place of x in g. Do the inner function first, and check the domain at the end.

Example. f(x) = 2x + 1, g(x) = x²: (g∘f)(3) = 49 and (f∘g)(3) = 19. (g∘f)(x) = (2x + 1)², (f∘g)(x) = 2x² + 1, and they are equal when 2x² + 4x = 0, that is x = 0 or x = −2.

Domain of a composite

dom(g∘f) = {x ∈ dom f : f(x) ∈ dom g}
simplifying the formula may hide a restriction
f∘f: apply f twice and keep both conditions

How to apply. Write the domain of the inner function first, then add the condition that its value lies in the domain of the outer function.

Example. f(x) = √x, g(x) = x² − 1: (f∘g)(x) = √(x² − 1), domain |x| ≥ 1. (g∘f)(x) = x − 1 but its domain is x ≥ 0, not all of ℝ.

Inverse function

f⁻¹ exists ⇔ f is a bijection
f⁻¹(f(x)) = x and f(f⁻¹(y)) = y
to find f⁻¹: put y = f(x), solve for x in y, then swap
dom(f⁻¹) = range(f), range(f⁻¹) = dom(f)
(f⁻¹)⁻¹ = f
(g∘f)⁻¹ = f⁻¹∘g⁻¹
the graph of f⁻¹ is the reflection of the graph of f in the line y = x
(a, b) on f ⇔ (b, a) on f⁻¹

How to apply. Solve y = f(x) for x. Restrict the domain and codomain if the function is not a bijection. In (g∘f)⁻¹ the order reverses.

Example. f(x) = (2x + 1)/(x − 3), x ≠ 3: y(x − 3) = 2x + 1 gives x = (3y + 1)/(y − 2), so f⁻¹(y) = (3y + 1)/(y − 2). Check: f(4) = 9 and f⁻¹(9) = 28/7 = 4.

Self-inverse functions

f∘f = identity ⇔ f = f⁻¹
f(x) = (ax + b)/(cx − a) is self-inverse
f(x) = a − x, f(x) = 1/x, f(x) = k/x are self-inverse
the graph is symmetric about y = x
for an increasing f: f(x) = f⁻¹(x) ⇔ f(x) = x
self-inverse functions on 3 elements: 4

How to apply. Test f(f(x)) = x. For x = f⁻¹(x) on an increasing function solve f(x) = x, which is easier.

Example. f(x) = (x + 1)/(x − 1): f(f(x)) = x, so f = f⁻¹. On {1, 2, 3} the identity and the three swaps give 4 functions with f(f(x)) = x.

Domain, range and graphs

Finding the domain

denominator ≠ 0
even root: expression under it ≥ 0
log: argument > 0 (and base > 0, ≠ 1)
sin⁻¹ x, cos⁻¹ x: −1 ≤ x ≤ 1
tan x: x ≠ (2k + 1)π/2
[x] and |x| are defined for all x
for a sum or product take the intersection of the domains
f(x)^(g(x)): base > 0

How to apply. List each condition as a separate inequality and take the common part. Factorise the quadratic and use a sign line.

Example. f(x) = √(x² − 5x + 6): (x − 2)(x − 3) ≥ 0, so x ≤ 2 or x ≥ 3.

Finding the range

method 1: put y = f(x), solve for x, and keep the y for which x is real and in the domain
method 2: use bounds such as sin x ∈ [−1, 1], x² ≥ 0, a sin x + b cos x ∈ [−√(a² + b²), √(a² + b²)]
complete the square for a quadratic
for a one-one continuous f read the range from the two ends

How to apply. Choose the shortest method: bounds for trigonometric and square expressions, discriminant for ratios of quadratics.

Example. x² − 4x + 7 = (x − 2)² + 3 has the range [3, ∞). The range of 3 sin x + 4 cos x is [−5, 5]. x²/(1 + x²) = 1 − 1/(1 + x²) has the range [0, 1).

Range by the discriminant

y = (ax² + bx + c)/(px² + qx + r): write it as a quadratic in x, with x real so the discriminant ≥ 0
check separately the y where the coefficient of x² is 0
for y = x/(1 + x²): yx² − x + y = 0, D = 1 − 4y² ≥ 0

How to apply. The coefficient of x² may vanish at some y; solve that case separately.

Example. y = x/(1 + x²): D = 1 − 4y² ≥ 0 gives −1/2 ≤ y ≤ 1/2, and y = 0 is reached at x = 0, so the range is [−1/2, 1/2].

Transformations of graphs

y = f(x) + a: up by a
y = f(x + a): left by a
y = −f(x): flip in the x-axis
y = f(−x): flip in the y-axis
y = |f(x)|: reflect the part below the x-axis upwards
y = f(|x|): keep x ≥ 0 and reflect it in the y-axis
y = af(x): stretch in y
y = f(ax): compress in x

How to apply. Move the vertex or a key point first, then draw the shape. The sign inside the bracket moves the graph the opposite way from what it seems.

Example. y = (x − 2)² + 3 is y = x² shifted 2 right and 3 up, with the vertex at (2, 3). y = |x − 1| has its vertex at (1, 0).

Standard graphs and their ranges

x²: range [0, ∞), even, many-one
x³: ℝ, odd, one-one
1/x: ℝ − {0}, odd, one-one
√x: [0, ∞), one-one
|x|: [0, ∞), even
eˣ: (0, ∞), one-one
ln x: ℝ, domain (0, ∞), one-one
sin x: [−1, 1], odd, periodic
sgn x: {−1, 0, 1}

How to apply. Learn the shapes with their symmetry and range; most questions ask for one of these properties.

Example. ln x and eˣ are inverses: the domain (0, ∞) of ln x is the range of eˣ, and the range ℝ of ln x is the domain of eˣ.

Even, odd, periodic, [x] and ｛x｝

Even and odd functions

even: f(−x) = f(x), the graph is symmetric about the y-axis
odd: f(−x) = −f(x), the graph is symmetric about the origin
the domain must be symmetric about 0
f(0) = 0 for an odd function defined at 0
even × even and odd × odd are even; even × odd is odd
every f = (f(x) + f(−x))/2 (even) + (f(x) − f(−x))/2 (odd)
the only function that is both is f = 0

How to apply. Replace x by −x and compare with f(x) and −f(x). For a sum of terms test each term.

Example. f(x) = x³ + x²: f(2) = 12 and f(−2) = −4, neither even nor odd; the even part is x² and the odd part is x³. f(x) = ln(x + √(x² + 1)) is odd, since f(x) + f(−x) = ln 1 = 0.

Periodic functions

f(x + T) = f(x), least positive T is the period
sin(ax + b), cos(ax + b), sec, cosec: 2π/|a|
tan(ax + b), cot(ax + b): π/|a|
|sin x|, |cos x|, sin²x, cos²x, |tan x|: π
period of f(x) + g(x) is the LCM of the periods (when it exists)
LCM of a/b and c/d = LCM(a, c)/HCF(b, d)
if f has the period T, f(ax + b) has the period T/|a|

How to apply. Find the period of each term, then the least common multiple. If the ratio of the periods is irrational the sum is not periodic.

Example. sin(x/2) + cos(x/3): periods 4π and 6π, so 12π. sin(πx) + cos(πx/2): periods 2 and 4, so 4. tan 3x has the period π/3.

Special periods

|sin x| + |cos x|: π/2
sin⁴x + cos⁴x = 1 − ½sin²2x: π/2
{x} = x − [x]: 1
constant function: every real number is a period, no least period
[x], x, x² are not periodic
f(x + a) = f(x − a) ⇒ period 2a

How to apply. Rewrite the expression with a double angle first when powers appear. Then use the standard rule.

Example. sin⁴x + cos⁴x repeats after π/2: for example the value at x = 0 is 1 and the value at x = π/2 is also 1, and sin²2x has the period π/2.

Greatest integer function [x]

[x] = n for n ≤ x < n + 1 (n an integer)
[x] ≤ x < [x] + 1
[x + n] = [x] + n for an integer n
[x] + [−x] = 0 if x is an integer, −1 otherwise
domain ℝ, range ℤ
the graph is a staircase (left end closed, right end open)
many-one, not onto ℝ, not periodic
[−2.3] = −3, [2.9] = 2

How to apply. Reduce an equation in [x] to an equation for the integer value n, then turn n ≤ x < n + 1 into an interval.

Example. [x]² − 5[x] + 6 = 0 ⇒ [x] = 2 or 3 ⇒ 2 ≤ x < 4. The range of [sin x] is {−1, 0, 1}: −1 for sin x < 0, 0 for 0 ≤ sin x < 1 and 1 at sin x = 1.

Fractional part ｛x｝

{x} = x − [x]
0 ≤ {x} < 1
range [0, 1)
period 1
{x} = 0 ⇔ x is an integer
{x} + {−x} = 1 if x is not an integer, 0 if it is
{x + n} = {x} for an integer n
[x] + {x} = x

How to apply. Write x as its integer part plus the fraction. For negative x do not drop the sign: {−1.3} = −1.3 − (−2) = 0.7.

Example. {−1.3} = 0.7 and {1.3} = 0.3, so {x} + {−x} = 1. {x} is not even or odd and it repeats with period 1.

Modulus, signum and combinations

|x| = x if x ≥ 0, −x if x < 0
|x| = √(x²)
|a + b| ≤ |a| + |b|
|a b| = |a||b|
sgn x = 1, 0, −1 for x > 0, = 0, < 0
f(x) = [x] + [−x] is an even function
[x] − x = −{x}
|x| is even, sgn x is odd

How to apply. To handle |x| split the line at the point where the expression inside is zero and write each part without the modulus.

Example. f(x) = [x] + [−x] gives f(−x) = [−x] + [x] = f(x), so f is even; it takes the values 0 and −1 only.

Most asked question types

How do I solve “Number of elements, subsets and power set” questions in JEE Main?

A set is given in roster form or by a rule. Find the number of its elements, subsets, proper subsets or sets between two given sets. 1. List the elements that satisfy the rule and count them. 2. A set of n elements has 2^n subsets, 2^n − 1 proper subsets. 3. For A ⊆ X ⊆ B the number is 2^(n(B) − n(A)). Example: {x ∈ ℤ : x² < 10} has 7 elements and 128 subsets.

How do I solve “Set operations and De Morgan’s laws” questions in JEE Main?

Two or three sets are given, or a set identity. Find A − B, A Δ B, a complement, or decide which identity is true. 1. Write A − B as A ∩ B′. 2. Use De Morgan to take a complement. 3. Check an identity with a small example or a Venn diagram. Example: A = {1, 2, 3, 4}, B = {3, 4, 5}: A Δ B = {1, 2, 5}.

How do I solve “Two-set counting (tea, coffee, cricket, football)” questions in JEE Main?

A group likes two things. Some numbers are given. Find those who like both, only one, or neither. 1. Write n(A ∪ B) = n(A) + n(B) − n(A ∩ B). 2. Find the unknown from the formula. 3. Neither = total − union. Example: 40, 30 and 55 in the union: 15 like both.

How do I solve “Three-set counting with exactly one, exactly two” questions in JEE Main?

Three sets with pair and triple intersections are given. Find the union, the number in exactly one or exactly two sets, or the number in none. 1. Union = Σ1 − Σ2 + T. 2. Exactly two = Σ2 − 3T. 3. Exactly one = Σ1 − 2Σ2 + 3T. Example: 30, 25, 20; 10, 8, 6; 3: union 54, exactly one 36.

How do I solve “Greatest and least value of an overlap” questions in JEE Main?

Sizes of two or three sets in a population are given. Find the least or the greatest number of elements in the intersection. 1. Least n(A ∩ B) = max(0, a + b − N). 2. Greatest = min(a, b). 3. Three sets: least triple = max(0, a + b + c − 2N). Example: N = 100, a = 70, b = 60: between 30 and 60.

How do I solve “Pairs of subsets and special subsets” questions in JEE Main?

S has n elements. Count ordered pairs (A, B) with A ⊆ B, A ∩ B = ∅ or A ∪ B = S, or subsets with a stated property. 1. Give each element its possible places. 2. Multiply over the n elements. 3. Fixed elements are not counted again. Example: n = 3, A ⊆ B: 3³ = 27 pairs.

How do I solve “Cartesian product and number of relations” questions in JEE Main?

Sizes of A and B are given. Find n(A × B), the number of relations, or a statement on products. 1. n(A × B) = mn. 2. Relations from A to B: 2^mn. 3. Use A × (B ∩ C) = (A × B) ∩ (A × C). Example: n(A) = 3, n(B) = 2: 6 pairs and 64 relations.

How do I solve “Domain and range of a relation” questions in JEE Main?

A relation is given by a rule on ℕ or on a small set. Write it as pairs, find the domain and range, or the inverse. 1. List the pairs. 2. Domain = first coordinates, range = second. 3. Inverse: swap each pair. Example: x + 2y = 8 on ℕ: domain {2, 4, 6}, range {1, 2, 3}.

How do I solve “Type of a given relation” questions in JEE Main?

A relation is given on a set of numbers or lines. Decide which of reflexive, symmetric, transitive, antisymmetric it is. 1. Test reflexive: is every (a, a) present? 2. Test symmetric: has every pair its reverse? 3. Test transitive: chain two pairs. Example: |x − y| ≤ 1 on ℤ: reflexive, symmetric, not transitive.

How do I solve “Proving an equivalence relation and finding the classes” questions in JEE Main?

A rule such as x − y divisible by m, or "same remainder", is given. Prove it is an equivalence relation or find the class of an element. 1. Show reflexive, symmetric and transitive. 2. The class of a is the set of all elements related to a. 3. The classes are disjoint and cover the set. Example: x − y divisible by 3 on ℤ: 3 classes.

How do I solve “Number of reflexive, symmetric or equivalence relations” questions in JEE Main?

A set with n elements is given. Count the relations that are reflexive, symmetric, reflexive and symmetric, antisymmetric or equivalence. 1. Split the n² pairs into n diagonal and n(n − 1)/2 mirror pairs. 2. Apply the formula for the type. 3. Equivalence: count the partitions. Example: n = 3: reflexive 64, symmetric 64, equivalence 5.

How do I solve “Combining relations” questions in JEE Main?

Two relations are given. Decide whether the union, intersection or inverse of equivalence relations is an equivalence relation. 1. Intersection of equivalence relations is one. 2. Union may fail transitivity: look for a chain. 3. The inverse of an equivalence relation is itself. Example: Classes {1,2},{3} and {1},{2,3}: the union is not transitive.

How do I solve “Is it a function, and its domain and range” questions in JEE Main?

A rule or a set of pairs is given. Decide if it is a function and find the domain and the range. 1. Each input needs exactly one output. 2. Domain: conditions on the rule. 3. Range: solve for x or use bounds. Example: √(9 − x²): domain [−3, 3], range [0, 3].

How do I solve “One-one and onto: type of a given function” questions in JEE Main?

A function f: A → B is given by a rule. Decide whether it is one-one, onto, both or neither. 1. One-one: f(a) = f(b) ⇒ a = b, or the horizontal line test. 2. Onto: range = codomain. 3. Change the codomain to make it onto. Example: x² on ℝ is neither; on [0, ∞) → [0, ∞) it is both.

How do I solve “Number of functions of a given kind” questions in JEE Main?

Two finite sets are given. Count all functions, one-one, onto, strictly increasing or functions with a stated condition. 1. All: n^m; one-one: n!/(n − m)!. 2. Onto: subtract the functions that miss an element. 3. Increasing: choose m values: ⁿC_m. Example: {1..5} → {1,2,3}: 243 in all, 150 onto.

How do I solve “Composition of two functions” questions in JEE Main?

Two rules f and g are given. Find (g∘f)(x), a value, the points where g∘f = f∘g, or its domain. 1. Put f(x) in place of x in g. 2. Simplify and keep the domain condition. 3. Solve g∘f = f∘g as an equation. Example: f = 2x + 1, g = x²: g∘f = f∘g at x = 0, −2.

How do I solve “Inverse of a function” questions in JEE Main?

A bijection is given. Find f⁻¹ or a value of it, or a relation between f⁻¹ and f. 1. Put y = f(x). 2. Solve for x in terms of y. 3. Swap x and y and state the domain. Example: f = (2x + 1)/(x − 3): f⁻¹(y) = (3y + 1)/(y − 2).

How do I solve “Domain of a function” questions in JEE Main?

A rule with a root, a fraction, a logarithm or an inverse trigonometric function is given. Find the domain. 1. List each restriction. 2. Solve each inequality. 3. Take the common part. Example: √(x² − 5x + 6): x ≤ 2 or x ≥ 3.

How do I solve “Range of a function” questions in JEE Main?

A rational, quadratic or trigonometric expression is given. Find the range. 1. Complete the square or use a known bound. 2. For a ratio solve for x and use D ≥ 0. 3. Check the end points and any excluded y. Example: 3 sin x + 4 cos x: [−5, 5].

How do I solve “Even, odd and periodic functions” questions in JEE Main?

A function is given. Say whether it is even, odd or neither, or find its period. 1. Compare f(−x) with f(x) and −f(x). 2. Find the period of each term. 3. Take the LCM for a sum. Example: sin(x/2) + cos(x/3): period 12π.

How do I solve “Greatest integer and fractional part” questions in JEE Main?

An equation, a graph or a range with [x] or {x} is given. Solve it, find the range or compare the values. 1. Put [x] = n and solve for the integer n. 2. Turn n ≤ x < n + 1 into an interval. 3. Use {x} = x − [x] ∈ [0, 1). Example: [x]² − 5[x] + 6 = 0: 2 ≤ x < 4.

Common myths and the truth

Myth: ∅ and {∅} are the same. Truth: ∅ has no element; {∅} has one element, namely ∅.
Myth: n(A ∪ B) = n(A) + n(B). Truth: this holds only for disjoint sets; in general subtract n(A ∩ B).
Myth: A relation that is symmetric and transitive is always reflexive. Truth: if some element has no pair it is not reflexive. On {1, 2}, R = {(1, 1)} is symmetric and transitive but misses (2, 2).
Myth: A − B and B − A are equal. Truth: they are disjoint parts of A Δ B and equal only when A = B.
Myth: Every function with the same rule is the same function. Truth: the domain and codomain count too; x² on ℝ and x² on [0, ∞) are different functions.
Myth: A one-one function is onto. Truth: that holds only for finite sets of equal size. eˣ: ℝ → ℝ is one-one and not onto.
Myth: g∘f = f∘g. Truth: composition is not commutative. With f = 2x + 1 and g = x², the two composites differ.
Myth: [−2.3] = −2. Truth: [x] is the greatest integer not above x, so [−2.3] = −3.
Myth: The period of a sum is always the product of the periods. Truth: it is the least common multiple, and only if one exists.
Myth: The number of relations on a 3-set is 3² = 9. Truth: a relation is any subset of the 9 pairs, so there are 2⁹ = 512.

Mistakes to avoid

The empty set is a subset of every set, but {∅} has one element.
n(A ∪ B) needs the overlap taken off; for three sets add the triple back.
Exactly two = Σ2 − 3T, not Σ2 − T.
A relation can be symmetric and transitive without being reflexive when an element has no pair.
One missing (a, a) is enough to make a relation not reflexive.
Check the codomain before you say a function is onto.
(g∘f)⁻¹ = f⁻¹∘g⁻¹: the order reverses.
[−2.3] = −3 and {−2.3} = 0.7: the integer part of a negative number goes down.

More: all JEE Main chapters.