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JEE Mains Maths · Formula sheet

Relations and Functions formulas

21 formulas and 21 common traps for JEE Mains Maths Relations and Functions, grouped by subtopic.

Full notes with worked examples

Sets and Counting Elements

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Sets given by conditions, then combined

Difference of sets

A−B=A∩B′A-B=A\cap B'

Counting overlapping groups by inclusion–exclusion

Three sets

n(A∪B∪C)=∑n(A)−∑n(A∩B)+n(A∩B∩C)n(A\cup B\cup C)=\textstyle\sum n(A)-\sum n(A\cap B)+n(A\cap B\cap C)

Counting subsets

Subsets meeting a k-element part

2n−2n−k2^n-2^{n-k}

Common traps

Strict or not at the ends

∣x∣<2|x|<2 leaves out ±2\pm2; ∣x∣≤2|x|\le2 keeps them. An option that differs from the truth only at an endpoint is usually the wrong one.

The least overlap is not zero

Two groups inside a fixed total must overlap by at least n(A)+n(B)−Tn(A)+n(B)-T. A bound of zero ignores the total.

Only the shared elements count

A subset of AA meets BB only through A∩BA\cap B. Elements of BB outside AA do not change the count.

Reflexive, Symmetric, Transitive and Equivalence Relations

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Testing reflexive, symmetric and transitive

Transitivity

aRb and bRc ⇒ aRcaRb\ \text{and}\ bRc\ \Rightarrow\ aRc

Equivalence relations and their classes

The usual shape

aRb  ⟺  g(a)=g(b)aRb\iff g(a)=g(b)

Common traps

A chain back to the start

Transitivity applies to aRbaRb and bRabRa too: together they force aRaaRa. A symmetric relation missing (a,a)(a,a) for such an aa is not transitive.

Reflexive and symmetric is not enough

Many relations here are reflexive and symmetric but fail transitivity, for example ∣a−b∣≤1|a-b|\le1. Test a chain that crosses the limit before calling it an equivalence.

Counting Relations and Their Elements

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Counting the pairs in a relation

Pairs in a relation

n(R)=∑a#{b:aRb}n(R)=\sum_{a}\#\{b: aRb\}

Fewest pairs to add: reflexive, symmetric, equivalence

Smallest equivalence containing R

∑iki2 pairs, ki=class sizes\sum_i k_i^2\ \text{pairs, } k_i=\text{class sizes}

Counting relations of a given type

Symmetric relations

2n(n+1)/22^{n(n+1)/2}

Common traps

Count both orders

(1,2)(1,2) and (2,1)(2,1) are two elements of a relation. A count done over unordered pairs must be doubled, except for pairs with equal entries.

Symmetry then transitivity brings the diagonal

Once (1,2)(1,2) and (2,1)(2,1) are both present, transitivity forces (1,1)(1,1) and (2,2)(2,2). A count for 'symmetric and transitive' must include them.

Equivalences are not a power of 2

Equivalence relations correspond to partitions, not to free yes/no choices. Count the ways to split the set into classes.

Domain of a Function

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The conditions that fix a domain

Inverse-sine domain

sin⁡−1u, cos⁡−1u: −1≤u≤1\sin^{-1}u,\ \cos^{-1}u:\ -1\le u\le1

Nested logarithms and greatest-integer parts

Two nested logarithms

log⁡a(log⁡bu) defined  ⟺  u>1(a,b>1)\log_a(\log_b u)\ \text{defined}\iff u>1\quad(a,b>1)

Domain of a composite function

Domain of f∘g

{x∈dom⁡g: g(x)∈dom⁡f}\{x\in\operatorname{dom}g:\ g(x)\in\operatorname{dom}f\}

Common traps

A logarithm in a denominator

1log⁡u\frac1{\log u} needs log⁡u≠0\log u\ne0 as well as u>0u>0: the argument u=1u=1 is excluded. This is the usual missing point in a domain written as 'an interval minus a point'.

Turn integers back into intervals

[x]≤−3[x]\le-3 means x<−2x<-2, not x≤−3x\le-3: every xx with integer part −3-3 lies in [−3,−2)[-3,-2).

The inner function's gaps stay

If gg is undefined at a point, so is f∘gf\circ g, whatever the simplified formula says.

Range, One-One and Onto

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Finding the range

Rational function

y=p(x)q(x) ⇒ Δx≥0y=\frac{p(x)}{q(x)}\ \Rightarrow\ \Delta_x\ge0

Deciding one-one and onto

The two tests

one-one: f(a)=f(b)⇒a=b;onto: f(A)=B\text{one-one: } f(a)=f(b)\Rightarrow a=b;\qquad \text{onto: } f(A)=B

Common traps

When the x² term vanishes

In the discriminant method, the value of yy that makes the x2x^2 coefficient zero turns the equation linear. Check it separately: it may or may not be in the range.

Read the codomain

The same formula can be onto one codomain and not another. A range of (−1,1)(-1,1) is not onto (−∞,1)(-\infty,1), even though every value is below 1.

Counting Functions

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Counting all functions and one-one functions

One-one functions

n!(n−m)!=n(n−1)⋯(n−m+1)\frac{n!}{(n-m)!}=n(n-1)\cdots(n-m+1)

Counting onto functions

Onto functions

nm−(n1)(n−1)m+(n2)(n−2)m−⋯n^m-\binom n1(n-1)^m+\binom n2(n-2)^m-\cdots

Common traps

More inputs than outputs

A one-one function needs at least as many outputs as inputs. From a larger set to a smaller one there are none.

Add back the double-counted

Subtracting 'misses aa' and 'misses bb' removes the functions that miss both twice. The +(n2)(n−2)m+\binom n2(n-2)^m term puts them back.

Composition, Inverse and Iterates

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Composing functions, and recovering one from the other

Composition

(f∘g)(x)=f(g(x))(f\circ g)(x)=f\big(g(x)\big)

Inverse functions and self-inverse maps

Self-inverse Möbius map

f(x)=ax+bcx+d,f∘f=id  ⟺  a+d=0f(x)=\frac{ax+b}{cx+d},\quad f\circ f=\mathrm{id}\iff a+d=0

Applying a function many times

A cycle

fk=id ⇒ fn=f n mod kf^k=\mathrm{id}\ \Rightarrow\ f^{n}=f^{\,n \bmod k}

Common traps

The inner function acts first

f∘gf\circ g means gg first. Reading it left to right gives g∘fg\circ f, a different function.

Only for increasing functions

A decreasing function can meet its inverse off the line y=xy=x. The shortcut f(x)=xf(x)=x is safe only when ff is increasing.

Find the cycle on a general x

A value that repeats at one particular xx proves nothing about the others. Show fk(x)=xf^k(x)=x as an identity before reducing nn.

Functional Equations

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Additive and multiplicative rules

Additive rule

f(x+y)=f(x)+f(y) ⇒ f(n)=n f(1)f(x+y)=f(x)+f(y)\ \Rightarrow\ f(n)=n\,f(1)

Substitute again, then solve the pair

Two equations, two unknowns

af(x)+bf(g(x))=h(x),af(g(x))+bf(x)=h(g(x))af(x)+bf\big(g(x)\big)=h(x),\quad af\big(g(x)\big)+bf(x)=h\big(g(x)\big)

Pairing terms when f(x) + f(a − x) is constant

The standard pair

f(x)=bxbx+b ⇒ f(x)+f(1−x)=1f(x)=\frac{b^x}{b^x+\sqrt b}\ \Rightarrow\ f(x)+f(1-x)=1

Common traps

A constant term changes f(0)

In f(x+y)=f(x)+f(y)+cf(x+y)=f(x)+f(y)+c, putting x=y=0x=y=0 gives f(0)=−cf(0)=-c, not 0. Find it before using any pattern.

A sum may need no solving

If the question asks for f(a)+f(b)f(a)+f(b) where bb is aa's partner, putting x=ax=a and x=bx=b and adding the two equations can give it directly, without finding ff.

Count the terms

An odd number of terms leaves a middle term unpaired. Add it separately; forgetting it is the usual gap between two options.

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