Math · Sets, Relations and Functions

Relations and Functions revision notes

A concise JEE revision summary of Relations and Functions.

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Key Concepts & Definitions

Relations

Relation:
A mathematical connection or link between two objects. A relation RRR from a set AAA to a set BBB is defined mathematically as an arbitrary subset of the Cartesian product A×BA \times BA×B. If (a,b)∈R(a, b) \in R(a,b)∈R, we say that aaa is related to bbb under the relation RRR, written as aRba R baRb.
Empty Relation:
A relation RRR in a set AAA where no element of AAA is related to any element of AAA, i.e., R=ϕ⊂A×AR = \phi \subset A \times AR=ϕ⊂A×A.
Universal Relation:
A relation RRR in a set AAA where each element of AAA is related to every element of AAA, i.e., R=A×AR = A \times AR=A×A.
Trivial Relations:
Both the empty relation and the universal relation are sometimes called trivial relations.
Identity Relation:
A relation where every element is related only to itself. I={(a,a):a∈A}I = \{(a,a) : a \in A\}I={(a,a):a∈A}.

Functions

  • Function: Functions are a special kind of relation. The concept evolved historically: Descartes first used the term to denote powers of xx; Leibnitz introduced the phrase "function of xx"; Euler introduced the standard notation f(x)f(x), F(x)F(x), etc.; and Dirichlet provided the rigorous foundation for the modern set-theoretic definition abstraction.
  • Domain, Co-domain, and Range: The set of all valid inputs is the domain, the set in which the outputs fall is the co-domain, and the set of actual outputs is the range.
  • Equal Functions: Two functions f:ABf: A \to B and g:ABg: A \to B are called equal functions if their domains and codomains are identical, and f(a)=g(a)f(a) = g(a) for all aAa \in A.

Types of Relations

To study equivalence, we classify relations in a set AA into three specific types:

  • Reflexive Relation: (a,a)R(a, a) \in R for every aAa \in A.JEE TIPIf even a single element aAa \in A does not have (a,a)R(a,a) \in R, the relation is NOT reflexive.
  • Symmetric Relation: (a,b)R(a, b) \in R implies that (b,a)R(b, a) \in R for all a,bAa, b \in A.
  • Transitive Relation: (a,b)R(a, b) \in R and (b,c)R(b, c) \in R implies that (a,c)R(a, c) \in R for all a,b,cAa, b, c \in A.JEE TIPTransitivity ONLY fails if (a,b)(a,b) and (b,c)(b,c) exist, but (a,c)(a,c) does NOT. If (a,b)(a,b) exists but there is no (b,c)(b,c) pairing to check, the relation is vacuously transitive!

Equivalence Relation & Equivalence Classes

  • Equivalence Relation: A relation RR in a set AA is an equivalence relation if it is reflexive, symmetric, and transitive.
  • Equivalence Classes (Partitions): Given an arbitrary equivalence relation RR in an arbitrary set XX, RR divides XX into mutually disjoint subsets AiA_i called partitions or subdivisions.
    • All elements of AiA_i are related to each other.
    • No element of AiA_i is related to any element of AjA_j (iji \neq j).
    • The union of all AiA_i is XX, and their intersection is empty (ϕ\phi).
    • The subset AiA_i containing an element aa is called an equivalence class and is denoted by [a][a].

Types of Functions

  • One-One (Injective) Function: A function f:XYf: X \to Y is one-one if the images of distinct elements of XX are distinct. Mathematically: for every x1,x2Xx_1, x_2 \in X, f(x1)=f(x2)f(x_1) = f(x_2) implies x1=x2x_1 = x_2. If it is not one-one, it is called many-one.
  • Onto (Surjective) Function: A function f:XYf: X \to Y is onto if every element of YY is the image of some element of XX under ff. This means for every yYy \in Y, there exists an xXx \in X such that f(x)=yf(x) = y.
    • Rule: f:XYf: X \to Y is onto if and only if Range of f=Yf = Y.
  • Bijective Function: A function f:XYf: X \to Y is bijective if it is both one-one and onto.

Finite vs. Infinite Sets Property

  • For an arbitrary finite set XX, a one-one function f:XXf: X \to X is necessarily onto, and an onto map f:XXf: X \to X is necessarily one-one.
  • JEE TIPThis is a characteristic difference between a finite and an infinite set. For an infinite set (like NN or RR), a function can be one-one but not onto (e.g., f(x)=2xf(x) = 2x on NN), or onto but not one-one.

Composition of Functions & Invertibility

  • Composition of Functions (gofgof): Let f:ABf: A \to B and g:BCg: B \to C be two functions. The composition of ff and gg, denoted by gofgof, is the function gof:ACgof: A \to C given by gof(x)=g(f(x)),xAgof(x) = g(f(x)), \forall x \in A.JEE TIPgoffoggof \neq fog in general. Matrix multiplication and function composition are highly analogous (both are associative but non-commutative).
  • Invertible Function: A function f:XYf: X \to Y is invertible if there exists a function g:YXg: Y \to X such that gof=IXgof = I_X and fog=IYfog = I_Y. Here, IXI_X and IYI_Y are identity functions.
  • Inverse: The function gg is called the inverse of ff and is denoted by f1f^{-1}.
  • Condition for Invertibility: If ff is invertible, then ff must be one-one and onto (bijective). Conversely, if ff is one-one and onto, then ff must be invertible.JEE TIPTo prove a function is invertible without finding its inverse, simply prove it is both one-one and onto.

Advanced Function Properties

  • Even and Odd Functions:
    • Even: f(x)=f(x)f(-x) = f(x). Graph is symmetric about the y-axis.
    • Odd: f(x)=f(x)f(-x) = -f(x). Graph is symmetric about the origin.
    • Every function can be uniquely expressed as the sum of an even and an odd function: f(x)=f(x)+f(x)2+f(x)f(x)2f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}.
  • Periodic Functions: A function f(x)f(x) is periodic if there exists a positive real number TT such that f(x+T)=f(x)f(x+T) = f(x) for all xx in the domain. The smallest such TT is the fundamental period.

Combinatorics of Relations & Functions

Let set AA have mm elements and set BB have nn elements.

  • Total number of relations from AA to BB is 2mn2^{mn}.
  • Total number of functions from AA to BB is nmn^m.
  • Total number of injective (one-one) functions from AA to BB is nPm^nP_m (where nmn \ge m). If n<mn < m, it is 00.
  • Total number of bijective functions from AA to AA is m!m! (this matches the permutation on symbols principle).
  • Total number of reflexive relations on set AA is 2m(m1)2^{m(m-1)}.
  • Total number of symmetric relations on set AA is 2m(m+1)/22^{m(m+1)/2}.
  • Specific Summations: The number of relations containing (1,2)(1,2) and (2,3)(2,3) which are reflexive and transitive but NOT symmetric on set {1,2,3}\{1, 2, 3\} is exactly three. The number of equivalence relations containing (1,2)(1, 2) and (2,1)(2, 1) on set {1,2,3}\{1, 2, 3\} is exactly two.

Formulae, Equations & Units

ConceptMathematical Condition/FormulaLimitations/Conditions
Reflexivity(a,a)R,aA(a, a) \in R, \forall a \in AMust hold for EVERY element in AA.
Symmetry(a,b)R    (b,a)R(a, b) \in R \implies (b, a) \in RMust hold for all pairs.
Transitivity(a,b)R(b,c)R    (a,c)R(a, b) \in R \land (b, c) \in R \implies (a, c) \in RVacuously true if (b,c)(b, c) doesn't exist.
Injectivity (1-1)f(x1)=f(x2)    x1=x2f(x_1) = f(x_2) \implies x_1 = x_2Can also be checked via monotonic derivatives.
Surjectivity (Onto)yY,xX\forall y \in Y, \exists x \in X st. f(x)=yf(x)=yTrue iff Range = Codomain.
Compositiongof(x)=g(f(x))gof(x) = g(f(x))Range of ff must be a subset of Domain of gg.
Invertibilitygof=IXfog=IY    g=f1gof = I_X \land fog = I_Y \implies g = f^{-1}Strictly requires ff to be bijective.

Conditions & Limitations

  • Finite versus Infinite Sets: The theorem "f:XXf: X \to X is one-one if and only if it is onto" strictly applies ONLY to finite sets. Do not apply this to sets like R,Z\mathbb{R}, \mathbb{Z}, or N\mathbb{N}.
  • Addition of Functions vs Co-Domain Limits: Let INI_N be the identity function on N\mathbb{N}. INI_N is onto. However, IN+INI_N + I_N (which evaluates to 2x2x) is NOT onto because odd numbers in the co-domain N\mathbb{N} lose their pre-image (e.g. 2x=32x = 3 has no solution in N\mathbb{N}).
  • Greatest Integer Function limitations: f(x)=[x]f(x) = [x] on RR\mathbb{R} \to \mathbb{R} is neither one-one nor onto.
  • Modulus & Signum Function limitations: Modulus function f(x)=xf(x) = \|x\| and Signum function f(x)=sgn(x)f(x) = \text{sgn}(x) defined on RR\mathbb{R} \to \mathbb{R} are strongly many-one and into (not onto), as they collapse multiple real values to single outputs and have heavily restricted ranges.
  • Domain restrictions in Inverse: Whenever solving f(x)=yf(x) = y to find the inverse g(y)g(y), you must check if the derived xx falls strictly within the defined domain of ff.

⚠️ COMMON MISCONCEPTIONS & SIGN CONVENTIONS

  • The Vacuous Truth of Transitivity: A relation where we have (1,2)(1,2) but no pairs starting with 22 is still transitive. You cannot say transitivity fails unless you explicitly find a broken chain (i.e., you have (a,b)(a,b) and (b,c)(b,c) but lack (a,c)(a,c)).
  • Intersection of Equivalence Relations: If R1R_1 and R2R_2 are equivalence relations in a set AA, then their intersection R1R2R_1 \cap R_2 is ALWAYS an equivalence relation.JEE TIPHowever, their union R1R2R_1 \cup R_2 is NOT necessarily an equivalence relation (it may fail transitivity).
  • Function Composition Domain Traps: gof(x)gof(x) is only defined if the range of f(x)f(x) is a subset of the domain of g(x)g(x). Blindly substituting algebraic formulas without checking this restriction leads to wrong domain answers.
  • Non-Commutative Composition: gofgof and fogfog are generally not equal (goffoggof \neq fog). For instance, if f(x)=cosxf(x) = \cos x and g(x)=3x2g(x) = 3x^2, gof(x)=3cos2xgof(x) = 3\cos^2x while fog(x)=cos(3x2)fog(x) = \cos(3x^2).

Previous Year JEE Topics

  • Checking Injectivity using Calculus: Differentiating f(x)f(x) to check if f(x)>0f'(x) > 0 or f(x)<0f'(x) < 0 continuously across the domain.
  • Number of Possible Functions/Relations: Combinatorics problems linking P&C with relations (e.g., number of reflexive and symmetric relations).
  • Equivalence Relations: Matrix/Coordinate-based equivalence conditions. Example: (x,y)R(u,v)    xv=yu(x, y) R (u, v) \iff xv = yu.
  • Finding the Inverse: Solving algebraically for x=f1(y)x = f^{-1}(y), and confirming the correct branch of a quadratic/trigonometric function based on domain boundaries.
  • Functional Equations: Guessing the function profile based on relationships like f(x+y)=f(x)f(y)f(x+y) = f(x)f(y) (implying exponential functions).

Memory Aids & JEE Traps

  • JEE TIPWhen checking onto functions, students look at the formula and forget to check the declared co-domain. f(x)=2xf(x) = 2x is onto if f:RRf: \mathbb{R} \to \mathbb{R}, but it is NOT onto if f:NNf: \mathbb{N} \to \mathbb{N} because there's no pre-image for odd numbers like 1.
  • JEE TIPf1(x)f^{-1}(x) means the inverse function, NOT 1f(x)\frac{1}{f(x)}. The inverse undoes the operation; the reciprocal divides it.
  • JEE TIPAn even function f(x)=f(x)f(x) = f(-x) maps two inputs to the same output. Therefore, it is many-one, and its global inverse never exists unless the domain is restricted to x0x \ge 0 or x0x \le 0.

Top 10 JEE MCQ Traps (Misconception → Correct Understanding)

  1. Misconception: If a relation is not symmetric, it must be anti-symmetric. → Correct Understanding: Symmetry and anti-symmetry are not polar opposites. A relation can easily be neither symmetric nor anti-symmetric (e.g., if it contains (1,2)(1,2) and (2,1)(2,1) but also contains (1,3)(1,3) without (3,1)(3,1)).
  2. Misconception: Transitivity fails if we only have the pair (a,b)(a,b) and nothing else. → Correct Understanding: Transitivity ONLY fails if we have (a,b)(a,b) AND (b,c)(b,c) but are missing (a,c)(a,c). If there is no second pair starting with bb, the relation is vacuously transitive.
  3. Misconception: Any function f:ABf: A \to B is onto if the cardinalities A=B|A| = |B|. → Correct Understanding: This is only true if AA and BB are FINITE sets AND the function is one-one. For infinite sets, A=B|A| = |B| does not guarantee an injective function is surjective.
  4. Misconception: The empty relation is an equivalence relation on an empty set. → Correct Understanding: The empty relation on a non-empty set is symmetric and transitive (vacuously) but it is NEVER reflexive because (a,a)(a,a) is missing for elements in the set.
  5. Misconception: gof(x)gof(x) and fog(x)fog(x) always share the same domain limits. → Correct Understanding: The domain of gof(x)gof(x) is strictly constrained by the domain of f(x)f(x), whereas the domain of fog(x)fog(x) is constrained by the domain of g(x)g(x). They are completely independent.
  6. Misconception: If f(x)f(x) is an even function, its inverse can be found via algebra. → Correct Understanding: Even functions are globally many-one (since f(x)=f(x)f(x) = f(-x)), meaning they are not bijective. Their global inverse DOES NOT exist.
  7. Misconception: f(x)=x2f(x) = x^2 is a strictly non-invertible function. → Correct Understanding: Invertibility strictly depends on the domain and codomain. While f:RRf: \mathbb{R} \to \mathbb{R} is neither 1-1 nor onto, f:[0,)[0,)f: [0, \infty) \to [0, \infty) is bijective and fully invertible.
  8. Misconception: Composition of functions is commutative. → Correct Understanding: Function composition is associative but generally NON-commutative (foggoffog \neq gof).
  9. Misconception: To find the inverse of f(x)f(x), just compute (f(x))1(f(x))^{-1} or 1f(x)\frac{1}{f(x)}. → Correct Understanding: The inverse f1(x)f^{-1}(x) is the function that reverses the mapping (reflecting the graph across the line y=xy=x). It is completely different from the algebraic reciprocal.
  10. Misconception: A reflexive relation is identical to the identity relation. → Correct Understanding: The identity relation contains only the pairs (a,a)(a,a). A reflexive relation MUST contain at least all (a,a)(a,a) pairs, but it is free to contain other cross-pairs (a,b)(a,b) as well.
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