Relations and Functions

2.1 Core concepts

• A relation R in a set A is any subset of A x A; the two extreme relations are the empty relation R = phi and the universal relation R = A x A — both called trivial relations (NCERT §1.2, p. 2). Relations encode any binary "is related to" structure on a set.
• Empty relation example: in A = {1,2,3,4}, R = {(a,b) : a - b = 10} contains no pair, so R = phi (NCERT §1.2, p. 2).
• Universal relation example: in the same A, R' = {(a,b) : |a - b| >= 0} equals A x A (NCERT §1.2, p. 2). Both extremes are technically equivalence relations.
• A relation R in A is reflexive if (a,a) belongs to R for every a in A; symmetric if (a1,a2) in R implies (a2,a1) in R; transitive if (a1,a2) in R and (a2,a3) in R imply (a1,a3) in R (NCERT §1.2, p. 2).
• A relation that is simultaneously reflexive, symmetric and transitive is called an equivalence relation (NCERT §1.2, p. 3). Equivalence relations capture the idea of "indistinguishable up to some criterion".
• "Congruence of triangles" on the set T of all triangles in a plane is the standard equivalence relation example, since every triangle is congruent to itself and congruence is symmetric and transitive (NCERT §1.2, Example 2, p. 3).
• "L1 is perpendicular to L2" on the set L of lines is symmetric but neither reflexive nor transitive — a line cannot be perpendicular to itself, and two perpendiculars to a common line are parallel, not perpendicular (NCERT §1.2, Example 3, p. 3). This is the standard "fails some properties" example.
• Every equivalence relation R on a set X partitions X into mutually disjoint equivalence classes A_i such that elements of A_i are related to each other, no element of A_i is related to an element of A_j (i != j), and union of A_i = X (NCERT §1.2, p. 4). The reverse is also true: every partition defines an equivalence relation.
• For R = {(a,b) : 2 divides a - b} on Z, equivalence classes are [0] = set of even integers and [1] = set of odd integers, and [0] = [2r], [1] = [2r+1] for r in Z (NCERT §1.2, p. 4).
• For R = {(a,b) : 3 divides a - b} on Z, the three equivalence classes are [0] = multiples of 3, [1] = numbers of form 3r + 1 and [2] = numbers of form 3r + 2 (NCERT §1.2, p. 4). In general, "n divides a − b" gives n equivalence classes — the residue classes mod n.
• A function f : X to Y is one-one (injective) if distinct elements of X have distinct images, i.e., f(x1) = f(x2) implies x1 = x2; otherwise it is many-one (NCERT §1.3, Definition 5, p. 7).
• A function f : X to Y is onto (surjective) if every y in Y is the image of some x in X — equivalently, Range of f = Y (NCERT §1.3, Definition 6 & Remark, pp. 7–8).
• f is bijective if it is both one-one and onto (NCERT §1.3, Definition 7, p. 8). Bijection is the precise mathematical sense of "perfect pairing".
• f : N to N given by f(x) = 2x is one-one but not onto (no x with 2x = 1); f : R to R given by f(x) = 2x is one-one AND onto (NCERT §1.3, Examples 8 & 9, pp. 8–9).
• f : R to R, f(x) = x^2 is neither one-one (f(-1) = f(1)) nor onto (no x with x^2 = -2) (NCERT §1.3, Example 11, p. 9).
• f : N to N defined as f(x) = x + 1 if x is odd and f(x) = x - 1 if x is even is both one-one and onto (NCERT §1.3, Example 12, pp. 9–10).
• Characteristic property of a finite set X: every one-one function f : X to X is onto, and every onto function f : X to X is one-one; this fails for infinite sets (NCERT §1.3, Remark after Examples 13–14, p. 10). The pigeonhole principle underlies this.
• Composition of functions: if f : A to B and g : B to C, then gof : A to C is given by gof(x) = g(f(x)) for all x in A (NCERT §1.4, Definition 8, p. 12). Read "gof" as "g composed with f" or "g after f".
• Composition is NOT commutative in general: for f(x) = cos x and g(x) = 3x^2 on R, gof(x) = 3 cos^2 x while fog(x) = cos(3x^2), and 3cos^2 x != cos 3x^2 at x = 0 (NCERT §1.4, Example 16, p. 12).
• A function f : X to Y is invertible if there exists g : Y to X with gof = I_X and fog = I_Y; this g is unique and is written f^{-1} (NCERT §1.4, Definition 9, p. 12).
• f is invertible iff f is one-one AND onto (bijective) (NCERT §1.4, p. 12). This is the central characterization of invertibility.
• Worked inverse example: f : N to Y, f(x) = 4x + 3 with Y = {y in N : y = 4x + 3 for some x in N} is invertible with inverse g(y) = (y - 3) / 4 (NCERT §1.4, Example 17, pp. 12–13).
• Miscellaneous result: intersection of two equivalence relations is an equivalence relation (NCERT Miscellaneous Example 18, p. 13).
• The number of one-one functions from {1,2,3} to itself is 3! = 6 (NCERT Miscellaneous Example 22, p. 14).
• Number of equivalence relations on {1,2,3} containing (1,2) and (2,1) is two (NCERT Miscellaneous Example 24, p. 14).
• These ideas underpin the rest of Class XII algebra and calculus: invertibility is used in inverse-trig and exponential-log work; equivalence relations underlie modular arithmetic and number theory; bijections appear in counting problems and group theory.
• A practical CUET technique: when the relation R is given as a set of ordered pairs, check each property by going through the list mechanically. When R is given as a rule (e.g., R = {(a, b) : 3 | a − b}), check the general assertion using algebraic manipulation.
• Standard examples to memorise for fast MCQ work: identity function I_X is bijective; constant function on a multi-element domain is many-one and not onto; characteristic functions are typically not invertible.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig 1.1 (p. 3): illustrates the "perpendicular lines" relation — useful visual for showing why perpendicularity is symmetric but not transitive (L1 perp L2 and L2 perp L3 forces L1 parallel L3, not perpendicular).
• Fig 1.2 (i)–(iv) (p. 8): the four arrow-diagrams for f1, f2, f3, f4 showing all four combinations of one-one/many-one with onto/not-onto; the canonical visual for testing injectivity and surjectivity from a mapping diagram.
• Fig 1.3 (p. 9): graph of f(x) = 2x from R to R — straight line through the origin demonstrating bijectivity (horizontal-line test passes; line meets every horizontal level exactly once).
• Fig 1.4 (p. 9): illustration of f : N to N with the odd/even swap rule — depicts how every natural number is hit exactly once (bijection on N).
• Fig 1.5 (p. 12): composition diagram showing A —f→ B —g→ C and the combined arrow A —gof→ C; reinforces order ("g of f", apply f first then g).
• Process — test reflexive/symmetric/transitive: (i) for each a, check (a, a) ∈ R; (ii) for each (a, b) ∈ R, check (b, a) ∈ R; (iii) for each chain (a, b), (b, c) ∈ R, check (a, c) ∈ R. Equivalence iff all three hold.
• Process — test injectivity: assume f(x₁) = f(x₂); derive x₁ = x₂. If derivation fails, find a counter-example.
• Process — test surjectivity: pick arbitrary y in co-domain; solve y = f(x) for x and check x lies in domain. If solvable for every y, function is onto.
• Process — find inverse: solve y = f(x) for x in terms of y; verify gof = I_X and fog = I_Y.
• Process — compute composition: identify inner and outer functions; substitute inner output as outer input.

2.4 Common confusions / NTA trap points

• Reversed composition: students write fog when gof is asked. Remember (gof)(x) = g(f(x)) — the function written closer to x acts first.
• Treating "onto" loosely: a function is onto only if Range = Co-domain. f : N to N, f(x) = 2x is NOT onto (1 has no pre-image); the same formula on f : R to R IS onto. The co-domain matters as much as the formula.
• Confusing reflexive with the universal relation: every element being related to itself is required, but other pairs may be absent. A reflexive relation need not be the universal relation.
• Symmetry vs antisymmetry: students mark "a <= b" as symmetric. It is reflexive and transitive but NOT symmetric (a <= b does not give b <= a).
• Forgetting the bijection condition for invertibility: a function with a "formula inverse" like sqrt is not automatically invertible unless domain and co-domain are restricted to make it bijective.
• For a finite set X, one-one and onto are equivalent for f : X to X — but this equivalence FAILS for infinite sets (e.g., f : N to N, f(x) = 2x is one-one but not onto).
• Confusing "many-one" with "many-many"; functions are never many-many.
• Mis-identifying the image of x² as all real numbers; image is [0, ∞), not R.
• Treating x → 1/x as a function on R; the natural domain is R \ {0}.
• Forgetting empty domain quirks; functions on the empty set are vacuously both one-one and onto.
• Confusing the inverse relation R⁻¹ (a set-theoretic concept) with the inverse function f⁻¹ (defined only when f is bijective).
• Mis-reading an arrow diagram: an arrow from x to y means f(x) = y; the function is many-to-one if two arrows arrive at the same y.
• Treating "identity function" as f(x) = 1; the identity is I_X(x) = x, not the constant 1.

2.5 Key formulas & theorems

2.6 Solved examples (NCERT-grounded)

Example A (NCERT Example 5, p. 3). Show R = {(a, b) : 2 divides (a − b)} on Z is an equivalence relation.

Step 1 — reflexive: 2 | (a − a) = 0 ⇒ (a, a) ∈ R. Step 2 — symmetric: 2 | (a − b) ⇒ 2 | −(a − b) = (b − a). Step 3 — transitive: 2 | (a − b), 2 | (b − c) ⇒ 2 | (a − b) + (b − c) = (a − c). Equivalence.

Example B (NCERT Example 11, p. 9). Show f(x) = x² on R is neither one-one nor onto.

Step 1 — not one-one: f(−1) = 1 = f(1) but −1 ≠ 1. Step 2 — not onto: there is no x ∈ R with x² = −2. Step 3 — conclude: neither injective nor surjective.

Example C (NCERT Example 12, pp. 9–10). Show f : N → N defined as f(x) = x + 1 (odd x) and f(x) = x − 1 (even x) is a bijection.

Step 1 — one-one: if f(x₁) = f(x₂) with both odd or both even, immediate cancellation; mixed parity is impossible because outputs differ in parity. Step 2 — onto: every natural y is f(y − 1) if y even and y + 1 odd, or f(y + 1) if y odd and y + 1 even. Step 3 — conclude: bijection on N.

Example D (NCERT Example 16, p. 12). f(x) = cos x, g(x) = 3x². Find gof and fog; check (gof)(0) = (fog)(0).

Step 1 — gof: g(f(x)) = g(cos x) = 3 cos² x. Step 2 — fog: f(g(x)) = f(3x²) = cos(3x²). Step 3 — at x = 0: gof(0) = 3·1 = 3; fog(0) = cos 0 = 1. Different; composition not commutative.

Example E (NCERT Example 17, pp. 12–13). Find inverse of f : N → Y, f(x) = 4x + 3, Y = {y ∈ N : y = 4x + 3, x ∈ N}.

Step 1 — solve for x: y = 4x + 3 ⇒ x = (y − 3)/4. Step 2 — verify gof: g(f(x)) = ((4x + 3) − 3)/4 = x. ✓ Step 3 — verify fog: f(g(y)) = 4·(y − 3)/4 + 3 = y. ✓ Inverse: g(y) = (y − 3)/4.