Sets, Relations and Functions — JEE Main & Advanced Notes

Build domain-range discipline, relation properties, composition and inverse-function thinking before moving into calculus.

Material signal: Mapped from repeated function and inverse-trigonometry revision assignments.

Priority: Must Do. Treat this as a foundation chapter inside the JEE Mathematics ladder.

• Sets and the inclusion-exclusion principle: A set is a well-defined collection of distinct objects. When two or more sets overlap, simply adding their sizes overcounts the intersection. The formula n(A∪B) = n(A)+n(B)−n(A∩B) corrects this. For three sets, triple overlaps are added back after double intersections are subtracted. Draw a Venn diagram and assign an unknown to each region before forming equations — this approach handles virtually every JEE set-counting problem.
• Relations and equivalence: A relation R on A is reflexive if (a,a)∈R for every a, symmetric if (a,b)∈R implies (b,a)∈R, and transitive if (a,b)∈R and (b,c)∈R together imply (a,c)∈R. An equivalence relation satisfies all three and partitions A into disjoint equivalence classes. The classic example: aRb iff (a−b) is divisible by n partitions ℤ into n residue classes.
• Types of functions: A function f: A→B is one-one (injective) if f(a₁)=f(a₂) ⇒ a₁=a₂. Test algebraically: set f(x₁)=f(x₂) and show x₁=x₂. Graphically, every horizontal line cuts the graph at most once. A function is onto (surjective) if every element of B is an image: for every y∈B, solve f(x)=y and show a real solution exists. A bijection is both one-one and onto; only bijections have inverses.
• Domain discipline: Finding the domain means simultaneously enforcing every restriction. List them all: square roots need non-negative arguments, logarithms need strictly positive arguments, denominators must be non-zero, and for inverse trig, arguments must lie in [−1,1]. Intersect all constraint sets. When a simplification cancels a factor (e.g. (x²−1)/(x−1) = x+1), the excluded point x=1 is still not in the domain.
• Finding the range: Two reliable methods. (1) Direct method: write y=f(x), express x as g(y), then find the domain of g(y) — that is the range of f. (2) Derivative method: find turning points and check the function’s behavior at boundaries/infinity to identify all achievable y-values. For polynomials on a closed interval, always evaluate at endpoints and critical points.
• Composition and its domain: (f∘g)(x)=f(g(x)) requires x to be in domain(g) and g(x) to be in domain(f). The domain of f∘g = {x∈domain(g) : g(x)∈domain(f)}. Composition is associative: f∘(g∘h)=(f∘g)∘h. It is generally NOT commutative: f∘g ≠ g∘f. Always work the inner function first when evaluating.
• Inverse functions: f⁻¹ exists only for bijections. To find f⁻¹: (1) write y=f(x), (2) solve for x in terms of y, (3) write f⁻¹(y)=x and rename the variable. The domain of f⁻¹ equals the range of f, and the range of f⁻¹ equals the domain of f. The graph of f⁻¹ is the reflection of the graph of f across y=x. Check: f(f⁻¹(x))=x on the range of f.
• Even and odd functions: Even functions (f(−x)=f(x)) are symmetric about the y-axis; examples include x², cos(x), |x|. Odd functions (f(−x)=−f(x)) pass through the origin and are symmetric about it; examples include x, x³, sin(x). The product of two odd functions or two even functions is even; the product of an odd and even function is odd. The zero function is both even and odd.

• (f ∘ g)(x) = f(g(x)) — apply g first, then apply f to the result
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• f⁻¹ exists iff f is bijective (both one-one and onto)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• n(A ∪ B ∪ C) = n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Number of functions from A to B = |B|^|A|
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Number of one-one functions from A to B = P(|B|,|A|) = |B|!/(|B|−|A|)! (if |B| ≥ |A|, else 0)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Number of relations from A to B = 2^(|A|×|B|)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Domain of √(f(x)): require f(x) ≥ 0; Domain of log(f(x)): require f(x) > 0; Domain of 1/f(x): require f(x) ≠ 0
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Even function: f(−x) = f(x) [y-axis symmetry]; Odd function: f(−x) = −f(x) [origin symmetry]
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.
• Decomposition: any f = [f(x)+f(−x)]/2 (even part) + [f(x)−f(−x)]/2 (odd part)
  Hint: do not memorise this in isolation; connect it to the definition or diagram that produces it.

• Finding inverse without restricting the domain: f(x)=x² is not one-one on ℝ, so its inverse doesn’t exist there. Restrict to [0,∞) first, then f⁻¹(x)=√x. Always state the domain restriction explicitly.
• Treating every relation as a function: a relation where one input maps to two outputs is not a function. E.g., x²+y²=1 is a relation, not a function of x on [−1,1], because each x-value gives two y-values. The vertical line test catches this graphically.
• Ignoring excluded values after simplification: (x²−4)/(x−2) simplifies to x+2, but x=2 is excluded from the domain. Writing domain = ℝ is wrong; it is ℝ−2.
• Confusing domain of f∘g with domain of f: the domain of f(g(x)) is the set of x in domain(g) such that g(x) lies in domain(f), not just any x in domain(f).
• Assuming f∘g = g∘f: this is almost never true. Always compute both and compare rather than assuming commutativity.

For JEE Main, aim for fast recognition and clean substitution. Finish the first pass of Sets, Relations and Functions questions in 60–90 seconds each. Prioritise standard formulas, short sign/domain checks and option elimination only after the setup is correct.

For JEE Advanced, expect combined conditions, hidden domains, multi-correct traps and integer-style answers. Build the solution from definitions, not memorised tricks. When a parameter appears, solve the general case and then filter using restrictions.

• Domain first, always. For each subexpression, list its restriction and intersect all constraints at the end.
• For inverse: restrict domain → confirm one-one → write y=f(x) → solve for x → state domain of f⁻¹ = range of f.
• Use horizontal line test for one-one. If any horizontal line meets the graph more than once, restrict the domain.
• Composition order: (f∘g)(x) = f(g(x)). The inner function g runs first; f acts on the output of g.
• For inclusion-exclusion with three sets: start from the Venn diagram, assign a variable to each of the seven regions, and build equations from the given totals.