Relations & Functions

Relations and Functions opens the CUET Mathematics (class-12) domain. It is definition-heavy: types of relations, types of functions, and composition/inverse. Getting the classifications right earns steady, low-calculation marks.

Types of relations

A relation R on set A can be:

• Reflexive: (a, a) in R for every a.
• Symmetric: (a, b) in R implies (b, a) in R.
• Transitive: (a, b) and (b, c) in R implies (a, c) in R.
• Equivalence: reflexive, symmetric AND transitive together.

Types of functions

• One-one (injective): different inputs give different outputs; f(a) = f(b) implies a = b.
• Onto (surjective): every element of the codomain is an image.
• Bijective: both one-one and onto - only a bijection has an inverse.

Composition and inverse

• Composite: (g of f)(x) = g(f(x)); apply f first, then g.
• Inverse f-inverse exists only if f is bijective; it satisfies f(f-inverse(x)) = x. To find it, set y = f(x) and solve for x in terms of y.

Exam Tricks & Tips

• 🎯 Test reflexive/symmetric/transitive separately; all three together make an equivalence relation.
• 🎯 A function is invertible if and only if it is bijective - check one-one and onto before inverting.
• 🎯 For one-one, either show f(a)=f(b) forces a=b, or that f' keeps one sign (strictly monotonic).
• 🎯 In composition, apply the inner function first: (g of f)(x) = g(f(x)).
• 🎯 Find an inverse by writing y = f(x) and solving for x, then swap variables.
• ❌ Common mistake: assuming symmetry implies reflexivity - a relation can be symmetric and transitive yet not reflexive, so check each property independently.

Expected exam pattern

Classify a given relation, decide one-one/onto, compute a composite, and find an inverse function. Moderate; mostly conceptual.

Quick recap

Equivalence = reflexive + symmetric + transitive; invertible = bijective; (g of f)(x) applies f first; find inverse by solving y = f(x). Verify each relation property on its own.

Cover the answer, recall, then check. 11 cards on relations and functions for CUET Mathematics.

Q1. When is a relation reflexive?
A1. When (a, a) belongs to R for every element a in the set.

Q2. Define a symmetric relation.
A2. If (a, b) is in R then (b, a) is also in R.

Q3. What three properties make a relation an equivalence relation?
A3. Reflexive, symmetric and transitive.

Q4. Define a one-one (injective) function.
A4. f(a) = f(b) implies a = b; different inputs give different outputs.

Q5. Define an onto (surjective) function.
A5. Every element of the codomain is the image of some element.

Q6. A function has an inverse only if it is what?
A6. Bijective (both one-one and onto).

Q7. What is the composite (g of f)(x)?
A7. g(f(x)) - apply f first, then g.

Q8. How do you find the inverse of a function?
A8. Set y = f(x), solve for x in terms of y, then interchange x and y.

Q9. How can calculus show a function is one-one?
A9. If its derivative keeps one sign (the function is strictly increasing or decreasing).

Q10. Does symmetry guarantee reflexivity?
A10. No - a relation can be symmetric and transitive without being reflexive.

Q11. What relation does f(f-inverse(x)) = x express?
A11. That f and f-inverse undo each other (definition of the inverse function).

Relations and functions are the vocabulary that the whole of Class 12 maths speaks. Get comfortable classifying a relation as reflexive, symmetric or transitive, and a function as one-one or onto, and later chapters (inverse trig, calculus) rest on solid ground.

What this tests / core idea: properties of relations (reflexive, symmetric, transitive, equivalence) and types of functions (injective/one-one, surjective/onto, bijective), plus composition and inverses.

Deep explanation

Beginner — what is a relation and a function

A relation R from set A to B is any subset of A × B (pairs (a, b)). A function f: A → B assigns each element of A exactly one image in B. So every function is a relation, but a relation is a function only if no input maps to two outputs.

Intermediate — properties of relations

On a set A, a relation R is:

• Reflexive if (a, a) ∈ R for every a.
• Symmetric if (a, b) ∈ R implies (b, a) ∈ R.
• Transitive if (a, b) and (b, c) ∈ R imply (a, c) ∈ R.
  A relation that is all three is an equivalence relation (e.g. "has the same remainder mod 3"), which partitions A into equivalence classes.

Advanced — types of functions and composition

• One-one (injective): different inputs give different outputs — f(x₁) = f(x₂) ⇒ x₁ = x₂.
• Onto (surjective): every element of the codomain is an image.
• Bijective: both one-one and onto — only bijections have an inverse f⁻¹.
• Composition: (g ∘ f)(x) = g(f(x)); apply f first, then g.

Worked example

Q. Is f: R → R, f(x) = 2x + 3, a bijection?
One-one: 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂, so injective. Onto: for any y, solve y = 2x + 3 ⇒ x = (y − 3)/2, a real number, so every y is hit — surjective. Both ⇒ bijection, with inverse f⁻¹(y) = (y − 3)/2. Answer: Yes, it is a bijection.

CUET relevance

The Mathematics section opens with relations-and-functions MCQs: classify a relation, decide one-one/onto, or compute a composition/inverse. Concept-heavy but formulaic (3–5 marks).

Speed tricks & shortcuts

• Test each property with a quick counter-example — one failing pair kills reflexive/symmetric/transitive.
• A linear function ax + b (a ≠ 0) over R is always a bijection.
• Only bijections are invertible; to invert, swap x and y and solve.

Key formulas

• f: domain → codomain; range = set of outputs.
• Even f(−x)=f(x) (symmetric about y-axis); odd f(−x)=−f(x) (origin symmetry).
• Composition (f∘g)(x) = f(g(x)); inverse exists iff f is bijective; graph of f⁻¹ reflects across y = x.
• Transformations: f(x)+c up, f(x+c) left, af(x) vertical stretch, f(ax) horizontal compress.
• |f(x)| reflects negatives up; number of functions A→B = |B|^|A|.
• Domain restrictions: denominators ≠ 0, non-negative under even roots, positive inside log.

Usage: check domain restrictions (roots, logs, denominators) before analysing a function.

Worked Example

Problem: Let f: R → R be defined by f(x) = 2x + 3. (a) Show that f is one-one (injective) and onto (surjective), hence a bijection. (b) Find f⁻¹.

Solution:
(a) One-one (injective): a function is one-one if f(a) = f(b) implies a = b (distinct inputs give distinct outputs).
Assume f(a) = f(b): 2a + 3 = 2b + 3 ⟹ 2a = 2b ⟹ a = b.
Since f(a) = f(b) forces a = b, f is one-one.

Onto (surjective): a function onto R is surjective if every y in the codomain R has some x with f(x) = y.
Take any y ∈ R and solve f(x) = y: 2x + 3 = y ⟹ x = (y − 3)/2. This x is a real number for every real y, and f((y−3)/2) = 2·(y−3)/2 + 3 = (y − 3) + 3 = y. So every y is achieved; f is onto.

Being both one-one and onto, f is a bijection, so f⁻¹ exists.

(b) To find f⁻¹, set y = f(x) and solve for x:
y = 2x + 3 ⟹ x = (y − 3)/2.
Therefore f⁻¹(y) = (y − 3)/2, or writing the variable as x, f⁻¹(x) = (x − 3)/2.
Verification: f(f⁻¹(x)) = 2·((x−3)/2) + 3 = (x − 3) + 3 = x, and f⁻¹(f(x)) = ((2x+3) − 3)/2 = x. Both compositions give x, confirming the inverse.

Answer: (a) f is one-one (2a+3 = 2b+3 ⟹ a = b) and onto (for any y, x = (y−3)/2 gives f(x) = y), so it is a bijection. (b) f⁻¹(x) = (x − 3)/2, verified by f(f⁻¹(x)) = x.