Relations and Functions ISC Class 12: Notes, Methods and Solved Questions

What Relations and Functions is about

Weightage: Relations and Functions is part of the unit Relations and Functions, which carries 10 of the 80 marks in the ISC Mathematics, shared with 1 other chapter. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Types of relations

For a set A, a relation R on A is a subset of A × A; we write a R b when (a, b) ∈ R. The empty relation contains no pair; the universal relation is A × A itself; the identity relation is $I_A=\{(a,a):a\in A\}$. R is reflexive if (a, a) ∈ R for every a ∈ A; symmetric if (a, b) ∈ R implies (b, a) ∈ R; transitive if (a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R. To show a property fails, one counter-example is enough; to show it holds, argue for general elements. Note that every identity relation is reflexive, but a reflexive relation may contain other pairs too.

Worked example

Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2)}. Is R reflexive, symmetric, transitive?

Try it first, then show the solutionSolution

1. Reflexive: (1, 1), (2, 2), (3, 3) are all in R. Yes.
2. Symmetric: (1, 2) ∈ R but (2, 1) ∉ R. No.
3. Transitive: the only chains are (1, 1), (1, 2) giving (1, 2) ∈ R and (1, 2), (2, 2) giving (1, 2) ∈ R. Yes.

Answer: Reflexive and transitive, but not symmetric

Equivalence relations and equivalence classes

A relation that is reflexive, symmetric and transitive is an equivalence relation. It partitions the set into disjoint equivalence classes: the class of a is $[a]=\{x\in A:xRa\}$. Two classes are either identical or disjoint, and together they make up the whole set. A standard example is congruence modulo n on the integers: a R b if n divides a − b. The smallest equivalence relation on A is the identity relation; the largest is the universal relation A × A.

Worked example

On Z define a R b if a − b is divisible by 3. Show R is an equivalence relation and write the equivalence class of 1.

Try it first, then show the solutionSolution

1. Reflexive: a − a = 0 is divisible by 3, so a R a.
2. Symmetric: if 3 | (a − b) then 3 | (b − a) = −(a − b), so b R a.
3. Transitive: if a − b = 3m and b − c = 3n, then a − c = 3(m + n), so a R c. Class of 1: {…, −5, −2, 1, 4, 7, …} = {3k + 1 : k ∈ Z}.

Answer: R is an equivalence relation; [1] = {3k + 1 : k ∈ Z}

Functions, domain and range

A function f from A to B assigns to every element x of A exactly one element y = f(x) of B. A is the domain, B the co-domain, and the set of images actually taken is the range (a subset of B). A real-valued function has its values in R. For a formula, the domain is the set of x for which it is defined: no division by zero, no square root of a negative number, no logarithm of a non-positive number. Vertical line test: a curve is the graph of a function of x if every vertical line cuts it at most once.

Worked example

Find the domain and range of $f(x)=\sqrt{9-x^2}$.

Try it first, then show the solutionSolution

1. Need $9-x^2\ge 0$, i.e. $-3\le x\le 3$: domain [−3, 3].
2. $9-x^2$ takes every value from 0 (at x = ±3) to 9 (at x = 0), so $\sqrt{9-x^2}$ takes every value from 0 to 3.

Answer: Domain [−3, 3]; range [0, 3]

One-one, many-one, onto and into functions

f: A → B is one-one (injective) if different elements have different images: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$. Otherwise it is many-one. f is onto (surjective) if every element of B is an image: range = co-domain. Otherwise it is into. To prove onto, take any y in B, solve y = f(x), and check that the x found lies in A. A function that is one-one and onto is bijective. Graphically, a one-one function passes the horizontal line test. Counting: from a set of m elements to one of n elements there are $n^m$ functions, and $n(n-1)\cdots (n-m+1)$ one-one functions if m ≤ n (none if m > n).

Worked example

Show that f: R → R, f(x) = 3x + 5 is one-one and onto, but g: R → R, g(x) = x² is neither.

Try it first, then show the solutionSolution

1. f: $3x_1+5=3x_2+5\Rightarrow x_1=x_2$, so one-one. For any real y, $x=\frac{y-5}{3}$ is real and f(x) = y, so onto.
2. g: g(−2) = g(2) = 4 with −2 ≠ 2, so not one-one.
3. g(x) ≥ 0, so a negative number such as −1 has no pre-image: not onto.

Answer: f is bijective; g is many-one and into

Composite functions

If f: A → B and g: B → C, the composite g∘f: A → C is (g∘f)(x) = g(f(x)): apply f first, then g. It exists only when the range of f lies in the domain of g. In general g∘f ≠ f∘g (composition is not commutative), but it is associative: h∘(g∘f) = (h∘g)∘f. If f and g are both one-one (or both onto), so is g∘f. For ISC, composition and inverses are tested on algebraic functions only.

Worked example

If f(x) = x² + 1 and g(x) = 2x − 3, find f∘g(x), g∘f(x), and their values at x = 2.

Try it first, then show the solutionSolution

1. f∘g(x) = f(2x − 3) = (2x − 3)² + 1 = 4x² − 12x + 10; at x = 2: 16 − 24 + 10 = 2.
2. g∘f(x) = g(x² + 1) = 2(x² + 1) − 3 = 2x² − 1; at x = 2: 7.

Answer: f∘g(x) = 4x² − 12x + 10, g∘f(x) = 2x² − 1; 2 and 7 (so f∘g ≠ g∘f)

Invertible functions and finding the inverse

f: A → B is invertible if there is g: B → A with g∘f = $I_A$ and f∘g = $I_B$; then g = $f^{-1}$, and it is unique. Condition of invertibility: f must be bijective (one-one and onto). Method: put y = f(x), solve for x in terms of y to get $x=f^{-1}(y)$, then replace y by x to write $f^{-1}(x)$. State the domain of $f^{-1}$ (the range of f). The graph of $f^{-1}$ is the reflection of the graph of f in the line y = x.

Worked example

Show that f: R − {2} → R − {3}, $f(x)=\frac{3x+1}{x-2}$ is invertible and find $f^{-1}$.

Try it first, then show the solutionSolution

1. One-one: $\frac{3a+1}{a-2}=\frac{3b+1}{b-2}\Rightarrow 3ab-6a+b-2=3ab-6b+a-2\Rightarrow 7b=7a$, so a = b.
2. Onto: $y=\frac{3x+1}{x-2}\Rightarrow yx-2y=3x+1\Rightarrow x=\frac{2y+1}{y-3}$, which is real and ≠ 2 for every y ≠ 3.
3. So f is bijective and $f^{-1}(x)=\frac{2x+1}{x-3}$, x ≠ 3.

Answer: $f^{-1}(x)=\frac{2x+1}{x-3}$

Maths in real life: Relations and Functions

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• One-One and Onto Functions
• Equivalence Relations
• Composite and Invertible Functions

Relations and Functions formula sheet

Relation on A

$R\subseteq A\times A$

Empty: ∅; universal: A × A; identity: {(a, a)}.

Reflexive, symmetric, transitive

$(a,a)\in R;(a,b)\in R\Rightarrow (b,a)\in R;(a,b),(b,c)\in R\Rightarrow (a,c)\in R$

All three: equivalence relation.

Equivalence class

$[a]=\{x\in A:xRa\}$

Classes are disjoint and cover A.

One-one

$f(x_1)=f(x_2)\Rightarrow x_1=x_2$

Horizontal line test.

Onto

$\text{Range of }f=\text{co-domain}$

Solve y = f(x) and check x lies in the domain.

Number of functions

$n^m;\text{one-one: }\frac{n!}{(n-m)!}(m\le n)$

From a set of m elements to a set of n elements.

Composition

$(g\circ f)(x)=g(f(x))$

Usually g∘f ≠ f∘g.

Inverse

$f^{-1}\circ f=I_A,f\circ f^{-1}=I_B$

Exists only for bijections; graph reflected in y = x.

Inverse of a composite

$(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$

Undo in reverse order.

How the board asks Relations and Functions

The question patterns to expect, and the method that works for each.

Tricks to remember Relations and Functions

Short memory aids. Say each one aloud once, then use it on a question the same day.

• RST

  Equivalence = Reflexive + Symmetric + Transitive: 'Really Smart Teachers'.

  Use it when: Checking all three properties.

• Perpendicular is only symmetric

  A line is not perpendicular to itself (not reflexive); l ⊥ m and m ⊥ n make l ∥ n (not transitive).

  Use it when: Line-relation MCQs.

• Vertical for function, horizontal for one-one

  Vertical line cuts once: function. Horizontal line cuts once: one-one.

  Use it when: Graph-based questions.

• Solve, then swap

  Inverse: solve y = f(x) for x, then swap the letters y → x.

  Use it when: Avoiding the '$f^{-1}(y)$' answer that examiners penalise.

• Socks and shoes

  $(g\circ f{)}^{-1}=f^{-1}\circ g^{-1}$: you put socks then shoes on, but take shoes off first.

  Use it when: Inverse of a composite.

Relations and Functions: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Relations and Functions

The slips examiners see most often, and what to write instead.

• ✗Mistake: Confusing symmetric with transitive

  CISCE examiners' comments, 2024: for the relation 'is perpendicular to' on lines in a plane, candidates confused symmetric and transitive properties.

  Wrong working

  Perpendicularity is symmetric and transitive.

  Right working

  l ⊥ m ⇒ m ⊥ l (symmetric), but l ⊥ m and m ⊥ n give l ∥ n, so not transitive; no line is perpendicular to itself, so not reflexive.

  ✓Do this: Test each property with a quick sketch or counter-example.

  Usually costs: 1 mark.

• ✗Mistake: Leaving the inverse as $f^{-1}(y)$

  CISCE examiners' comments, 2024: candidates found $f^{-1}(y)$ and stopped, instead of replacing y by x to give $f^{-1}(x)$.

  Wrong working

  $f^{-1}(y)=\frac{5-y}{3y}$ (final answer).

  Right working

  Then write $f^{-1}(x)=\frac{5-x}{3x}$.

  ✓Do this: The last line must be in terms of x, as the question asks.

  Usually costs: 1 mark.

• ✗Mistake: Claiming onto without justification

  CISCE examiners' comments, 2024: candidates solved y = f(x) for x and declared f onto without checking that x belongs to the domain.

  Wrong working

  $x=\frac{4y+7}{2}$, hence onto.

  Right working

  For every real y, $x=\frac{4y+7}{2}$ is real (in the domain) and f(x) = y; hence onto.

  ✓Do this: Always add the sentence 'x ∈ domain and f(x) = y'.

  Usually costs: 1 mark.

• ✗Mistake: Not knowing the vertical line test

  CISCE examiners' comments, 2024: asked which of two graphs is a function of x, candidates did not know the vertical line test.

  Wrong working

  Chooses a circle-like curve as a function.

  Right working

  A curve is a function of x only if every vertical line meets it at most once.

  ✓Do this: Draw a vertical line across each graph before choosing.

  Usually costs: 1 mark.

• ✗Mistake: Misjudging one-one functions between sets of different sizes

  CISCE examiners' comments, 2025: in an assertion-reason item on one-one functions from a larger set to a smaller one, candidates did not understand mappings between sets of unequal size; others could not write the smallest equivalence relation on {1, 2, 3} because they could not list A × A.

  Wrong working

  With m > n there are still some one-one functions.

  Right working

  If m > n, two elements must share an image (pigeonhole), so there are no one-one functions. The smallest equivalence relation on A is the identity relation.

  ✓Do this: Count with small sets; list A × A when needed.

  Usually costs: 1 mark.

Practice questions on Relations and Functions

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   Statement I: Every equivalence relation on a set is reflexive.
   Statement II: Every reflexive relation on a set is symmetric.
   (a) Both statements are true.
   (b) Both statements are false.
   (c) Statement I is true and Statement II is false.
   (d) Statement I is false and Statement II is true.

   Hint

   Try R = {(1,1), (2,2), (1,2)} on {1, 2}.

   Answer

   (c)

2. Q2 · 1 mark · MCQBoard level

   The number of one-one functions from {a, b, c} to {1, 2, 3, 4} is:
   (a) 12
   (b) 24
   (c) 64
   (d) 81

   Hint

   4 × 3 × 2.

   Answer

   (b) 24

3. Q3 · 1 mark · MCQBoard level

   The domain of $f(x)=\frac{1}{\sqrt{x-2}}$ is:
   (a) [2, ∞)
   (b) (2, ∞)
   (c) (−∞, 2)
   (d) R − {2}

   Hint

   The square root must be positive, not zero.

   Answer

   (b)

4. Q4 · 1 mark · Very short answerBoard level

   Write the smallest equivalence relation on A = {a, b}.

   Hint

   Only what reflexivity forces.

   Answer

   {(a, a), (b, b)}

5. Q5 · 2 marks · Short answerBoard level

   If f(x) = x² + 1 and g(x) = 2x − 3, find (f∘g)(2) and (g∘f)(2).

   Hint

   Inner function first.

   Answer

   (f∘g)(2) = f(1) = 2; (g∘f)(2) = g(5) = 7.

6. Q6 · 2 marks · Short answerHarder board level

   Show that f: R → R, $f(x)=\frac{2x-3}{5}$ is one-one and onto.

   Hint

   Justify that the pre-image is real.

   Answer

   One-one: $\frac{2a-3}{5}=\frac{2b-3}{5}\Rightarrow a=b$. Onto: for any real y, $x=\frac{5y+3}{2}$ is real and f(x) = y.

7. Q7 · 2 marks · Short answerHarder board level

   Check whether R = {(a, b) : a ≤ b²} on R is reflexive, symmetric or transitive.

   Hint

   Try a = 1/2; try 3, 2, 1.5.

   Answer

   Not reflexive: 1/2 ≤ 1/4 is false. Not symmetric: 1 ≤ 4 but 2 ≤ 1 is false. Not transitive: 3 ≤ 2² and 2 ≤ 1.5², but 3 ≤ 1.5² = 2.25 is false.

8. Q8 · 2 marks · Short answerHOTS / competency

   If $f(x)=\frac{1}{1-x}$ (x ≠ 0, 1), find f∘f(x) and hence $f^{-1}(x)$.

   Hint

   f∘f∘f = I.

   Answer

   f∘f(x) = (x − 1)/x; since f∘(f∘f) = I, $f^{-1}(x)=\frac{x-1}{x}$.

9. Q9 · 3 marks · Short answerHarder board level

   On N × N define (a, b) R (c, d) ⇔ a + d = b + c. Show that R is an equivalence relation.

   Hint

   Transitive: add the two equations.

   Answer

   Reflexive: a + b = b + a. Symmetric: a + d = b + c ⇒ c + b = d + a. Transitive: a + d = b + c and c + f = d + e; adding, a + f = b + e, so (a, b) R (e, f).

10. Q10 · 3 marks · Short answerHarder board level

    Show that f: [0, ∞) → [0, ∞), f(x) = x² is invertible, find $f^{-1}$, and describe how the two graphs are related.

    Hint

    Restricting the domain removes negative inputs.

    Answer

    On [0, ∞), a² = b² ⇒ a = b (one-one); every y ≥ 0 has pre-image √y (onto). $f^{-1}(x)=\sqrt{x}$. The graphs are mirror images in the line y = x.

11. Q11 · 3 marks · Short answerHarder board level

    Is f: R → R, f(x) = |x| + x one-one? onto? Justify.

    Hint

    Write f piecewise.

    Answer

    f(x) = 2x for x ≥ 0 and 0 for x < 0. Not one-one: f(−1) = f(−2) = 0. Not onto: negative numbers are never images (range [0, ∞)).

12. Q12 · 5 marks · Long answerHarder board level

    (a) Show that f: R − {3} → R − {1}, $f(x)=\frac{x+2}{x-3}$ is bijective. [3]
    (b) Find $f^{-1}$ and verify $f^{-1}$∘f(x) = x. [2]

    Hint

    Solve y(x − 3) = x + 2.

    Answer

    (a) One-one: $\frac{a+2}{a-3}=\frac{b+2}{b-3}\Rightarrow -3a+2b=-3b+2a\Rightarrow a=b$. Onto: $x=\frac{3y+2}{y-1}$, real for y ≠ 1, and it never equals 3 (that would need 2 = −3). (b) $f^{-1}(x)=\frac{3x+2}{x-1}$; $f^{-1}(f(x))=\frac{3\frac{x+2}{x-3}+2}{\frac{x+2}{x-3}-1}=\frac{5x}{5}=x$.

13. Q13 · 2 marks · Short answerHarder board level

    Find the domain and range of $f(x)=\sqrt{9-x^2}$. Is f: [−3, 3] → [0, 3] one-one?

    Hint

    Symmetric about x = 0.

    Answer

    Domain [−3, 3], range [0, 3]. Not one-one: f(−1) = f(1) = √8.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Define an equivalence relation.Show answer

   A relation that is reflexive, symmetric and transitive.

2. 2.What is the identity relation on A?Show answer

   $I_A=\{(a,a):a\in A\}$.

3. 3.When is f one-one?Show answer

   When $f(x_1)=f(x_2)$ implies $x_1=x_2$.

4. 4.When is f onto?Show answer

   When its range equals its co-domain.

5. 5.Condition for f to be invertible?Show answer

   f must be one-one and onto (bijective).

6. 6.How are the graphs of f and $f^{-1}$ related?Show answer

   Reflections of each other in the line y = x.

7. 7.Is composition commutative?Show answer

   No; in general g∘f ≠ f∘g.

8. 8.$(g\circ f{)}^{-1}$ = ?Show answer

   $f^{-1}\circ g^{-1}$.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you test a relation for reflexive, symmetric and transitive properties?
• ☐Can you prove an equivalence relation and write its classes?
• ☐Can you find the domain and range of a function with a square root or fraction?
• ☐Can you prove one-one and onto with justification?
• ☐Can you count one-one functions between two finite sets?
• ☐Can you find f∘g and g∘f and show they differ?
• ☐Can you find an inverse and write it as $f^{-1}(x)$ with its domain?
• ☐Can you explain why f and $f^{-1}$ are reflections in y = x?

Relations and Functions: frequently asked questions