Exam Weightage & Blueprint
Total: ~10 MarksRelations and Functions is a fundamental chapter that appears in various forms throughout the Board exam. Concepts from this chapter are also used in Calculus and other areas.
| Question Type | Marks | Frequency | Focus Topic |
|---|---|---|---|
| MCQ | 1 | Very High | Types of Relations, One-One/Onto identification |
| Short Answer (2M) | 2 | High | Proving relations (reflexive/symmetric/transitive) |
| Short Answer (3M) | 3 | Very High | Equivalence Relations, One-One/Onto proofs |
| Long Answer | 4 | Medium | Composition of Functions, Invertible Functions |
Last 24-Hour Checklist
Types of Relations
Basic Definitions
Empty Relation
$R = \phi \subset A \times A$
No element is related to any element
Universal Relation
$R = A \times A$
Every element is related to every element
Equivalence Relation
Reflexive, Symmetric & Transitive
Most important for exams!
Properties of Relations - Simply Explained
1. Reflexive (Selfie)
Cue: "Everyone takes a selfie."
Every single person in the group must be related to themselves.
$(a, a) \in R$ for ALL $a$
2. Symmetric (Friends)
Cue: "If I choose you, you choose me."
If A is related to B, then B must be related back to A.
$(a, b) \in R \Rightarrow (b, a) \in R$
3. Transitive (Chain)
Cue: "Pass it on."
If A connects to B, and B connects to C, does A connect to C?
$(a, b), (b, c) \in R \Rightarrow (a, c) \in R$
- Reflexive: $(a, a) \in R$ for every $a \in A$.
- Symmetric: If $(a_1, a_2) \in R$, then $(a_2, a_1) \in R$.
- Transitive: If $(a_1, a_2) \in R$ and $(a_2, a_3) \in R$, then $(a_1, a_3) \in R$.
Equivalence Relation
Reflexive AND Symmetric AND Transitive
- "is equal to" in any set
- "is congruent to" for triangles
- "is similar to" for triangles
- "has same number of pages as" for books
- $R = \{(a, b) : |a - b|$ is even$\}$ in integers
Equivalence Class $[a]$
Denoted by $[a] = \{x \in A : (x, a) \in R\}$.
- All elements of $A$ related to each other belong to the same equivalence class.
- No element of one equivalence class is related to any element of another equivalence class.
- Two equivalence classes are either identical or disjoint (no overlapping elements).
- The union of all equivalence classes gives the original set $A$ (Partition of a set).
Quick Check: If $R = \{(1,1), (2,2), (1,2), (2,1)\}$ on $\{1,2\}$, what is $[1]$?
Types of Functions - Simply Explained
1. One-One (Injective)
Cue: "Reserved Seating"
Every passenger gets a unique seat. No two passengers share the same seat.
$f(x_1) = f(x_2) \Rightarrow x_1 = x_2$
Check: Does any horizontal line cut the graph more than once? If yes, it's NOT One-One.
2. Onto (Surjective)
Cue: "Full House"
Every seat is taken. No empty seats left in the bus (Co-domain).
Range = Co-domain
Check: Is the Range equal to the Co-domain?
It's a "Perfect Match"! Everyone has a unique seat (One-One) AND all seats are full (Onto).
Only Bijective functions have an Inverse ($f^{-1}$).
One-One Examples
• $f(x) = 2x$ (Linear is usually 1-1)
• $f(x) = x^3$ (Odd powers usually 1-1)
Not One-One (Many-One)
• $f(x) = x^2$ (Parabola: Two inputs give same output)
• $f(x) = |x|$ (Modulus: Positive & negative give same result)
• $f(x) = \sin x$ (Periodic functions repeat values)
Quick Reference: Function Types
| Function Type | Condition | Inverse Exists? |
|---|---|---|
| One-One (Injective) | $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ | No |
| Onto (Surjective) | Range = Co-domain | No |
| Bijective | Both One-One and Onto | Yes |
| Many-One | Different inputs can give same output | No |
Composition of Functions
- Read $gof$ as "g circle f" or "g composed with f"
- First apply $f$, then apply $g$ to the result
- $gof \neq fog$ (composition is NOT commutative)
- $gof$ is defined only if Range of $f \subseteq$ Domain of $g$
Invertible Functions
1. Replace $f(x)$ with $y$
2. Solve for $x$ in terms of $y$
3. Replace $y$ with $x$ to get $f^{-1}(x)$
4. Verify: $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$
- Reversal Rule: $(gof)^{-1} = f^{-1}og^{-1}$
- The inverse of a bijection is also a bijection.
- $(f^{-1})^{-1} = f$
How to Solve Problems (Step-by-Step)
Proving Relations (The 3-Step Check)
- Step 1 (Reflexive?):
Check if $(a,a)$ works for every $a$.
Pass: Yes, for all.
Fail: Find just ONE example where it fails. - Step 2 (Symmetric?):
Assume $(a,b) \in R$. Check if $(b,a) \in R$.
Pass: If equation stays same after swapping. - Step 3 (Transitive?):
Assume $(a,b)$ and $(b,c) \in R$. Check $(a,c)$.
Pass: If you can derive a relation between a & c.
Proving Functions (The 2-Step Proof)
- Step 1 (One-One?):
Start with: Let $f(x_1) = f(x_2)$
Solve the equation.
Pass: If you get ONLY $x_1 = x_2$.
Fail: If you get $x_1 = \pm x_2$ etc. - Step 2 (Onto?):
Put $f(x) = y$.
Find $x$ in terms of $y$.
Pass: If this $x$ is valid in Domain for ALL $y$.
Solved Examples (Board Marking Scheme)
Q1. Show that the relation $R$ in $\mathbb{Z}$ given by $R = \{(a, b): 2$ divides $a - b\}$ is an equivalence relation. (3 Marks)
For any $a \in \mathbb{Z}$, we have $a - a = 0 = 2 \times 0$.
Therefore, 2 divides $(a - a)$, so $(a, a) \in R$.
Hence, $R$ is reflexive.
Let $(a, b) \in R$. Then 2 divides $(a - b)$.
So, $a - b = 2k$ for some integer $k$.
Therefore, $b - a = -2k = 2(-k)$, where $-k$ is an integer.
Hence, 2 divides $(b - a)$, so $(b, a) \in R$.
Therefore, $R$ is symmetric.
Let $(a, b) \in R$ and $(b, c) \in R$.
Then 2 divides $(a - b)$ and 2 divides $(b - c)$.
So, $a - b = 2m$ and $b - c = 2n$ for integers $m$ and $n$.
Adding: $(a - b) + (b - c) = 2m + 2n \Rightarrow a - c = 2(m + n)$.
Hence, 2 divides $(a - c)$, so $(a, c) \in R$.
Therefore, $R$ is transitive.
Conclusion: $R$ is reflexive, symmetric, and transitive. Hence, $R$ is an equivalence relation.
Q2. Show that $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = 2x$ is one-one and onto. (3 Marks)
Let $f(x_1) = f(x_2)$ where $x_1, x_2 \in \mathbb{R}$.
Then $2x_1 = 2x_2$
Dividing both sides by 2: $x_1 = x_2$
Therefore, $f$ is one-one.
Let $y \in \mathbb{R}$ (co-domain) be arbitrary.
We need to find $x \in \mathbb{R}$ such that $f(x) = y$.
From $f(x) = y$, we have $2x = y \Rightarrow x = \frac{y}{2}$.
Since $y \in \mathbb{R}$, we have $\frac{y}{2} \in \mathbb{R}$ (domain).
Therefore, for every $y$ in co-domain, there exists $x = \frac{y}{2}$ in domain such that $f(x) = y$.
Hence, $f$ is onto.
Q3. Let $f: \{2, 3, 4, 5\} \to \{3, 4, 5, 9\}$ and $g: \{3, 4, 5, 9\} \to \{7, 11, 15\}$ be given by $f(2) = 3, f(3) = 4, f(4) = f(5) = 5$ and $g(3) = g(4) = 7, g(5) = g(9) = 11$. Find $gof$. (2 Marks)
$(gof)(x) = g(f(x))$
$(gof)(2) = g(f(2)) = g(3) = 7$
$(gof)(3) = g(f(3)) = g(4) = 7$
$(gof)(4) = g(f(4)) = g(5) = 11$
$(gof)(5) = g(f(5)) = g(5) = 11$
Therefore: $gof = \{(2, 7), (3, 7), (4, 11), (5, 11)\}$
Previous Year Questions (PYQs)
(A) $\{1, 2, 3, ...\}$ (B) $\{2, 3\}$ (C) $\{1, 2, 3\}$ (D) $\{1, 2, 3, 4\}$
Ans: (C) $\{1, 2, 3\}$. For $x + 2y = 8$: when $y = 1, x = 6$; when $y = 2, x = 4$; when $y = 3, x = 2$. So range = $\{1, 2, 3\}$.
Ans:
• Not Reflexive: For $a = -1$, we have $-1 \not\leq (-1)^2 = 1$ is false. So $(-1, -1) \notin R$.
• Not Symmetric: $(1, 2) \in R$ since $1 \leq 4$, but $(2, 1) \notin R$ since $2 \not\leq 1$.
• Not Transitive: $(3, 2) \in R$ and $(2, 1.5) \in R$, but $(3, 1.5) \notin R$ since $3 \not\leq 2.25$.
Ans:
• Not One-One: $f(1) = 1 = f(2)$, but $1 \neq 2$. Hence not one-one.
• Onto: For any $y \in \mathbb{N}$, if $y = 1$, then $f(1) = 1$. If $y > 1$, choose $x = y + 1$, then $f(y + 1) = (y + 1) - 1 = y$. Hence onto.
Solution Outline:
1. Prove one-one by assuming $f(x_1) = f(x_2)$ and showing $x_1 = x_2$
2. Prove onto by taking arbitrary $y \in B$ and finding $x = \frac{3y - 2}{y - 1} \in A$
3. Since bijective, $f^{-1}$ exists: $f^{-1}(y) = \frac{3y - 2}{y - 1}$
Exam Strategy & Mistake Bank
Common Mistakes
Scoring Tips
Practice Problems (Self-Assessment)
Level 1: Basic (1-2 Marks Each)
Q1. Check whether the relation $R$ in $\{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ is reflexive, symmetric, or transitive.
We need $(1,1), (2,2), (3,3)$.
Are they present? Yes.
$\therefore$ Reflexive.
We have $(1,2)$. Is $(2,1)$ present? No.
We have $(2,3)$. Is $(3,2)$ present? No.
$\therefore$ Not Symmetric.
We have $(1,2)$ and $(2,3)$.
Transitive requires $(1,3)$ to be present.
Is $(1,3) \in R$? No.
$\therefore$ Not Transitive.
Q2. Show that the function $f: \mathbb{N} \to \mathbb{N}$ given by $f(x) = x^2$ is one-one but not onto.
Let $f(x_1) = f(x_2)$
$\Rightarrow x_1^2 = x_2^2$
$\Rightarrow x_1 = x_2$ (Since $x \in \mathbb{N}$, no negative numbers allowed)
$\therefore$ One-One.
Co-domain is $\mathbb{N} = \{1, 2, 3, 4, ...\}$.
Range is $\{1^2, 2^2, 3^2, ...\} = \{1, 4, 9, ...\}$.
Are numbers like 2 or 3 in the range? No.
$\therefore$ Range $\neq$ Co-domain.
$\therefore$ Not Onto.
Q3. If $f(x) = x + 1$ and $g(x) = 2x$, find $(gof)(x)$ and $(fog)(x)$. Are they equal?
$(gof)(x) = g(f(x)) = g(x + 1)$
$= 2(x + 1) = 2x + 2$
$(fog)(x) = f(g(x)) = f(2x)$
$= (2x) + 1 = 2x + 1$
$2x + 2 \neq 2x + 1$
$\therefore$ They are Not Equal.
Level 2: Intermediate (3 Marks Each)
Q4. Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5, 6\}$ given by $R = \{(a, b): |a - b|$ is even$\}$ is an equivalence relation.
$|a - a| = |0| = 0$, which is even.
$\therefore (a, a) \in R$. Reflexive.
Let $|a - b|$ be even.
$|b - a| = |-(a - b)| = |a - b|$, which is also even.
$\therefore (b, a) \in R$. Symmetric.
Let $|a - b|$ (even) and $|b - c|$ (even).
Sum of two even numbers is even: $(a - b) + (b - c) = a - c$.
So $a - c$ is even $\Rightarrow |a - c|$ is even.
$\therefore (a, c) \in R$. Transitive.
Q5. Show that the function $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^3$ is bijective.
Let $x_1^3 = x_2^3$. Taking cube root on both sides.
$x_1 = x_2$. (No $\pm$ issue for odd powers)
$\therefore$ One-One.
Let $y \in \mathbb{R}$. We need $x^3 = y \Rightarrow x = y^{1/3}$.
Is cube root of any real number real? Yes (e.g., $\sqrt[3]{-8} = -2$).
$\therefore$ Range = $\mathbb{R}$. Onto.
Since One-One & Onto, it is Bijective.
Level 3: Advanced (4-5 Marks Each)
Q6. Let $f: \mathbb{R} - \{-\frac{4}{3}\} \to \mathbb{R}$ be a function defined as $f(x) = \frac{4x}{3x + 4}$. Show that $f$ is one-one. Find the inverse of $f$.
Let $f(x_1) = f(x_2)$
$\frac{4x_1}{3x_1 + 4} = \frac{4x_2}{3x_2 + 4}$
Cross-multiply: $4x_1(3x_2 + 4) = 4x_2(3x_1 + 4)$
$12x_1x_2 + 16x_1 = 12x_1x_2 + 16x_2$
$16x_1 = 16x_2 \Rightarrow x_1 = x_2$
$\therefore$ One-One.
Let $y = \frac{4x}{3x + 4}$. We need to find $x$.
$y(3x + 4) = 4x$
$3xy + 4y = 4x$
$4y = 4x - 3xy = x(4 - 3y)$
$x = \frac{4y}{4 - 3y}$
$\therefore$ Inverse is $f^{-1}(y) = \frac{4y}{4 - 3y}$.
Q7. Consider $f: \{1, 2, 3\} \to \{a, b, c\}$ and $g: \{a, b, c\} \to \{apple, ball, cat\}$ defined as $f(1) = a, f(2) = b, f(3) = c$ and $g(a) = apple, g(b) = ball, g(c) = cat$. Show that $f, g$ and $gof$ are invertible. Find $(gof)^{-1}$.
For $f$: 1-to-a, 2-to-b, 3-to-c (Bijective) $\Rightarrow$ Invertible.
For $g$: a-to-apple, b-to-ball, c-to-cat (Bijective) $\Rightarrow$ Invertible.
$(gof)(1) = g(f(1)) = g(a) = apple$
$(gof)(2) = g(f(2)) = g(b) = ball$
$(gof)(3) = g(f(3)) = g(c) = cat$
$\therefore gof$ is also Bijective $\Rightarrow$ Invertible.
Just reverse the arrows:
$(gof)^{-1} = \{(apple, 1), (ball, 2), (cat, 3)\}$
Formula Sheet (Must Remember!)
Relations
1. Empty Relation: $R = \phi \subset A \times A$2. Universal Relation: $R = A \times A$
3. Reflexive: $(a, a) \in R$ for all $a \in A$
4. Symmetric: $(a, b) \in R \Rightarrow (b, a) \in R$
5. Transitive: $(a, b), (b, c) \in R \Rightarrow (a, c) \in R$
6. Equivalence Relation: Reflexive + Symmetric + Transitive
Functions
7. One-One: $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$8. Onto: Range = Co-domain
9. Bijective: One-One + Onto
10. Composition: $(gof)(x) = g(f(x))$
11. Invertible: $f$ is invertible $\Leftrightarrow$ $f$ is bijective
12. Inverse Property: $(gof)^{-1} = f^{-1}og^{-1}$
13. Identity: $fof^{-1} = I_Y$ and $f^{-1}of = I_X$
Key Theorems:
• Intersection of two equivalence relations is also an equivalence relation.
• For finite set $A$, $f: A \to A$ is one-one $\Leftrightarrow$ $f$ is onto.