Practice Misc. Exercise

Overview

This page provides an Overview of Chapter 1 Relations And Functions for CBSE Class 12. Key topics covered: Q2. Injectivity, Q3. Relations on Sets, Q4. Counting Functions, Q5. Equality of Functions, Q6. Counting Relations (Reflexive), Q7. Counting Equivalence Relations. All content is aligned with the NCERT syllabus for Board Exam preparation.

Advanced concepts of Functions & Relations: Bijective mappings, Equivalence relations, and Function Equality.

Chapter Index Start Solving

Q1 One-One Q2 Injective Q3 Equivalence Q4 Bijective Q5 Equal Func Q6 Relation#1 Q7 Relation#2

Q1. Bijective Check

Practice Question 1

Let f: [0, ∞) → [0, 1) be defined by f(x) = x / (1 + x). Show that f is one-one and onto.

View Step-by-Step Solution

Step 1: Check One-One (Injectivity)

Let x₁, x₂ ∈ [0, ∞) such that f(x₁) = f(x₂).
x₁/(1+x₁) = x₂/(1+x₂)
x₁(1+x₂) = x₂(1+x₁) ⇒ x₁ + x₁x₂ = x₂ + x₂x₁
Subtract x₁x₂ from both sides: x₁ = x₂.
Conclusion: f is one-one.

Step 2: Check Onto (Surjectivity)

Let y ∈ [0, 1). We need to find x.
y = x / (1+x) ⇒ y(1+x) = x ⇒ y + xy = x
y = x - xy ⇒ y = x(1 - y) ⇒ x = y / (1 - y)
Since 0 ≤ y < 1, the denominator (1-y) is positive and ≤ 1.
Also y ≥ 0, so x ≥ 0. Thus x ∈ [0, ∞) exists for every y.
Conclusion: f is onto.

Answer: The function is One-One and Onto (Bijective).

Q2. Injectivity

Practice Question 2

Show that the function f: R → R given by f(x) = x³ + 5 is injective (one-one).

View Step-by-Step Solution

Step 1: Set up Condition

Let x₁, x₂ ∈ R such that f(x₁) = f(x₂).
x₁³ + 5 = x₂³ + 5

Step 2: Simplify

Subtract 5 from both sides: x₁³ = x₂³
Since the cube function preserves signs and magnitude uniquely for real numbers, taking the cube root gives:
x₁ = x₂

Answer: Since f(x₁) = f(x₂) implies x₁ = x₂, f is Injective.

Q3. Relations on Sets

Practice Question 3

Given a non-empty set X, consider P(X), the set of all subsets of X. Define relation R in P(X) as: A R B if and only if A ∪ B = B. Is R an equivalence relation?

View Step-by-Step Solution

Step 1: Understand Relation

A ∪ B = B is mathematically equivalent to A ⊆ B (A is a subset of B).

Step 2: Check Reflexive

Is A ⊆ A? Yes, every set is a subset of itself.
Reflexive: Yes.

Step 3: Check Symmetric

If A ⊆ B, does it imply B ⊆ A? No.
Example: A={1}, B={1,2}. A ⊆ B but B ⊈ A.
Symmetric: No.

Step 4: Check Transitive

If A ⊆ B and B ⊆ C, then A ⊆ C. This is a standard property of sets.
Transitive: Yes.

Answer: No, R is not an Equivalence relation (it is not Symmetric).

Q4. Counting Functions

Practice Question 4

Find the number of bijective (one-one and onto) functions from the set A = {1, 2, 3, 4} to itself.

View Step-by-Step Solution

Step 1: Concept

A bijective function from a finite set to itself is essentially a permutation of its elements.
For a set with n elements, the number of bijections is n! (n factorial).

Step 2: Calculation

Here, n = 4 (elements are 1, 2, 3, 4).
Number of functions = 4! = 4 × 3 × 2 × 1.

Answer: 24.

Q5. Equality of Functions

Practice Question 5

Let A = {-1, 1}. Let f, g: A → R be defined by f(x) = x² and g(x) = |x|. Are f and g equal? Justify.

View Step-by-Step Solution

Step 1: Check Domain & Codomain

Both functions have the same domain A = {-1, 1} and same codomain R.

Step 2: Check Values

For x = 1: f(1) = 1² = 1. g(1) = |1| = 1. (Match)
For x = -1: f(-1) = (-1)² = 1. g(-1) = |-1| = 1. (Match)

Step 3: Conclusion

Since f(x) = g(x) for all x ∈ A, the functions are equal.

Answer: Yes, f and g are equal functions.

Q6. Counting Relations (Reflexive)

Practice Question 6

Find the number of relations on the set A = {1, 2, 3} containing (1, 2) which are reflexive and symmetric but NOT transitive.

View Step-by-Step Solution

Step 1: Mandatory Pairs

For Reflexive: Must contain {(1,1), (2,2), (3,3)}.
For Symmetric & containing (1,2): Must contain {(1,2), (2,1)}.

Step 2: Building Relations

Current Base: R₁ = {(1,1), (2,2), (3,3), (1,2), (2,1)}. (This is transitive).
We need to add pairs to break transitivity but keep symmetry.
Option A: Add (2,3) and (3,2).
R₂ = Base ∪ {(2,3), (3,2)}.
Check Transitivity: We have (1,2) and (2,3). Is (1,3) in R₂? No.
So R₂ is Reflexive, Symmetric, Not Transitive.

Step 3: Other Options

Option B: Add (1,3) and (3,1).
R₃ = Base ∪ {(1,3), (3,1)}.
Check Transitivity: We have (2,1) and (1,3). Is (2,3) in R₃? No.
So R₃ is also Reflexive, Symmetric, Not Transitive.
Note: Adding all pairs creates the Universal relation, which IS transitive.

Answer: There are 2 such relations.

Q7. Counting Equivalence Relations

Practice Question 7

Let A = {1, 2, 3}. Find the number of equivalence relations containing (1, 2) and (2, 3).

View Step-by-Step Solution

Step 1: Construct Smallest Relation

Reflexive: {(1,1), (2,2), (3,3)}.
Given: (1,2), (2,3).
Symmetric: Must add (2,1), (3,2).
Transitive: (1,2) & (2,3) ⇒ Must add (1,3). Symmetry ⇒ Must add (3,1).

Step 2: Analyze Result

The relation now contains: {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2), (1,3), (3,1)}.
This includes ALL possible pairs in A × A.
This is the Universal Relation.

Step 3: Count

There are no smaller equivalence relations containing both (1,2) and (2,3) because transitivity forces all other pairs to be included.

Answer: Only 1 such relation exists.