Practice Exercise 1.2

Revise Chapter Notes

Overview

This page provides an Overview of Chapter 1 Relations And Functions for CBSE Class 12. Key topics covered: Q2. Injectivity & Surjectivity Check, Q3 - Q5. Special Functions, Q6 - Q8. Sets & Functions, Q9 - Q12. Advanced Functions. All content is aligned with the NCERT syllabus for Board Exam preparation.

Functions: One-one (Injective), Onto (Surjective) & Bijective

Chapter Index Start Solving
Q1 Reciprocal Q2 Basic Types Q3 GIF Q4 Modulus Q5 Signum Q6 Sets Q7 Linear/Quad Q8 Product Q9 Natural Q10 Rational Q11 MCQ Q12 MCQ

Q1. Reciprocal Function

Similar to Q1
Show that f: R* → R* defined by f(x) = 2/x is one-one and onto, where R* is the set of all non-zero real numbers. Is the result true if the domain R* is replaced by N with co-domain being same as R*?
View Step-by-Step Solution
Step 1: Check One-One (Injectivity)
Let x₁, x₂ ∈ R* such that f(x₁) = f(x₂).
2/x₁ = 2/x₂
⇒ x₁ = x₂
Conclusion: f is one-one.
Step 2: Check Onto (Surjectivity)
Let y ∈ R*. We need to find x ∈ R* such that f(x) = y.
2/x = y ⇒ x = 2/y.
Since y ≠ 0, 2/y is a defined non-zero real number.
So, x ∈ R* exists.
Conclusion: f is onto.
Step 3: Domain Change Case
New function g: N → R* where g(x) = 2/x.
One-One: Still true (same logic).
Onto: Take y = 3 (which is in R*).
We need x ∈ N such that 2/x = 3 ⇒ x = 2/3.
But 2/3 is NOT a natural number (∉ N).
Conclusion: Not onto.
Answer: Original f is bijective. Modified g is one-one but NOT onto.

Q2. Injectivity & Surjectivity Check

Similar to Q2(i)
f: N → N given by f(x) = x² + 1
View Solution
Injectivity
Let f(x₁) = f(x₂). x₁² + 1 = x₂² + 1 ⇒ x₁² = x₂².
Since domain is N, x₁, x₂ > 0, so x₁ = x₂.
One-One: Yes.
Surjectivity
Range of f = {2, 5, 10, 17...}.
Numbers like 3, 4 ∈ N are not images of any x ∈ N (e.g., x² + 1 = 3 ⇒ x² = 2 ⇒ x = √2 ∉ N).
Onto: No.
One-One but Not Onto.
Similar to Q2(ii)
f: Z → Z given by f(x) = x² + 1
View Solution
Injectivity
Take x₁ = 1, x₂ = -1. Both are in Z.
f(1) = 1² + 1 = 2.
f(-1) = (-1)² + 1 = 2.
Different elements have same image.
One-One: No.
Surjectivity
Range only includes positive integers ≥ 1. Negative integers in Z are not covered.
Onto: No.
Neither One-One nor Onto.
Similar to Q2(iii)
f: R → R given by f(x) = x² + 1
View Solution
Analysis
Similar to Z case, f(1) = f(-1) = 2. Not One-One.
Range is [1, ∞). Negative numbers in Co-domain R are not covered. Not Onto.
Neither One-One nor Onto.
Similar to Q2(iv)
f: N → N given by f(x) = x³ + 1
View Solution
Injectivity
x₁³ + 1 = x₂³ + 1 ⇒ x₁³ = x₂³ ⇒ x₁ = x₂.
One-One: Yes.
Surjectivity
Range = {2, 9, 28...}. Many natural numbers (e.g., 3, 4) are skipped.
Onto: No.
One-One but Not Onto.
Similar to Q2(v)
f: Z → Z given by f(x) = x³ + 1
View Solution
Injectivity
Cube function preserves sign. Distinct integers have distinct cubes.
One-One: Yes.
Surjectivity
We need x ∈ Z such that x³ + 1 = y. So x = ³√(y-1).
If y = 2, x = ³√(1) = 1 (Integer).
If y = 3, x = ³√(2) (Not an Integer).
Onto: No.
One-One but Not Onto.

Q3 - Q5. Special Functions

Similar to Q3
Prove f: R → R defined by f(x) = [x] (Greatest Integer Function) is neither one-one nor onto.
View Solution
Step 1: One-One Check
f(1.2) = [1.2] = 1.
f(1.9) = [1.9] = 1.
Different inputs (1.2 ≠ 1.9) give same output.
Not One-One.
Step 2: Onto Check
Range of [x] is always Z (Integers).
Non-integers in Co-domain R (e.g., 2.5) have no pre-image.
Not Onto.
Neither One-One nor Onto.
Similar to Q4
Show f: R → R defined by f(x) = |x| + 1 is neither one-one nor onto.
View Solution
One-One Check
f(-2) = |-2| + 1 = 3.
f(2) = |2| + 1 = 3.
Two inputs map to 3.
Not One-One.
Onto Check
Since |x| ≥ 0, f(x) = |x| + 1 ≥ 1.
Range is [1, ∞). Numbers < 1 in Co-domain R are not covered.
Not Onto.
Neither One-One nor Onto.
Similar to Q5
Show that Signum Function f: R → R given by f(x) = 1 if x > 0, 0 if x = 0, -1 if x < 0 is neither one-one nor onto.
View Solution
Step 1: One-One Check
f(5) = 1.
f(100) = 1.
Many inputs map to single output '1'.
Not One-One (Many-One).
Step 2: Onto Check
Range = {-1, 0, 1}.
Any real number other than these (e.g., 2, 5.5) has no pre-image.
Not Onto.
Neither One-One nor Onto.

Q6 - Q8. Sets & Functions

Similar to Q6
Let A = {1, 2}, B = {3, 4, 5}. f = {(1, 3), (2, 4)}. Show f is one-one.
View Solution
Check Mappings
f(1) = 3.
f(2) = 4.
Distinct elements of A have distinct images in B.
Conclusion: f is one-one.
Proven: f is One-One.
Similar to Q7
Check if bijective: (i) f: R → R, f(x) = 2x + 5. (ii) f: R → R, f(x) = 3 - x².
View Solution
Part (i): f(x) = 2x + 5
One-One: 2x₁ + 5 = 2x₂ + 5 ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂. (Yes)
Onto: y = 2x + 5 ⇒ x = (y-5)/2. For any real y, x is real. (Yes)
Result: Bijective.
Part (ii): f(x) = 3 - x²
One-One: f(1) = 2, f(-1) = 2. Not one-one.
Onto: Range is (-∞, 3]. Numbers > 3 not covered. Not onto.
Result: Not Bijective.
(i) Bijective, (ii) Not Bijective.
Similar to Q8
Let A = {1, 2}, B = {3, 4}. f: A × B → B × A defined by f(a, b) = (b, a). Is f bijective?
View Solution
One-One Check
Let f(a₁, b₁) = f(a₂, b₂).
(b₁, a₁) = (b₂, a₂) ⇒ b₁ = b₂ and a₁ = a₂.
So (a₁, b₁) = (a₂, b₂). Yes.
Onto Check
For any (b, a) in Co-domain B × A, there exists (a, b) in Domain A × B such that f(a, b) = (b, a).
Yes.
f is a Bijective function.

Q9 - Q12. Advanced Functions

Similar to Q9
f: N → N defined by f(n) = (n+1)/2 if n is odd, n/2 if n is even. State if bijective.
View Solution
One-One Check
f(1) = (1+1)/2 = 1.
f(2) = 2/2 = 1.
f(1) = f(2), but 1 ≠ 2.
Not One-One.
Onto Check
For any natural number y, consider 2y (even). f(2y) = 2y/2 = y.
So every y has a pre-image.
Is Onto.
Not Bijective (Many-One Onto).
Similar to Q10
A = R - {2}, B = R - {1}. f: A → B defined by f(x) = (x-1)/(x-2). Is f one-one and onto?
View Solution
One-One
(x₁-1)/(x₁-2) = (x₂-1)/(x₂-2)
Cross multiply: x₁x₂ - 2x₁ - x₂ + 2 = x₁x₂ - x₁ - 2x₂ + 2
-x₁ = -x₂ ⇒ x₁ = x₂.
Yes.
Onto
Let y = (x-1)/(x-2). Solve for x.
y(x-2) = x-1 ⇒ xy - 2y = x - 1
x(y-1) = 2y - 1 ⇒ x = (2y-1)/(y-1).
Since y ∈ B (y ≠ 1), x is defined. Also check if x ≠ 2 (if x=2, then 2y-1 = 2y-2 ⇒ -1=-2 impossible).
So x exists in A.
Yes.
f is One-One and Onto.
Similar to Q11
Let f: R → R be defined as f(x) = x⁴. Choose correct answer:
(A) One-one onto (B) Many-one onto (C) One-one not onto (D) Neither.
View Solution
Analysis
One-One: f(1) = 1, f(-1) = 1. No.
Onto: x⁴ cannot be negative. Range [0, ∞) ≠ R. No.
Option (D) Neither one-one nor onto.
Similar to Q12
Let f: R → R be defined as f(x) = 5x. Choose correct answer:
(A) One-one onto (B) Many-one onto (C) One-one not onto (D) Neither.
View Solution
Analysis
One-One: 5x₁ = 5x₂ ⇒ x₁ = x₂. Yes.
Onto: y = 5x ⇒ x = y/5. For any real y, y/5 is real. Yes.
Option (A) One-one onto.
Exercise 1.1 Chapter 1 Miscellaneous