Ncert Solutions Class 12 Maths Chapter 1 Relations And Functions Ex 1 2

The Relations and Functions Class 12 NCERT Solutions for Class 12 Mathematics Chapter 1 Relations and Functions Exercise 1.2 present a full step-by-step solutions to every question in this set. Each step in the Relations and Functions Class 12 NCERT Solutions states the rule used and shows the calculation in line, in agreement with the 2026-27 NCERT method.

• CBSE Weightage: 8-10 marks (with Ch 1 + Inverse Trig combined)
• JEE Main Weightage: 3-4% (functions block)
• Exercise 1.2 Question Count: 12 questions

The Exercise 1.2 PDF is built so that each of the 12 questions is solved in under one page, with the injective check and surjective check separated visually. Students preparing for CBSE Class 12 boards 2026-27 should attempt Q1, Q7, and Q9 first - these match the phrasing of the most-repeated board-style "prove one-one and onto" question.

How will Collegedunia's NCERT Solutions for Exercise 1.2 help you?

The Relations and Functions Class 12 NCERT Solutions address this in the same order as the NCERT textbook.

Exercise 1.2 is where students first meet formal proofs of injective and surjective behaviour. Most learners can quote the definitions but stumble when asked to write the proof on paper. Collegedunia's solutions PDF fixes that by following a three-line proof skeleton on every question:

assume f(x1) = f(x2) -> simplify -> conclude x1 = x2 for one-one; and take arbitrary y in codomain -> solve for x -> check x lies in domain for onto.

You will see the difference inside two weeks of practice. the the PDF also flags every place where the codomain choice (R, R*, N, Z) changes the answer, which is the single most common mistake in board-paper Q3-style questions.

Relations and Functions Exercise 1 2 Video Walkthrough

Source: Magnet Brains on YouTube

Topics Covered in Class 12 Maths Exercise 1.2

The this chapter address this in the same order as the NCERT textbook.

Exercise 1.2 sits under section 1.2 of the NCERT textbook (Types of Functions). The table below maps each block of questions to the underlying concept, so you can revise selectively before a class test.

Treat Q7, Q9, Q10 and Q12 as your priority four; these four templates cover every question style you will face in school exams and the CUET-UG functions section.

Class 12 Maths Exercise 1.2 Solution Approach Summary

These notes address this in the same order as the NCERT textbook.

Below is the compressed solution map for the 12 questions. Use it as a self-check after attempting the exercise on your own. The full step-by-step working is inside the downloadable PDF.

Notice the pattern: every odd-degree polynomial on R is bijective; every even-degree polynomial on R fails both injective and surjective tests because of symmetry around y-axis and non-negative range. Internalise this and you will answer Q2, Q8 and Q11 in under two minutes each.

Important Definitions Used in Exercise 1.2

The Collegedunia PDF places these three definitions at the top of every solution page, so you never have to flip back to section 1.2 of the this Class 12 page while practising.

Common Mistakes Students Make in Exercise 1.2

The the resource are written in formal mathematical notation, line by line, in the same convention as the official NCERT print.

• Ignoring the codomain. The same formula f(x) = x^2 is one-one on N but not on Z. Students who skip the "given codomain" line in the question lose 1-2 marks every time.
• Skipping the algebraic step. Writing "clearly one-one" is not a proof. The marker expects f(x1) = f(x2) followed by algebraic simplification to x1 = x2.
• Forgetting to verify x lies in the domain after solving f(x) = y. In Q10, the solved x must explicitly avoid x = 3.
• Confusing range with codomain. Range is the set of actually-achieved outputs; codomain is the declared target set. Onto means the two are equal.
• Treating piecewise functions as a single rule in Q9. You must split into the odd-n branch and even-n branch and check injectivity across both.

The PDF's expert solution callouts tag each of these traps right next to the relevant question, so you build the habit of avoiding them.

Related Exercises in Chapter 1 Relations and Functions

Exercise 1.2 sits between the relation-properties of Exercise 1.1 and the composition / inverse content of Miscellaneous Exercise. Use the cross-links below to jump to the rest of the chapter notes on this resource.

• Exercise 1.1: Types of relations - reflexive, symmetric, transitive, equivalence relations
• Exercise 1.2: Types of functions - one-one, onto, bijective (the the PDF)
• Miscellaneous Exercise: Mixed relations + functions, composition

Full chapter solutions: NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Exam Relevance of Exercise 1.2 in Class 12 Boards 2026-27

The functions block of Chapter 1 routinely contributes a 3-mark or 5-mark short answer question in CBSE Class 12 Maths board papers. Exercise 1.2-style "prove one-one and onto" questions appeared in 2024, 2023 and 2020 board papers, typically as Question 27 or 29 of the standard 38-question paper.

For JEE Main 2026 aspirants, the same concepts are tested as MCQ traps - the option set usually includes "one-one but not onto" as a distractor. Practise Q2 sub-parts carefully; that is the JEE Main bait pattern.

this resource Class 12 Maths Resources for Chapter 1

Pair the Exercise 1.2 solutions with the rest of the this resource Chapter 1 library for a complete preparation cycle.

Students who use the Solutions PDF together with the Handwritten Notes for last-week revision tend to score 1-2 marks higher on the Chapter 1 short-answer questions.

this chapter: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.

Exercise-wise Breakdown of the Relations and Functions Chapter

The Relations and Functions chapter splits into 2 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.

PDF Download Formats and Languages for the Relations and Functions Chapter

The Relations and Functions Class 12 PDF on this page is available in three formats - each suited to a different revision style. The table below summarises what each format is best for:

The relations and functions class 12 ncert pdf and the parallel Hindi-medium edition both follow the same notation and equation numbering as the printed NCERT 2026-27 release. Key points students should know:

• NCERT-faithful: Every definition, theorem and exercise on the relations and functions class 12 ncert pdf matches the printed textbook line for line.
• Hindi-medium edition: The relations and functions class 12 pdf is also available in Hindi - same page numbering, same equation labels.
• Formula PDF separate: The relations and functions class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
• Solutions PDF separate: The relations and functions class 12 solutions pdf gives every NCERT exercise worked out step by step.
• State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this relations and functions class 12 pdf - only the exercise numbers differ.

Tip: Many toppers keep two parallel copies - a printed formula sheet on A4 for desk revision (the relations and functions class 12 formulas pdf), and the full relations and functions class 12 pdf on a phone for commute revision. Both files are free and linked above.

Important Questions and Previous Year Trends for the Relations and Functions Chapter

The most repeated question patterns in CBSE Class 12 Maths for the Relations and Functions chapter have settled into a stable cluster across 2019 to 2024 boards. Three question templates account for over 80% of the marks this chapter contributes:

Walking through one example of each template before the exam covers most of the predictable relations and functions class 12 important questions you will see on board day.

• relations and functions class 12 previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
• The relations and functions class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
• For NCERT Exemplar practice, the matching relations and functions class 12 extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
• The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.

Year-wise PYQ Distribution

The table below maps the dominant question type asked from the Relations and Functions chapter across recent CBSE Class 12 Maths boards:

The full relations and functions class 12 important questions with solutions set (every year, every paper, every question type) is linked from the PYQ page at the bottom of this article.

How the Relations and Functions Notes Pair with NCERT Solutions and the Formula Sheet

The Relations and Functions Class 12 notes work best when paired with two sister resources from the Class 12 Maths hub. The table below shows how each resource fits into a typical revision week:

Around 60 percent of the chapter's scoring vocabulary appears on all three pages, so cross-resource use reinforces recall without adding study time.

• The this chapter cover every back-of-chapter exercise plus the miscellaneous exercise.
• The relations and functions class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
• The relations and functions class 12 formulas reference sheet is the same A4 file students sometimes refer to as relations and functions class 12 all formulas - it lists every identity used in the chapter.
• State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board relations and functions class 12 textbook PDF all share the same core definitions.
• For class-first search phrasings - class 12 relations and functions solutions, class 12 relations and functions ncert solutions, ncert class 12 relations and functions solutions - the same files cover the request.

Reference Books and State-Board Mapping

Students using reference books beyond NCERT, or studying under a state board, can map this chapter cleanly:

NCERT Solutions for Class 12 Mathematics: All Chapters

Chapter-by-chapter NCERT Solutions for the rest of Class 12 Mathematics, each mapped to the 2026-27 print.

All NCERT Solutions for Relations and Functions Ex 1.2 with Step-by-Step Working

Every NCERT textbook question for Class 12 Mathematics Chapter 1 Relations and Functions Ex 1.2 is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.

Questions

Q 1.1

Show that the function f : R_∗ → R_∗ defined by f(x) = 1x 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_∗?

Check SolutionExpert Solution

Concept used. A function f : X → Y is one-one (injective) when f(x₁) = f(x₂) implies x₁ = x₂ for every x₁, x₂ ∈ X. It is onto (surjective) when, for every y ∈ Y, there exists at least one x ∈ X with f(x) = y. Here R_∗ = R 0.

1. One-one on R_∗. Suppose f(x₁) = f(x₂). Then 1x₁ = 1x₂. Cross-multiplying (valid because x₁, x₂ ≠ 0): x₂ = x₁. Hence x₁ = x₂. So f is one-one.
2. Onto on R_∗. Take any y ∈ R_∗. We seek x ∈ R_∗ with f(x) = y. Solve 1x = y: x = 1y. Since y ≠ 0, 1y is a non-zero real number, so x ∈ R_∗. Verify: f(x) = 1x = 11/y = y. So every y has a pre-image. f is onto.
3. Now replace domain by N, keeping codomain R_∗. Define g : N → R_∗ by g(n) = 1n.

   Injective? If g(n₁) = g(n₂) then 1n₁ = 1n₂ ⇒ n₁ = n₂. Yes, injective.

   Onto? Take y = 12. We need n ∈ N with 1n = 12, i.e. n = 2. That works. But take y = 2 (which is in R_∗). We need n ∈ N with 1n = 2, i.e. n = 12. But 12 ∉ N. So y = 2 has no pre-image. g is not onto.

f : R_∗ → R_∗ is bijective. With domain replaced by N, the map is one-one but not onto.

KR

Karan Reddy

M.Sc Mathematics, IIT Madras

Verified Expert

Picture-first. The graph of y = 1/x on R_∗ is a hyperbola with two branches. Every horizontal line y = c ≠ 0 cuts the graph in exactly one point, witnessing bijectivity.

[See diagram in the PDF version]

1. For R_∗ → R_∗: pick any y ≠ 0, the line y = c hits the hyperbola at exactly one x = 1/c. Injective and surjective.
2. For N → R_∗: the image becomes the discrete set 1, 12, 13, …, a countable subset of R_∗. The value 2 ∈ R_∗ has no pre-image.

Why this matters. Restricting domain changes the image; restricting codomain affects only surjectivity. Surjective and injective behaviours can be independently turned off.

Bijective on R_∗; one-one but not onto when the domain shrinks to N.

Q 1.2

Check the injectivity and surjectivity of the following functions:
(i) f : N → N given by f(x) = x²;   (ii) f : Z → Z given by f(x) = x²;
(iii) f : R → R given by f(x) = x²;   (iv) f : N → N given by f(x) = x³;
(v) f : Z → Z given by f(x) = x³.

Check SolutionExpert Solution

Concept used. Squaring on R is even (f(-x) = f(x)) and non-negative, so it loses information about sign. Cubing is odd (f(-x) = -f(x)) and strictly increasing, so it is injective on R. Restricting the domain to N removes the sign issue but leaves a range-codomain mismatch.

1. (i) f(x) = x², N → N. Injective? f(x₁) = f(x₂) ⇒ x₁² = x₂² ⇒ x₁ = ± x₂. On N both are positive, so x₁ = x₂. Injective.
   Surjective? Need every n ∈ N to be a perfect square. But 2 ∈ N has no integer x with x² = 2. Not surjective.
2. (ii) f(x) = x², Z → Z. Injective? f(-1) = 1 = f(1), but -1 ≠ 1. Not injective.
   Surjective? -1 ∈ Z has no integer x with x² = -1 (squares are non-negative). Not surjective.
3. (iii) f(x) = x², R → R. Injective? f(-2) = 4 = f(2), -2 ≠ 2. Not injective.
   Surjective? -1 ∈ R has no real x with x² = -1. Not surjective.
4. (iv) f(x) = x³, N → N. Injective? f(x₁) = f(x₂) ⇒ x₁³ = x₂³ ⇒ x₁ = x₂ (cubing is one-one on R, hence on N). Injective.
   Surjective? 2 ∈ N has no x ∈ N with x³ = 2 (would need x = 2^1/3, irrational). Not surjective.
5. (v) f(x) = x³, Z → Z. Injective? Same as (iv); injective.
   Surjective? 2 ∈ Z has no integer x with x³ = 2. Not surjective.

(i) Injective, not surjective. (ii) Neither. (iii) Neither. (iv) Injective, not surjective. (v) Injective, not surjective.

AM

Aanya Mehta

M.Sc Mathematics, ISI Kolkata

Verified Expert

Pattern table. Stitch the five subcases into a 2 × 5 table by domain and rule.

1. For squaring on N, the codomain is too big (image = n² : n ∈ N, misses 2, 3, 5, 6, …). Negative codomain values are absent in N, so injectivity is restored.
2. For squaring on Z or R, negative pre-images appear: -x and +x collide. Both injectivity and surjectivity fail.
3. For cubing: strict monotonicity (x₁ < x₂ ⇒ x₁³ < x₂³) on R gives injectivity automatically. Surjectivity from N or Z to N or Z fails because non-cube integers (like 2, 3, 4) lie in the codomain but not the image.

Why this matters. The lesson: behaviour of f depends on three things simultaneously: rule, domain, codomain. Changing any one can flip injectivity or surjectivity.

Only cubing maps are injective; none of the five are surjective.

Q 1.3

Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

Check SolutionExpert Solution

Concept used. The greatest integer function [x] (also called the floor function) rounds x down to the nearest integer: [1.7] = 1, [1] = 1, [-1.3] = -2, [-1] = -1. Its image is exactly Z, a proper subset of R.

1. Not one-one. Choose x₁ = 1.2 and x₂ = 1.7. f(1.2) = [1.2] = 1, f(1.7) = [1.7] = 1. So f(1.2) = f(1.7) = 1, but 1.2 ≠ 1.7. Hence f is not one-one.
2. Not onto. Pick y = 0.5 ∈ R. We need x ∈ R with [x] = 0.5. But [x] always returns an integer, never 0.5. So no x exists. f is not onto.

[See diagram in the PDF version]

f(x) = [x] is neither one-one nor onto.

IV

Ishaan Verma

M.Sc Mathematics, IIT Kanpur

Verified Expert

Graph reading. The graph of y = [x] is a staircase: a flat segment over every half-open interval [k, k+1) at height k. Two readings of the staircase reveal both failures.

1. Many points on the same flat segment map to the same integer: e.g. every x ∈ [1, 2) maps to 1. Hence injectivity fails throughout.
2. Horizontal lines y = c with c non-integer never hit the staircase. So values like 0.5, 1.7, -π have no pre-image. Surjectivity fails everywhere off Z.
3. The image of f is exactly Z. To restore surjectivity, restrict the codomain to Z. To restore injectivity, restrict the domain to Z (then f(n) = n, the identity).

Why this matters. The floor function is the foundation of integer-part identities in number theory: x + x + 12 = 2x , and Beatty's theorem on irrational ratios.

Neither one-one nor onto.

Q 1.4

Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is -x, if x is negative.

Check SolutionExpert Solution

Concept used. The modulus or absolute value of x is |x| = cases x & if x ≥ 0 -x & if x < 0 cases It is always non-negative; its image is [0, ∞), a proper subset of R.

1. Not one-one. Pick x₁ = -1, x₂ = 1. f(-1) = |-1| = 1, f(1) = |1| = 1. So f(-1) = f(1) = 1 but -1 ≠ 1. Not one-one.
2. Not onto. Pick y = -3 ∈ R. We need x with |x| = -3. But |x| ≥ 0 always, so |x| = -3 has no solution. Hence y = -3 has no pre-image. Not onto.

[See diagram in the PDF version]

f(x) = |x| is neither one-one nor onto.

TN

Tara Nair

M.Sc Mathematics, IIT Madras

Verified Expert

Symmetry argument. The modulus function is even: f(-x) = f(x). Even non-constant functions are never one-one. The image is the non-negative reals, so any negative codomain value is missed.

1. Even symmetry: f(-1) = f(1), f(-2) = f(2), … a fresh collision at every non-zero x. Injectivity fails everywhere except at x = 0.
2. Image = [0, ∞): half the codomain R is missed. Surjectivity fails for every negative number.
3. Restricting domain to [0, ∞) restores injectivity (then f acts as identity). Restricting codomain to [0, ∞) restores surjectivity.

Why this matters. Triangle inequality, distance, and norm are all built on the modulus. Recognising its evenness keeps you from mistakenly invertings it without restriction.

Neither one-one nor onto.

Q 1.5

Show that the Signum Function f : R → R, given by f(x) = cases 1, & if x > 0 0, & if x = 0 -1, & if x < 0 cases is neither one-one nor onto.

Check SolutionExpert Solution

Concept used. The signum function returns +1 for positive inputs, -1 for negative inputs, and 0 for x = 0. Its image is the three-element set -1, 0, 1.

1. Not one-one. Pick x₁ = 2, x₂ = 5. Both are positive, so f(2) = 1 = f(5), but 2 ≠ 5. Not one-one.
2. Not onto. Pick y = 2 ∈ R. The image is -1, 0, 1, which doesn't include 2. So no x has f(x) = 2. Not onto.

The signum function is neither one-one nor onto.

AB

Aditi Banerjee

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Image-set argument. Whenever the image of a function is finite but the codomain is infinite, surjectivity is automatic to disprove (pick any codomain value outside the image). Whenever the image is finite but the domain is infinite, injectivity is automatic to disprove (pigeonhole: infinite many inputs share finitely many outputs).

1. Image of sgn is -1, 0, 1, three elements; domain R is uncountable. By pigeonhole, infinitely many inputs collide. Injectivity fails.
2. Codomain R has more than three values. Any y ∉ -1, 0, 1 has no pre-image. Surjectivity fails.

Why this matters. Pigeonhole-style arguments based on cardinality of image are a quick and clean way to disprove both injectivity and surjectivity in one stroke.

Neither one-one nor onto.

Q 1.6

Let A = 1, 2, 3, B = 4, 5, 6, 7 and let f = (1, 4), (2, 5), (3, 6) be a function from A to B. Show that f is one-one.

Check SolutionExpert Solution

Concept used. A function on a finite set is one-one iff distinct domain elements have distinct images. Equivalently, when written as a set of pairs, all second coordinates are distinct.

1. Read off the images: f(1) = 4, f(2) = 5, f(3) = 6.
2. Check pairwise: 4, 5, 6 are three different elements of B.
3. Equivalently: if f(x₁) = f(x₂), the pair x₁, x₂ must lie in 1, 2, 3 with f(x₁) = f(x₂). From the list, the only way two domain elements have the same image would be for two distinct rows of f to share a second coordinate, which they don't.

f is one-one.

YK

Yash Kapoor

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Pair-listing. On finite sets, the cleanest injectivity check is to list the images and verify they are mutually distinct.

1. Images: 4, 5, 6 has three distinct elements.
2. Domain has three elements; the function delivers three distinct images, so no two inputs share an output.
3. As a bonus, f is not onto: the codomain B = 4, 5, 6, 7 has four elements; 7 has no pre-image.

f is one-one (and not onto, since the image excludes 7).

Q 1.7

In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 - 4x.
(ii) f : R → R defined by f(x) = 1 + x².

Check SolutionExpert Solution

Concept used. A non-constant linear map f(x) = ax + b with a ≠ 0 is a bijection on R (strictly monotonic and ranging over all of R). A quadratic f(x) = 1 + x² is even and bounded below by 1, so neither one-one nor onto on R.

1. (i) f(x) = 3 - 4x. One-one? Suppose f(x₁) = f(x₂): 3 - 4x₁ = 3 - 4x₂ ⇒ -4x₁ = -4x₂ ⇒ x₁ = x₂. One-one.
   Onto? Take any y ∈ R. Solve y = 3 - 4x for x: 4x = 3 - y, x = 3 - y4. This x is a real number. Check: f(x) = 3 - 4 · 3-y4 = 3 - (3 - y) = y. So f is onto.
   Bijective.
2. (ii) f(x) = 1 + x². One-one? f(-1) = 1 + 1 = 2 and f(1) = 1 + 1 = 2, but -1 ≠ 1. Not one-one.
   Onto? The minimum value of 1 + x² over R is 1 (at x = 0), so 1 + x² ≥ 1. Take y = 0. Solve 1 + x² = 0 ⇒ x² = -1, no real solution. So y = 0 has no pre-image. Not onto.

(i) Bijective; (ii) Neither one-one nor onto.

DS

Diya Singh

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Inverse construction. If you can write down an explicit inverse, the function is bijective; if you cannot, diagnose where the inverse breaks.

1. For (i): solve y = 3 - 4x to get x = (3 - y)/4. The formula is defined for every y ∈ R and returns one and only one x. Hence injective and surjective.
2. For (ii): solving y = 1 + x² gives x² = y - 1, so x = ±√y - 1. Two pre-images for y > 1 (injectivity fails); no real pre-image for y < 1 (surjectivity fails).

Why this matters. Construction of an inverse is the canonical test for bijectivity. When the inverse becomes multi-valued or undefined, you read off which property is missing.

(i) Bijective; (ii) Neither.

Q 1.8

Let A and B be sets. Show that f : A × B → B × A such that f(a, b) = (b, a) is bijective function.

Check SolutionExpert Solution

Concept used. The map (a, b) ↦ (b, a) is the coordinate swap. We show both injectivity and surjectivity directly from the definition.

1. One-one. Suppose f(a₁, b₁) = f(a₂, b₂). Then (b₁, a₁) = (b₂, a₂). Two ordered pairs are equal iff their components are equal, so b₁ = b₂ and a₁ = a₂. Hence (a₁, b₁) = (a₂, b₂). One-one.
2. Onto. Take any (b, a) ∈ B × A. We seek (a', b') ∈ A × B with f(a', b') = (b, a). Setting a' = a and b' = b works: f(a, b) = (b, a) = (b, a). So every element of B × A is in the image. Onto.
3. Therefore f is bijective.

f(a, b) = (b, a) is a bijection from A × B to B × A.

SP

Sneha Patel

M.Sc Mathematics, IIT Madras

Verified Expert

Inverse-is-itself. Notice that applying f twice returns to the start: f(f(a, b)) = f(b, a) = (a, b). A function that is its own inverse is automatically a bijection (since it admits a two-sided inverse).

1. Compose: f ∘ f (a, b) = (a, b), so f ∘ f = I_{A × B}. Symmetrically, f ∘ f = I_{B × A} (on the codomain side).
2. A function with an inverse is bijective. Inverse here is f itself.
3. Therefore f : A × B → B × A is bijective.

Why this matters. Involutions (functions that are their own inverses) are common bijections; examples include complex conjugation, matrix transpose, and reflection across any line through the origin.

f is bijective with f^-1 = f.

Q 1.9

Let f : N → N be defined by f(n) = cases n+12, & if n is odd 4pt] n2, & if n is even cases) for all (n ∈ N). State whether the function (f) is bijective. Justify your answer.

Check SolutionExpert Solution

Concept used. A piecewise function on (N): odd inputs land on (n+12) (an integer since (n) is odd), even inputs on (n2). Check injectivity by trying small inputs, and surjectivity by inverting on the image.

1. Compute a few values: [aligned f(1) &= 1+12 = 1, & f(2) &= 22 = 1,
   f(3) &= 3+12 = 2, & f(4) &= 42 = 2,
   f(5) &= 5+12 = 3, & f(6) &= 62 = 3. aligned
2. Not one-one. (f(1) = 1 = f(2) with 1 ≠ 2. So f is not injective.
3. Onto. Given any m ∈ N, we want n with f(n) = m. Take n = 2m (even). Then f(2m) = 2m2 = m. So every m ∈ N has a pre-image. Onto.
4. One-one fails, onto holds. So f is not bijective.

f is not bijective. (Onto but not one-one.)

RP

Rohit Pillai

M.Sc Mathematics, IIT Bombay

Verified Expert

Pairing intuition. Group N in pairs 1, 2, 3, 4, 5, 6, …. Both elements of the k-th pair map to k. So f collapses every pair to its index, giving a 2 → 1 map onto N.

1. Pair k has elements (2k - 1, 2k). f(2k - 1) = 2k2 = k and f(2k) = 2k2 = k, the same value.
2. Injectivity fails on every pair.
3. Surjectivity holds: the index k is hit by both elements of pair k.

Why this matters. f is the typical example of a many-to-one onto map; its fibres f^-1(k) = 2k-1, 2k have size 2 each, so |N| = 2 |N| in cardinal arithmetic, the foundational identity for countable infinity.

Onto but not one-one; hence not bijective.

Q 1.10

Let A = R - 3 and B = R - 1. Consider the function f : A → B defined by f(x) = x - 2x - 3. Is f one-one and onto? Justify your answer.

Check SolutionExpert Solution

Concept used. A rational function x - 2x - 3 is a M"obius transformation. The point x = 3 is excluded from the domain because the denominator vanishes there. Algebraically, we test injectivity by solving f(x₁) = f(x₂) and surjectivity by inverting.

1. One-one. Suppose f(x₁) = f(x₂): x₁ - 2x₁ - 3 = x₂ - 2x₂ - 3. Cross-multiply (denominators non-zero): (x₁ - 2)(x₂ - 3) = (x₂ - 2)(x₁ - 3). Expand both sides: aligned LHS &= x₁ x₂ - 3 x₁ - 2 x₂ + 6,
   RHS &= x₁ x₂ - 3 x₂ - 2 x₁ + 6. aligned Subtract x₁ x₂ + 6 from both: -3 x₁ - 2 x₂ = -3 x₂ - 2 x₁ ⇒ -3 x₁ + 2 x₁ = -3 x₂ + 2 x₂ ⇒ -x₁ = -x₂. So x₁ = x₂. One-one.
2. Onto. Take any y ∈ B = R - 1. Solve y = x - 2x - 3 for x: aligned y (x - 3) &= x - 2,
   yx - 3y &= x - 2,
   yx - x &= 3y - 2,
   x (y - 1) &= 3y - 2,
   x &= 3y - 2y - 1. aligned Since y ≠ 1, the denominator is non-zero, so x is a well-defined real number. Check x ≠ 3: x = 3 ⇔ 3y - 2y - 1 = 3 ⇔ 3y - 2 = 3y - 3 ⇔ -2 = -3, false. So x ∈ A.

   Verify f(x) = y: f(x) = x - 2x - 3 = 3y - 2y - 1 - 23y - 2y - 1 - 3 = 3y - 2 - 2(y - 1)y - 13y - 2 - 3(y - 1)y - 1 = 3y - 2 - 2y + 23y - 2 - 3y + 3 = y1 = y. Every y ∈ B has a pre-image. Onto.

f is both one-one and onto; hence a bijection from A to B.

PD

Pranav Desai

M.Sc Mathematics, IIT Bombay

Verified Expert

M"obius perspective. Every map of the form f(x) = ax + bcx + d with ad - bc ≠ 0 is a bijection between R - -d/c and R - a/c. Compute the discriminant: ad - bc = (1)(-3) - (-2)(1) = -3 + 2 = -1 ≠ 0. So f is automatically bijective.

1. The excluded points are -d/c = 3 from domain (denominator zero) and a/c = 1 from codomain (horizontal asymptote).
2. Inverse f^-1(y) = dy - b-cy + a = -3y - (-2)-y + 1 · -1-1 = 3y - 2y - 1, matching the algebraic answer above.
3. Composition f ∘ f^-1 and f^-1 ∘ f are both identities on their respective sets.

Why this matters. M"obius maps are the bijections of the Riemann sphere; they form a group under composition (the group PGL₂), foundational to complex analysis and projective geometry.

f is bijective, with f^-1(y) = 3y - 2y - 1.

Q 1.11

Let f : R → R be defined as f(x) = x⁴. Choose the correct answer.
(A) f is one-one onto.   (B) f is many-one onto.
(C) f is one-one but not onto.   (D) f is neither one-one nor onto.

Check SolutionExpert Solution

Concept used. x⁴ is an even function with image [0, ∞).

1. One-one? f(-1) = (-1)⁴ = 1 and f(1) = 1⁴ = 1. Two distinct inputs share the same output. Not one-one.
2. Onto? Pick y = -1 ∈ R. Solve x⁴ = -1. Since x⁴ ≥ 0 for every real x, no real solution exists. So -1 has no pre-image. Not onto.
3. Neither property holds, matching (D).

(D) f is neither one-one nor onto.

KB

Krishna Bhat

M.Sc Mathematics, IIT Bombay

Verified Expert

Even power rule. Any even integer power x^2k on R collapses sign and skips negatives in the codomain. This kills both injectivity (via the even symmetry) and surjectivity (via the non-negative image).

1. Sign collapse: f(-c) = c⁴ = f(c) for any c ≠ 0. Infinitely many collisions ⇒ not one-one.
2. Non-negative image: x⁴ ≥ 0 for all real x, so negative reals are never images. Not onto.

Why this matters. The same reasoning applies to x², x⁴, x⁶, … on R; only odd powers x, x³, x⁵, … on R are bijective.

(D).

Q 1.12

Let f : R → R be defined as f(x) = 3x. Choose the correct answer.
(A) f is one-one onto.   (B) f is many-one onto.
(C) f is one-one but not onto.   (D) f is neither one-one nor onto.

Check SolutionExpert Solution

Concept used. f(x) = 3x is linear with non-zero slope, hence a bijection on R.

1. One-one. Suppose f(x₁) = f(x₂). Then 3x₁ = 3x₂, divide both sides by 3 (legal because 3 ≠ 0): x₁ = x₂. One-one.
2. Onto. Take any y ∈ R. Solve y = 3x: x = y/3, a real number. Verify f(x) = 3 · y3 = y. So f is onto.
3. Both properties hold, matching (A).

(A) f is one-one onto.

MC

Meera Chatterjee

M.Sc Mathematics, ISI Kolkata

Verified Expert

Inverse-by-inspection. The inverse f^-1(y) = y/3 is a well-defined function on all of R, confirming bijectivity in one move.

1. Compose: f(f^-1(y)) = 3 · (y/3) = y and f^-1(f(x)) = (3x)/3 = x.
2. Two-sided inverse exists ⇒ bijection.

Why this matters. Multiplication by a non-zero constant is the simplest non-trivial bijection of R; it scales the real line uniformly.

(A).

How to Use the Relations and Functions Notes Page Most Effectively

The recommended study plan for the Relations and Functions Class 12 chapter splits across three sittings. The table below outlines what to do in each:

For students preparing for both CBSE board and JEE Main:

• 60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
• 40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
• The relations and functions class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
• For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.