Class 12 Math Chapter 1 Relations and Functions is foundational for understanding higher-level mathematics. This chapter builds upon concepts studied in Class 11 and explores different types of relations—reflexive, symmetric, transitive—and functions such as one-one, onto, and bijective.

EXERCISE1.2
EXERCISE1.3
Class 12 Math Chapter 1 Relations and Functions

The NCERT solutions provided here for Exercise 1.1 offer step-by-step explanations to help students master these concepts with clarity and confidence. These are essential for board exam preparation and competitive exams like JEE.

EXERCISE 1.1 – Class 12 Math Chapter 1 Relations and Functions

Question 1: Determine whether the following relations are reflexive, symmetric, and transitive.

(i) R = {(x, y): 3x – y = 0} on A = {1, 2, …, 14}
Answer: This relation means y = 3x. It is not reflexive because x ≠ 3x in general. Not symmetric because if (1, 3) ∈ R, then (3, 1) ∉ R. Not transitive because (x, y) and (y, z) may not lead to (x, z) unless z = 9x.

(ii) R = {(x, y): y = x + 5 and x < 4} on N
Answer: Not reflexive since x ≠ x + 5. Not symmetric because (1, 6) ∈ R but (6, 1) ∉ R. Not transitive as addition of 5 does not lead to direct link between x and z.

(iii) R = {(x, y): y is divisible by x} on A = {1, 2, 3, 4, 5, 6}
Answer: Reflexive because any number divides itself. Not symmetric: (2, 4) ∈ R but (4, 2) ∉ R. Transitive: if x divides y and y divides z, then x divides z.

(iv) R = {(x, y): x – y is an integer} on Z
Answer: Reflexive (x – x = 0), symmetric (if x – y is integer, then y – x is also integer), transitive (x – y and y – z implies x – z is integer). So it’s an equivalence relation.

(v)(a) R = {(x, y): x and y work at same place}
Answer: Reflexive (anyone works at same place as self), symmetric, and transitive. Hence equivalence relation.

(v)(b) R = {(x, y): x and y live in same locality}
Answer: Same explanation as (v)(a). It’s an equivalence relation.

(v)(c) R = {(x, y): x is exactly 7 cm taller than y}
Answer: Not reflexive (x not 7 cm taller than self), not symmetric (if x taller than y, y not taller than x), not transitive.

(v)(d) R = {(x, y): x is wife of y}
Answer: Not reflexive, symmetric in usual social context (if x is wife of y, y is husband of x), but not transitive.

(v)(e) R = {(x, y): x is father of y}
Answer: Not reflexive, not symmetric (father-son relation not reversible), not transitive.

Question 2: R = {(a, b): a <= b^2} on R
Answer: Not reflexive because a ≤ a² is not always true (e.g. a = -1). Not symmetric, not transitive.

Question 3: R = {(a, b): b = a + 1} on {1, 2, 3, 4, 5, 6}
Answer: Not reflexive (a ≠ a+1), not symmetric ((1,2) ∈ R but (2,1) ∉ R), not transitive ((1,2) and (2,3) ∈ R but (1,3) ∉ R).

Question 4: R = {(a, b): a <= b} on R
Answer: Reflexive (a ≤ a), not symmetric (a ≤ b does not imply b ≤ a), transitive (a ≤ b and b ≤ c ⇒ a ≤ c).

Question 5: R = {(a, b): a <= b^3} on R
Answer: Reflexive for limited a (not for a > a³). Not symmetric, not transitive always.

Question 6: R = {(1, 2), (2, 1)} on {1, 2, 3}
Answer: Symmetric (both pairs included), but not reflexive (missing (1,1), (2,2), (3,3)), not transitive.

Question 7: R = {(x, y): x and y have same number of pages} on set of books
Answer: Reflexive (book has same pages as itself), symmetric and transitive. Hence equivalence relation.

Question 8: R = {(a, b): |a – b| is even} on A = {1, 2, 3, 4, 5}
Answer: Reflexive (|a – a| = 0, which is even), symmetric, and transitive. Equivalence relation.

Question 9:
(i) R = {(a, b): |a – b| is a multiple of 4} on A = {x in Z: 0 <= x <= 12}
Answer: Reflexive (|a – a| = 0), symmetric (|a – b| = |b – a|), transitive (if difference of a-b and b-c divisible by 4, then a-c also). Equivalence relation.

(ii) R = {(a, b): a = b}
Answer: This is the identity relation. Clearly reflexive, symmetric, and transitive.

Question 10: Examples:
(i) Symmetric but neither reflexive nor transitive: R = {(1, 2), (2, 1)}
(ii) Transitive but neither reflexive nor symmetric: R = {(1, 2), (2, 3), (1, 3)}
(iii) Reflexive and symmetric but not transitive: R = {(1, 1), (2, 2), (1, 2), (2, 1)}
(iv) Reflexive and transitive but not symmetric: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
(v) Symmetric and transitive but not reflexive: R = {(1, 2), (2, 1)}

Question 11: R = {(P, Q): distance from origin is same} on points in plane
Answer: All points lying on same circle are related. Reflexive (same point), symmetric (dist(P,O)=dist(Q,O)), transitive. Equivalence relation.

Question 12: R = {(T1, T2): T1 is similar to T2} on triangles
Answer: Any triangle is similar to itself (reflexive), if T1 ~ T2 then T2 ~ T1 (symmetric), and similarity is transitive. Equivalence relation.

Question 13: R = {(P1, P2): P1 and P2 have same number of sides} on polygons
Answer: Polygons with same number of sides are related. Reflexive, symmetric, transitive. Equivalence relation.

Question 14: R = {(L1, L2): L1 is parallel to L2} on set of lines
Answer: Reflexive (line parallel to itself), symmetric and transitive. Equivalence relation.

Question 15: R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
Answer: Contains all (x,x) so reflexive. From (1,2), (2,3) ⇒ (1,3) included, and so on. Not symmetric since (1,2) exists but (2,1) missing. Correct option: (B)

Question 16: R = {(a, b): a = b – 2, b > 6} on N
Answer: a = b – 2 ⇒ b = a + 2. Check which options satisfy this and b > 6. Only (6, 8) is valid. Correct option: (C)

EXERCISE 1.2 – Class 12 Math Chapter 1 Relations and Function NCERT Solutions Ex 1.1 to Ex 1.3

Question 1: Show that the function f: R* → R* defined by f(x) = 1/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?
Answer: To check injectivity (one-one), assume f(x1) = f(x2) ⇒ 1/x1 = 1/x2 ⇒ x1 = x2. So the function is injective. For surjectivity (onto), take any y ∈ R*, then x = 1/y ∈ R*, and f(x) = y. Hence f is bijective on R* → R*. However, if domain is N (natural numbers), then 1/x is not a natural number for x > 1. So the function is not onto in that case.

Question 2: Check injectivity and surjectivity: (i) f: N → N, f(x) = x²
Answer: One-one: Yes, square of each natural number is unique. Onto: No, since not every natural number is a perfect square.

(ii) f: Z → Z, f(x) = x²
Answer: One-one: No, since f(1) = f(-1) = 1. Onto: No, negative numbers are not squares.

(iii) f: R → R, f(x) = x²
Answer: One-one: No. f(2) = f(-2). Onto: No. All negative numbers are excluded.

(iv) f: N → N, f(x) = x³
Answer: One-one: Yes, every cube is unique in N. Onto: No. Only cubes appear in range.

(v) f: Z → Z, f(x) = x³
Answer: One-one: Yes. Onto: Yes. Every integer has a cube root in Z.

Question 3: Prove that the Greatest Integer Function f(x) = [x] is neither one-one nor onto.
Answer: Not one-one: Values like f(1.2), f(1.5), f(1.9) all map to 1. Not onto: The function never takes non-integer or fractional values.

Question 4: Show that the Modulus Function f(x) = |x| is neither one-one nor onto.
Answer: Not one-one: f(2) = f(-2) = 2, so different inputs yield the same output. Not onto: The range is only non-negative real numbers, so negative numbers are excluded.

Question 5: Show that the Signum Function is neither one-one nor onto.
Answer: Not one-one: All positive numbers give output 1. Not onto: Only values -1, 0, 1 are present in the range; real numbers like 2 or -3 are excluded.

Question 6: Let A = {1, 2, 3}, B = {4, 5, 6, 7}, and f = {(1, 4), (2, 5), (3, 6)}. Show f is one-one.
Answer: Each element of A maps to a unique and distinct value in B. No two elements of A have the same image. So f is injective (one-one).

Question 7: Determine whether the following are one-one, onto, or bijective. (i) f(x) = 3 – 4x
Answer: One-one: Yes, since the function is linear and decreasing. Onto: Yes, for every y ∈ R, there exists x ∈ R. Hence, it is bijective.

(ii) f(x) = 1 + x²
Answer: One-one: No, f(1) = f(-1) = 2. Onto: No, minimum value is 1, so numbers <1 are not in range.

Question 8: Show that f: A × B → B × A, f(a, b) = (b, a) is bijective.
Answer: One-one: f(a1, b1) = f(a2, b2) implies (b1, a1) = (b2, a2), so a1 = a2 and b1 = b2. Onto: Any element (b, a) ∈ B × A has a preimage (a, b) ∈ A × B. So the function is bijective.

Question 9: f: N → N defined as:

  • f(n) = n+1 if n is odd
  • f(n) = n–1 if n is even
    Answer: One-one: Yes, all odd numbers go to even and vice versa with no repetition. Onto: Every natural number is either n+1 or n–1 of some other number. So it’s bijective.

Question 10: f: A → B, f(x) = (2x – 3)/(x – 3), A = R – {3}, B = R – {1}
Answer: One-one: Assume f(x1) = f(x2), it simplifies to x1 = x2. Onto: Solve for x in terms of y, gives x = (3y – 3)/(y – 2), which lies in domain A. So it’s bijective.

Question 11: f(x) = x⁴ on R
Answer: One-one: No, f(2) = f(-2) = 16. Onto: No, negative numbers are not in range. Correct option: (D).

Question 12: f(x) = 3x on R
Answer: One-one: Yes, since each x maps uniquely. Onto: Yes, any real y = 3x has x = y/3 ∈ R. So f is bijective. Correct option: (A).

Introduction to Miscellaneous Exercise: The Miscellaneous Exercise of Class 12 Math Chapter 1 Relations and Functions combines all the concepts learned in the chapter, including types of relations, types of functions, one-one and onto mappings, composition of functions, and invertibility. These questions are excellent for revision and application-based understanding, making them vital for both board exams and entrance tests like JEE.

MISCELLANEOUS EXERCISE – Class 12 Math Chapter 1 Relations and Function

Question 1: Show that the function f: R → R given by f(x) = 2x + 3 is bijective.

Answer:

  • One-one: Assume f(x1) = f(x2) ⇒ 2×1 + 3 = 2×2 + 3 ⇒ x1 = x2 ⇒ Injective
  • Onto: For any y ∈ R, let x = (y – 3)/2 ⇒ x ∈ R ⇒ f(x) = y ⇒ Surjective
  • Hence, f is bijective.

Question 2: Let f: R → R be defined by f(x) = x². Show that f is not invertible.

Answer:

  • f is not one-one as f(2) = f(-2) = 4.
  • A function is invertible only if it is bijective.
  • Since f is not one-one, it is not invertible.

Question 3: Find the inverse of the bijective function f: R → R defined by f(x) = 3x – 4.

Answer:

  • Let y = f(x) = 3x – 4 ⇒ x = (y + 4)/3
  • So inverse is f⁻¹(y) = (y + 4)/3
  • Hence, f⁻¹(x) = (x + 4)/3.

Question 4: If f(x) = 2x – 1 and g(x) = x², find (g ∘ f)(x) and (f ∘ g)(x).

Answer:

  • (g ∘ f)(x) = g(f(x)) = g(2x – 1) = (2x – 1)²
  • (f ∘ g)(x) = f(g(x)) = f(x²) = 2x² – 1.

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Class 12 Math Chapter 1 Relations and Functions is a vital topic that lays the groundwork for abstract mathematics. By mastering the concepts of reflexive, symmetric, and transitive relations through these NCERT solutions, students gain confidence and clarity.

This set of Class 12 Math Chapter 1 Relations and Functions solutions helps students prepare effectively for board exams.

Class 12 Math Chapter 1 Relations and Functions also strengthens the logic required in competitive exams.

With these step-by-step Class 12 Math Chapter 1 Relations and Functions NCERT Solutions Ex 1.1, learners can practice and revise key ideas.

In summary, understanding Class 12 Math Chapter 1 Relations and Functions thoroughly ensures a strong start to the academic year.