1. Mathematics

2. Session Set, Relation & Function Session - 3

3. Session Objectives

4. 1.Function definition 2.Domain, codomain and range 3.Standard real functions 4.Types of functions 5.Number of various types of functions 6.Composition of functions 7.Inverse of element 8.Inverse of function

5. Function Let A and B be two non-empty sets having m and n elements respectively, then the number of relations possible from A to B is 2 mn. Out of these 2 mn relations some are called function (or mappings) from A to B provided following two conditions hold in the relation: (i)All the elements of A are associated to elements of B. (ii)Each element of A is associated to one and only one element of B, i.e. no element of A is associated to two (or more) elements of B.

6. Function ‘f’ from set A to set B associates each element of A to unique (i.e. one and only one) element of B denoted by (read as ‘f from A to B’) Definition Observations: (i)A relation from A to B is not a function if it either violates condition 1 or 2 or both, i.e. either some element of A is not associated to element of B or some element of A is associated to more than one elements of B or both. (ii) In a function from A to B, two elements of A can be associated to one element of B (examples R 7, R 10 )

7. Observations (i)A relation from A to B is not a function if it either violates condition 1 or 2 or both, i.e. either some element of A is not associated to element of B or some element of A is associated to more than one elements of B or both. (ii) In a function from A to B, two elements of A can be associated to one element of B (examples R 7, R 10 ) (iii) If be the function, then o(f) = o(A) and Dom(f) = A.

8. Domain, Codomain and Range of a Function Let be the function, then set ‘A’ is called the domain of f and set ‘B’ is called the codomain of f. The set of those elements of B which are related by elements of A is called range of f or image of set A under f and is denoted by f(A), i.e. Range of f. Clearly,

9. Domain, Codomain and Range of a Function For example: Dom (R 7 ) = {a, b}, Codomain = {1, 2} Range (R 7 ) = {1} Domain (R 8 ) = {a, b} Codomain (R 8 ) = {1, 2} Range (R 8 ) = {1, 2} = Codomain (R 8 )

10. Domain, Codomain and Range of a Function For example: Dom (R 7 ) = {a, b}, Codomain = {1, 2} Range (R 7 ) = {1} Domain (R 8 ) = {a, b} Codomain (R 8 ) = {1, 2} Range (R 8 ) = {1, 2} = Codomain (R 8 )

11. Equal Functions Two functions f and g are said to be equal iff (i) Dom (f) = Dom (g) (ii) Codom (f) = Codom (g) If all these three conditions holds, then we can write f = g. (iii) or Dom (g)

12. Mathematical Way to Prove a Relation to be a Function If A and B be two non-empty sets, f be the relation from A to B (i.e. ), then f is function from A to B if (i) and (ii)

13. Some Standard Real Functions and Their Graphs Real functions Functions in which both domain and codomain are the subsets of R, i.e. set of real numbers. Domain of f is [a, c] Codomain is R Range is [u, w]

14. Some Standard Real Functions and Their Graphs Constant function: f(x) = c Let f : R R be the real function defined as Dom (f) = R, Codomain (f) = R, Range (f) = {c} Note:f : A B is constant function if f(a) = c for some

15. Some Standard Real Functions and Their Graphs Identity function: f(x) = x Let f : R R be the real function defined as Dom (f) = R Codomain (f) = R Range (f) = R Note:f : A A given by f(a) = a is identity function denoted by (same as identity relation)

16. Some Standard Real Functions and Their Graphs Modulus function: f(x) = |x| Let f : R R be the real function defined as Domain (f) = R, Codomain (f) = R, Range (f) = (as ) = set of non-negative real numbers.

17. Some Standard Real Functions and Their Graphs Greatest integer function: f(x) = [x] Let f : R R be the real function defined as = greatest integer less than or equal to x. For example: [2.1] = 2, i.e. greatest integer less than or equal to 2.1 is 2, similarly [–2.1] = –3 [2] = 2 [3. 9] = 3 [–3. 9] = –4

18. Some Standard Real Functions and Their Graphs Hence [x] = 0 = 1 = 2 and so on. Also [x] = –1 = –2 and so on. Combining we get [x] = n for

19. Some Standard Real Functions and Their Graphs Filled circle means, point is on the graph. unfilled circle means, point is not on the graph. Dom(f) = R, Codomain (f) = R Range (f) = z

20. Some Standard Real Functions and Their Graphs Exponential function: f(x) = a x Let f : R R be the real function defined as Case I: Let 0 < a < 1 Say, then

21. Some Standard Real Functions and Their Graphs Case II: Let a > 1 Say a = 2, then y = a x = 2 x

22. Some Standard Real Functions and Their Graphs Domain (f) = R, Codomain (f) = R, Range (f) = Special case a = e > 1 y = f(x) = e x

23. Some Standard Real Functions and Their Graphs Logarithmic function: y = log b x Let be the real function defined as Case 1: 0 < b < 1

24. Some Standard Real Functions and Their Graphs Case 2: 1 < b Domain (f) = Codomain (f) = R Range (f) = R

25. Types of Functions One-one function (or injective) A function is said to be one-one function or injective if different elements of A have different images in B, i.e. if Thus iff

26. Types of Functions For example: Let be the function given by (i) (ii) (iii) (iv)

27. Types of Functions Here only (ii) and (iii) are one-one functions Hence (i) and (iv) are not one-one functions. In (i) In (iv) (i) To check injectivity of functions Let f(x) = f(y) if it gives x = y only, then f is a one-one function. Observation: (ii) is true for all the functions (condition 2) but its converse is true for one-one function.

28. Types of Functions (iii) Injectivity of f(x) can also be checked by its graph. If all lines parallel to x-axis cut f(x) in not more than one point, then f(x) is one-one function, i.e. f(x) is not a one-one function if at least one line parallel to x-axis cuts f(x) in more than one point. (iv) If

29. Types of Functions Alternative: f(1) = f(–1) = 1 but Using graphs (i) Clearly, any line parallel to x-axis cuts f(x) = 2x only at one point, thus f(x) is one-one.

30. Types of Functions (ii) Clearly, y = 4 cuts y = x 2 in two points hence f(x) = x 2 is not one-one (iii) Clearly, y = 1 cuts y = |x| in two points hence f(x) = |x| is not one-one.

31. Types of Functions Many-one function A function which is not one-one is many-one function, i.e. at least two different elements of A have same image in B or s.t. But f(x) = f(y). For example: given by, are both many-one functions as.

32. Types of Functions Onto function (or surjective) A function is said to be onto function or subjective if all the elements of B have preimage in A, i.e. for each i.e. A function is not onto if s.t. there is no for which f(a) = b.

33. Types of Functions For example: Let be the function given by (i) (ii) (iii) (iv)

34. Types of Functions Here functions (ii) and (iii) are onto functions as all the elements of B have pre-image in A. In (i) have no pre-image in A and in (iv) have no pre-image in A, thus not onto functions. Observations: (i) is onto function if Range (f) = Codomain (f) = B Proof: (by definition) let, then if f is onto it has pre-image in A (As range (f) contains those elements of B which have pre-image in A).

35. Types of Functions Hence codomain (f) Range (f) = Codomain (f) (ii) To check surjectivity of function take some and follow the steps given below: Step 1: Let f(x) = y Step 2: Express x in terms of y from above equation say x = g(y) Step 3: Now find the domain of g, if Dom (g) = Codomain (f) (i.e. B) then f is onto (or surjective) (iii) If is onto, then

36. Types of Functions Into function A function which is not onto is into function, i.e. at least one element of B have no pre-image in A or such that there is no for which f(a) = b. So in example given above, functions, (i) and (iv) are into functions. Bijective function (or one-one and onto) A function is said to be bijective if it is injective as well as surjective, i.e. one-one as well as onto.

37. Types of Functions In other words is bijective if (i) f is one-one, i.e. and (ii) f is onto, i.e. for each some st f(a) = b or Range (f) = Codomain (f)

38. Types of Functions Observations: (i) A function is not bijective if it is either not injective or not surjective or not both. (ii) If f is bijective, then it is injective and surjective Hence o(A) = o(B).

39. Number of Functions of Various Types Let A and B be two non-empty sets, let o(A) = m, o(B) = n and be the function from A to B Number of functions from A to B Each element of A can be associated to n elements of B, so total number of functions that can be formed from A to B is n × n ×... × n (m times), i.e. n m. Hence total number of functions from A to B =

40. Number of Functions of Various Types Number of functions from A to B (i)Every function is a relation but not vice versa. (ii)If A and B are two non-empty sets such that o(A) = m, o(B) = n, then number of relations possible from A to B is 2 mn and number of functions possible from A to B is n m. (iii) Number of relations from A to B which are not functions is 2 mn – n m or.

41. Number of Functions of Various Types Number of one-one functions from A to B Out of n m functions, from A to B some are one-one functions. Now if we order the elements of A from 1 to m say first, second,..., mth then for first element of A we have n choices from set B, for second we have (n – 1) choices from set B (as function has to be one-one) and so on. Thus total number of one-one functions possible from A to B is n (n – 1) (n – 2)... (n – m + 1), i.e. Here note that m has to be less than or equal to n, i.e. otherwise if m is greater than n, i.e. no 1-1 function is possible from A to B (as in that case first n elements of A will be associated to n elements of B and still m – n elements of A remains to be associated). Hence number of 1-1 functions from A to B.

42. Number of Functions of Various Types Observations: (i)Out of functions from A to B, functions are one-one (provided ). (ii) If functions from A to B are many-one functions. (iii) If m > n, then all the n m functions are many-one functions.

43. Number of Functions of Various Types Number of bijective functions from A to B A function is bijective iff function is 1-1 as well as onto. This implies o(A) = o(B), i.e. m = n. Hence for first element of A we have n options, for second we have (n – 1) options and so on, for last we have only one option. Therefore, total number of bijective functions from A to B is n (n – 1)... 2.1 = n!

44. Number of Functions of Various Types Show all bijective functions from where A = {a, b, c} and B = {x, y, z} Hence number of bijective functions

45. Number of Functions of Various Types Observations: (i)If o(A) = o(B) and function is 1-1, then function is onto also and hence bijective. (ii)If o(A) = o(B) and function is onto, then function is 1-1 also and hence bijective.

46. Composition of Functions Let be two functions, then function gof : defined as is called the composition of f and g Observations: (i) For gof to exist, range of f must be subset of domain of g. (ii) Similarly for fog to exit, range of g must be subset of domain of f.

47. Composition of Functions Properties of composition of functions (i) Composition of functions is non commutative, i.e. Note that its possible that fog may not exist even if gof exists. (ii)Composition of functions is associative, i.e. f, g, h be three functions. Then (fog) oh = fo (goh) (provided they exist) (iii)If f and g are bijections, then gof is also a bijection (provided exist) (iv) Composition of identity function with any function f : is f itself, i.e..

48. Inverse of Element Let be the function from A to B, then for any if f(a) = b, then a is called the pre-image of ‘b’ or inverse of ‘b’ denoted by f –1 (b). For example: Let be given by, Note: Inverse of an element may not be unique.

49. Inverse of Function If be a bijective function, then we can define a new function from B to A as inverse of f denoted by given by i.e. each element of B is associated (or mapped) to its pre-image under f

50. Inverse of Function How to find f –1 If is a bijective function, then f –1 can be obtained using following steps: (i) Let y = f(x) (ii) Express x in terms of y, say x = g(y) (iii) Interchanging x with y, i.e., we get y = g(x). Then g = f –1 Note that before finding f –1, you have to prove that f is a bijective function (i.e. 1-1 as well as onto) by using the rules given before.

51. Inverse of Function Properties of inverse function (i)Function is invertible iff it is bijective. (ii)Inverse of bijection is unique. (iii)Inverse of bijection is also bijection. (iv) If is a bijection, then where are identity functions on A and B respectively, i.e.

52. Inverse of Function Corollary: If is bijection, then. (v) If be two bijections, then inverse of gof : is given by (vi) If be two functions such that, then f and g are bijections and g = f –1 (vii) Domain (f) = Range (f –1 ) and Range (f) = Domain (f –1 )

53. Class Test

54. Class Exercise - 1 f and h are relation from A to B where A = {a, b, c, d}, B = {s, t, u} defined as follows: f(a) = t, f(b) = s, f(c) = s f(d) = u, h(a) = s, h(b) = t h(c) = s, h(a) = u, h(d) = u Which one of the following statements is true? (a) f and h are functions (b) f is a function and h is not a function (c) f and h are not functions (d) None of these

55. Solution Each element of A is associated to unique element of B, hence f is a function. Each element of A is associated to elements of B but is associated to two elements of B namely s and u violating second condition. Hence h is not a function. Ans. (b)

56. Class Exercise - 2 The function (N is the set of natural numbers) defined by f(n) = 2n + 3 is (a) surjective (b) injective (c) bijective (d) None of these

57. Solution Check one-one Let f(m) = f(n) 2m + 3 = 2n + 3 m = n Hence one-one If f(n) = 2 There is no pre-image of. Hence not onto. f is injective only. Note that 2n + 3 is always odd and range = {5, 7, 9,...}. Hence, answer is (b).

58. Class Exercise - 3 The function defined by f(x) = (x – 1) (x – 2) (x – 3) is (a) one-one but not onto (b) onto but not one-one (c) both one-one and onto (d) neither one-one nor onto

59. Solution Check for one-one Let f(x) = f(y), i.e. (x – 1) (x – 2) (x – 3) = (y – 1) (y – 2) (y – 3) (x – 1) (x 2 – 5x + 6) = (y – 1) (y 2 – 5y + 6)

60. Solution contd.. Not one-one Check for onto Let f(a) = b. i.e. (a – 1) (a – 2) (a – 3) = b This is cubic in ‘a’ hence there exist at least one real root of a (as complex roots occur in pair) f is onto but not one-one

61. Solution contd.. Alternate method: Draw the graph of f(x) = (x – 1) (x – 2) (x – 3) Clearly f is not one-one as line parallel to x-axis cuts the curve in more than one point and any line parallel to x-axis (in the codomain) must cuts the curve.

62. Class Exercise - 4 Let defined as check f for one-one, onto.

63. Solution Check for one-one. Let f(x) = f(y) for some Case 1: x, y both even Case 2: x even, y odd If x is even, y cannot be odd. Case 3: x odd, y even Not possible Case 4: x odd, y odd Hence x = y f is 1–1

64. Solution contd.. Check for onto f : {0, 1, 2,...} {0, 1, 2,...} f(0) = 0 + 1 = 10 is even f(1) = 1 – 1 = 01 is odd f(2) = 2 + 1 = 32 is even f(3) = 3 – 1 = 23 is even Range = {0, 1, 2,...} = Codomain Hence f is one-one, onto

65. Class Exercise - 5 Show that defined by f((a, b)) = (b, a) is bijection

66. Solution Check one-one Hence one-one Check onto Let where, then f((a, b)) = (b, a) Hence f is bijective.

67. Class Exercise - 6 The number of surjections from A = {1, 2,..., n}, onto B = {a, b} is (a) n P 2 (b) 2 n – 2 (c) 2 n – 1 (d) None of these

68. Solution O(A) = n, O(B) = 2 Hence number of surjections (onto) functions

69. Solution contd.. Alternative method: Total number of functions from A to B is 2 n. Number of functions which are not onto, i.e. all the elements of A are either associated to ‘a’ or ‘b’ is 2. or Onto functions = 2 n – 2.

70. Class Exercise - 7 If the functions f and g are defined from the set of real numbers R to R such that, then find functions fog and gof. Also find

71. Solution (fog) (x) = f(g(x)) = f (3x – 2) (gof) (x) = g(f(x)) To find

72. Class Exercise - 8 If f(x) = sin 2 x + sin 2, then (gof) (x) = (a) 1(b) 0 (c) sinx(d) None of these

73. Solution (gof) (x) = g(f(x)) As g(x) is defined only for hence we cannot find g(f(x)) in general, until and unless f(x) turns out to be for all x. Now simplify f(x).

74. Solution contd.. Hence, answer is (a).

75. Class Exercise - 9 If f: {1, 2, 3,...} is defined by y = f(x) =, then (a) 100 (b) 199 (c) 201(d) 200

76. Solution i.e. f(x) = 100 If x is even, then f(x) = x – 1 = –200 Hence, answer is (d).

77. Class Exercise - 10 If such that, then (a) (b) (c) (d)

78. Solution then Domain = Codomain = R Range = R = Codomain Let f(x) = f(y) Hence f is one-one. Step 1: Step 2: Step 3:

79. Thank you