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X std mathematics - Relations and functions (Ex 1.4), Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, representation of functions, set or ordered pair, table form, arrow diagram, graph, vertical line test, types of function, one -one function, many- one function, onto function, surjection, into function, horizontal line test, special cases of function,

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Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.3, Set Language, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, 1c. Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.3

1c. Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.3Dr. I. Uma Maheswari Maheswari

Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.3, Set Language, Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, 1c. Pedagogy of Mathematics (Part II) - Set language introduction and Ex.1.3

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- 1. PEDAGOGY OF MATHEMATICS – PART II BY Dr. I. UMA MAHESWARI Principal Peniel Rural College of Education,Vemparali, Dindigul District iuma_maheswari@yahoo.co.in
- 2. X STD Ex 1.4
- 38. Solution: (i) It is not a function. The graph meets the vertical line at more than one points. (ii) It is a function as the curve meets the vertical line at only one point. (iii) It is not a function as it meets the vertical line at more than one points. (iv) It is a function as it meets the vertical line at only one point.
- 39. Solution: f: A → B A = {2,4, 6, 10, 12}, B = {0,1, 2, 4,5,9}
- 40. (i) Set of ordered pairs = {(2, 0), (4, 1), (6, 2), (10, 4), (12, 5)} (ii) a table (iii) an arrow diagram; (iv) a graph
- 41. Solution: f = {(1,2), (2, 2), (3, 2), (4, 3), (5, 4)} (i) An arrow diagram.
- 42. (ii) a table form (iii) A graph representation.
- 43. Solution: f: N → N f(x) = 2x – 1 N = {1,2, 3, 4, 5,…} f(1) = 2(1) – 1 = 1 f(2) = 2(2) – 1 = 3 f(3) = 2(3) – 1 = 5 f(4) = 2(4) – 1 = 7 f(5) = 2(5) – 1 = 9 Hence f : N → N is a one-one function A function f: N → N is said to be onto function if the range of f is equal to the co-domain of f Range = {1,3, 5, 7, 9,…} Co-domain = {1, 2,3,..} But here the range is not equal to co-domain. Therefore it is one-one but not onto function.
- 44. Solution: f: N → N f(m) = m2 + m + 3 N = {1,2, 3,4, 5…..}m ∈ N f{m) = m2 + m + 3 f(1) = 12 + 1 + 3 = 5 f(2) = 22 + 2 + 3 = 9 f(3) = 32 + 3 + 3 = 15 f(4) = 42 + 4 + 3 = 23 In the figure, for different elements in the (X) domain, there are different images in f(x). Hence f: N → N is a one to one but not onto function as the range of f is not equal to co-domain. Hence it is proved.
- 45. Solution: A = {1,2,3,4} B = N f: A → B,f(x) = x3 (i) f(1) = 13 = 1 f(2) = 23 = 8 f(3) = 33 = 27 f(4) = 43 = 64 (ii) The range of f = {1,8,27,64 ) (iii) It is one-one and into function.
- 46. Solution: (i) f : R → R f(x) = 2x + 1 f(1) = 2(1) + 1 = 3 f(2) = 2(2) + 1 = 5 f(-1) = 2(-1) + 1 = -1 f(0) = 2(0) + 1 = 1 It is a bijective function. Distinct elements of A have distinct images in B and every element in B has a pre-image in A. (ii) f: R → R; f(x) = 3 – 4x2 f(1) = 3 – 4(12) = 3 – 4 = -1 f(2) = 3 – 4(22) = 3 – 16 = -13 f(-1) = 3 – 4(-1)2 = 3 – 4 = -1 It is not bijective function since it is not one-one
- 47. Solution: A= {-1, 1},B = {0,2} f: A → B, f(x) = ax + b f(-1) = a(-1) + b = -a + b f(1) = a(1) + b = a + b Since f(x) is onto, f(-1) = 0 ⇒ -a + b = 0 …(1) & f(1) = 2 ⇒ a + b = 2 …(2) -a + b = 0
- 48. Solution: (i) f(3) ⇒ f(x) = x + 2 ⇒ 3 + 2 = 5 (ii) f(0) ⇒ 2 (iii) f (- 1.5) = x – 1 = -1.5 – 1 = -2.5
- 49. Solution: f : [-5, 9] → R (i) f(-3) + f(2) f(-3) = 6x + 1 = 6(-3) + 1 = -17 f(2) = 5×2 – 1 = 5(22) – 1 = 19 ∴ f(-3) + f(2) = -17 + 19 = 2
- 50. (ii) f(7) – f(1) f(7) = 3x – 4 = 3(7) – 4 = 17 f(1) = 6x + 1 = 6(1) + 1 = 7 f(7) – f(1) = 17 – 7 = 10 (iii) 2f(4) + f(8) f(4) = 5x2 – 1 = 5 × 42 – 1 = 79 f(8) = 3x – 4 = 3 × 8 – 4 = 20 ∴ 2f(4) + f(8) = 2 × 79 + 20 = 178
- 51. Solution: Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.
- 52. Solution: