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### X std maths - Relations and functions (ex 1.4)

• 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
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• 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.
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