Tiêu đề 1. RELATIONS AND FUNCTIONS.pdf ✅

Relations and Functions OR Check if the relation R in the set R of real numbers defined as R = {(a, b) : a < b} is (i) symmetric, (ii) transitive. R&U [Delhi Set-1, 2, 3, 2020] Concept Applied Symmetric and Transitive Sol. A = {1, 2, 3, 4, 5, 6} R = {(x, y) : y is divisible by x} (i) Symmetric Let (x, y) Î R y is divisible by x \ x is not necessarily divisible by y (y, x) Ï R e.g., (1, 2) Î R 2 is divisible by 1 but 1 is not divisible by 2 (2, 1) Ï R Hence, given relation is not symmetric. 1 (ii) Transitive Let (x, y) ÎR y is divisible by x ...(i) and (y, z) Î R z is divisible by y ...(ii) From eq(i) and eq(ii), z is divisible by x \ (x, z) Î R e.g., (1, 2) Î R 2 is divisible by 1 ...(i) (2, 4) Î R 4 is divisible by 2 ...(ii) From eq(i) and eq(ii), 4 is divisible by 1 (1, 4) Î R Hence, given relation is transitive. 1 [Marking Scheme OD, 2020] Commonly Made Error Some students take the relation as “is a factor of” and go wrong. ‘is divisible by’ should be taken as ‘is a multiple of’ Answering Tip 3. How many equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) are there in all ? Justify your answer. U [SQP 2016-17] Sol. Equivalence relations could be the following : R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)} and 1 R2 = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)} 1 So, only two equivalence relations. [Marking Scheme SQP 2016-17] Short Answer Type Questions-II (3 marks each) 1. Check whether the relation R in the set Z of integers defined as R = {(a, b) : a + b is "divisible by 2"} is reflexive, symmetric or transitive. Write the equivalence class containing 0 i.e. [0]. R&U [SQP 2020-21] OR Prove that the relation R on Z, defined by R = {(x, y) : (x − y) is divisible by 5} is an equivalence relation. [Outside Delhi Set- 1, 2020] Concept Applied Equivalence Relation Sol. (i) Reflexive: Since, a + a = 2a which is even \ (a, a) Î R " a Î Z Hence R is reflexive. 1⁄2 (ii) Symmetric: If (a, b) Î R, then a + b = 2l Þ b + a = 2l Þ (b, a) Î R. Hence R is symmetric. 1 (iii) Transitive: If (a, b) Î R and (b, c) Î R then a + b = 2l ...(i) and b + c = 2m ...(ii) Adding (i) and (ii) we get a + 2b + c = 2(l + m) Þ a + c = 2(l + m – b) Þ a + c = 2k where l + m – b = k Þ (a, c) Î R Hence R is transitive Thus, R is an equivalence relation. [0] = {...–4, –2, 0, 2, 4...} 11⁄2 [Marking Scheme SQP 2020-21] Commonly Made Error Equivalence class of 0 is the set of all elements related to 0. Mostly students go wrong in finding the equivalence class. Some students forget to write 0 in the equivalence class. Answering Tip 2. Show that the relation R on R defined as R = {(a, b) : a £ b}, is reflexive, and transitive but not symmetric. U [Delhi Set-3, 2019] Sol. Topper's Answer, 2019