(ii) If (x + 1, 1) = (3y, y \u2013 1), find the values of x and y.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n (a\/3 + 1, b \u2013 2\/3) = (5\/3, 1\/3)<\/p>\n\n\n\n By the definition of equality of ordered pairs,<\/p>\n\n\n\n Let us solve for a and b<\/p>\n\n\n\n a\/3 + 1 = 5\/3 and b \u2013 2\/3 = 1\/3<\/p>\n\n\n\n a\/3 = 5\/3 \u2013 1 and b = 1\/3 + 2\/3<\/p>\n\n\n\n a\/3 = (5-3)\/3 and b = (1+2)\/3<\/p>\n\n\n\n a\/3 = 2\/3 and b = 3\/3<\/p>\n\n\n\n a = 2(3)\/3 and b = 1<\/p>\n\n\n\n a = 2 and b = 1<\/p>\n\n\n\n \u2234 Values of a and b are, a = 2 and b = 1<\/p>\n\n\n\n (ii) <\/strong>If (x + 1, 1) = (3y, y \u2013 1), find the values of x and y.<\/p>\n\n\n\n Given:<\/p>\n\n\n\n (x + 1, 1) = (3y, y \u2013 1)<\/p>\n\n\n\n By the definition of equality of ordered pairs,<\/p>\n\n\n\n Let us solve for x and y<\/p>\n\n\n\n x + 1 = 3y and 1 = y \u2013 1<\/p>\n\n\n\n x = 3y \u2013 1 and y = 1 + 1<\/p>\n\n\n\n x = 3y \u2013 1 and y = 2<\/p>\n\n\n\n Since, y = 2 we can substitute in<\/p>\n\n\n\n x = 3y \u2013 1<\/p>\n\n\n\n = 3(2) \u2013 1<\/p>\n\n\n\n = 6 \u2013 1<\/p>\n\n\n\n = 5<\/p>\n\n\n\n \u2234 Values of x and y are, x = 5 and y = 2<\/p>\n\n\n\n 2. If the ordered pairs (x, \u2013 1) and (5, y) belong to the set {(a, b): b = 2a \u2013 3}, find the values of x and y.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n The ordered pairs (x, \u2013 1) and (5, y) belong to the set {(a, b): b = 2a \u2013 3}<\/p>\n\n\n\n Solving for first order pair<\/p>\n\n\n\n (x, \u2013 1) = {(a, b): b = 2a \u2013 3}<\/p>\n\n\n\n x = a and -1 = b<\/p>\n\n\n\n By taking b = 2a \u2013 3<\/p>\n\n\n\n If b = \u2013 1 then 2a = \u2013 1 + 3<\/p>\n\n\n\n = 2<\/p>\n\n\n\n a = 2\/2<\/p>\n\n\n\n = 1<\/p>\n\n\n\n So, a = 1<\/p>\n\n\n\n Since x = a, x = 1<\/p>\n\n\n\n Similarly, solving for second order pair<\/p>\n\n\n\n (5, y) = {(a, b): b = 2a \u2013 3}<\/p>\n\n\n\n 5 = a and y = b<\/p>\n\n\n\n By taking b = 2a \u2013 3<\/p>\n\n\n\n If a = 5 then b = 2\u00d75 \u2013 3<\/p>\n\n\n\n = 10 \u2013 3<\/p>\n\n\n\n = 7<\/p>\n\n\n\n So, b = 7<\/p>\n\n\n\n Since y = b, y = 7<\/p>\n\n\n\n \u2234 Values of x and y are, x = 1 and y = 7<\/p>\n\n\n\n 3. If a \u2208 {- 1, 2, 3, 4, 5} and b \u2208 {0, 3, 6}, write the set of all ordered pairs (a, b) such that a + b = 5.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: a \u2208 {- 1, 2, 3, 4, 5} and b \u2208 {0, 3, 6},<\/p>\n\n\n\n To find: the ordered pair (a, b) such that a + b = 5<\/p>\n\n\n\n Then the ordered pair (a, b) such that a + b = 5 are as follows<\/p>\n\n\n\n (a, b) \u2208 {(- 1, 6), (2, 3), (5, 0)}<\/p>\n\n\n\n 4. If a \u2208 {2, 4, 6, 9} and b \u2208 {4, 6, 18, 27}, then form the set of all ordered pairs (a, b) such that a divides b and a<b.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n a \u2208 {2, 4, 6, 9} and b \u2208{4, 6, 18, 27}<\/p>\n\n\n\n Here,<\/p>\n\n\n\n 2 divides 4, 6, 18 and is also less than all of them<\/p>\n\n\n\n 4 divides 4 and is also less than none of them<\/p>\n\n\n\n 6 divides 6, 18 and is less than 18 only<\/p>\n\n\n\n 9 divides 18, 27 and is less than all of them<\/p>\n\n\n\n \u2234 Ordered pairs (a, b) are (2, 4), (2, 6), (2, 18), (6, 18), (9, 18) and (9, 27)<\/p>\n\n\n\n 5. If A = {1, 2} and B = {1, 3}, find A x B and B x A.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2} and B = {1, 3}<\/p>\n\n\n\n A \u00d7 B = {1, 2} \u00d7 {1, 3}<\/p>\n\n\n\n = {(1, 1), (1, 3), (2, 1), (2, 3)}<\/p>\n\n\n\n B \u00d7 A = {1, 3} \u00d7 {1, 2}<\/p>\n\n\n\n = {(1, 1), (1, 2), (3, 1), (3, 2)}<\/p>\n\n\n\n 6. Let A = {1, 2, 3} and B = {3, 4}. Find A x B and show it graphically<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2, 3} and B = {3, 4}<\/p>\n\n\n\n A x B = {1, 2, 3} \u00d7 {3, 4}<\/p>\n\n\n\n = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}<\/p>\n\n\n\n Steps to follow to represent A \u00d7 B graphically,<\/p>\n\n\n\n Step 1: One horizontal and one vertical axis should be drawn<\/p>\n\n\n\n Step 2: Element of set A should be represented in horizontal axis and on vertical axis elements of set B should be represented<\/p>\n\n\n\n Step 3: Draw dotted lines perpendicular to horizontal and vertical axes through the elements of set A and B<\/p>\n\n\n\n Step 4: Point of intersection of these perpendicular represents A \u00d7 B<\/p>\n\n\n\n 7. If A = {1, 2, 3} and B = {2, 4}, what are A x B, B x A, A x A, B x B, and (A x B) \u2229 (B x A)?<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2, 3} and B = {2, 4}<\/p>\n\n\n\n Now let us find: A \u00d7 B, B \u00d7 A, A \u00d7 A, (A \u00d7 B) \u2229 (B \u00d7 A)<\/p>\n\n\n\n A \u00d7 B = {1, 2, 3} \u00d7 {2, 4}<\/p>\n\n\n\n = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}<\/p>\n\n\n\n B \u00d7 A = {2, 4} \u00d7 {1, 2, 3}<\/p>\n\n\n\n = {(2, 1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)}<\/p>\n\n\n\n A \u00d7 A = {1, 2, 3} \u00d7 {1, 2, 3}<\/p>\n\n\n\n = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}<\/p>\n\n\n\n B \u00d7 B = {2, 4} \u00d7 {2, 4}<\/p>\n\n\n\n = {(2, 2), (2, 4), (4, 2), (4, 4)}<\/p>\n\n\n\n Intersection of two sets represents common elements of both the sets<\/p>\n\n\n\n So,<\/p>\n\n\n\n (A \u00d7 B) \u2229 (B \u00d7 A) = {(2, 2)}<\/p>\n\n\n\n EXERCISE 2.2 PAGE NO: 2.12<\/p>\n\n\n\n 1. Given A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, find (A x B) \u2229 (B x C).<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}<\/p>\n\n\n\n Let us find: (A \u00d7 B) \u2229 (B \u00d7 C)<\/p>\n\n\n\n (A \u00d7 B) = {1, 2, 3} \u00d7 {3, 4}<\/p>\n\n\n\n = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}<\/p>\n\n\n\n (B \u00d7 C) = {3, 4} \u00d7 {4, 5, 6}<\/p>\n\n\n\n = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)}<\/p>\n\n\n\n \u2234 (A \u00d7 B) \u2229 (B \u00d7 C) = {(3, 4)}<\/p>\n\n\n\n 2. If A = {2, 3}, B = {4, 5}, C = {5, 6} find A x (B \u222a C), (A x B) \u222a (A x C).<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given: A = {2, 3}, B = {4, 5} and C = {5, 6}<\/p>\n\n\n\n Let us find: A x (B \u222a C) and (A x B) \u222a (A x C)<\/p>\n\n\n\n (B \u222a C) = {4, 5, 6}<\/p>\n\n\n\n A \u00d7 (B \u222a C) = {2, 3} \u00d7 {4, 5, 6}<\/p>\n\n\n\n = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n (A \u00d7 B) = {2, 3} \u00d7 {4, 5}<\/p>\n\n\n\n = {(2, 4), (2, 5), (3, 4), (3, 5)}<\/p>\n\n\n\n (A \u00d7 C) = {2, 3} \u00d7 {5, 6}<\/p>\n\n\n\n = {(2, 5), (2, 6), (3, 5), (3, 6)}<\/p>\n\n\n\n \u2234 (A \u00d7 B) \u222a (A \u00d7 C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n A \u00d7 (B \u222a C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n 3.<\/strong> If A = {1, 2, 3}, B = {4}, C = {5}, then verify that: Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2, 3}, B = {4} and C = {5}<\/p>\n\n\n\n (i)<\/strong> A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)<\/p>\n\n\n\n Let us consider LHS: (B \u222a C)<\/p>\n\n\n\n (B \u222a C) = {4, 5}<\/p>\n\n\n\n A \u00d7 (B \u222a C) = {1, 2, 3} \u00d7 {4, 5}<\/p>\n\n\n\n = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}<\/p>\n\n\n\n Now, RHS<\/p>\n\n\n\n (A \u00d7 B) = {1, 2, 3} \u00d7 {4}<\/p>\n\n\n\n = {(1, 4), (2, 4), (3, 4)}<\/p>\n\n\n\n (A \u00d7 C) = {1, 2, 3} \u00d7 {5}<\/p>\n\n\n\n = {(1, 5), (2, 5), (3, 5)}<\/p>\n\n\n\n (A \u00d7 B) \u222a (A \u00d7 C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}<\/p>\n\n\n\n \u2234 LHS = RHS<\/p>\n\n\n\n (ii)<\/strong> A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n\n\n\n Let us consider LHS: (B \u2229 C)<\/p>\n\n\n\n (B \u2229 C) = \u2205 (No common element)<\/p>\n\n\n\n A \u00d7 (B \u2229 C) = {1, 2, 3} \u00d7 \u2205<\/p>\n\n\n\n = \u2205<\/p>\n\n\n\n Now, RHS<\/p>\n\n\n\n (A \u00d7 B) = {1, 2, 3} \u00d7 {4}<\/p>\n\n\n\n = {(1, 4), (2, 4), (3, 4)}<\/p>\n\n\n\n (A \u00d7 C) = {1, 2, 3} \u00d7 {5}<\/p>\n\n\n\n = {(1, 5), (2, 5), (3, 5)}<\/p>\n\n\n\n (A \u00d7 B) \u2229 (A \u00d7 C) = \u2205<\/p>\n\n\n\n \u2234 LHS = RHS<\/p>\n\n\n\n (iii)<\/strong> A \u00d7 (B \u2212 C) = (A \u00d7 B) \u2212 (A \u00d7 C)<\/p>\n\n\n\n Let us consider LHS: (B \u2212 C)<\/p>\n\n\n\n (B \u2212 C) = \u2205<\/p>\n\n\n\n A \u00d7 (B \u2212 C) = {1, 2, 3} \u00d7 \u2205<\/p>\n\n\n\n = \u2205<\/p>\n\n\n\n Now, RHS<\/p>\n\n\n\n (A \u00d7 B) = {1, 2, 3} \u00d7 {4}<\/p>\n\n\n\n = {(1, 4), (2, 4), (3, 4)}<\/p>\n\n\n\n (A \u00d7 C) = {1, 2, 3} \u00d7 {5}<\/p>\n\n\n\n = {(1, 5), (2, 5), (3, 5)}<\/p>\n\n\n\n (A \u00d7 B) \u2212 (A \u00d7 C) = \u2205<\/p>\n\n\n\n \u2234 LHS = RHS<\/p>\n\n\n\n 4. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}<\/p>\n\n\n\n (i) <\/strong>A x C \u2282 B x D<\/p>\n\n\n\n Let us consider LHS A x C<\/p>\n\n\n\n A \u00d7 C = {1, 2} \u00d7 {5, 6}<\/p>\n\n\n\n = {(1, 5), (1, 6), (2, 5), (2, 6)}<\/p>\n\n\n\n Now, RHS<\/p>\n\n\n\n B \u00d7 D = {1, 2, 3, 4} \u00d7 {5, 6, 7, 8}<\/p>\n\n\n\n = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}<\/p>\n\n\n\n Since, all elements of A \u00d7 C is in B \u00d7 D.<\/p>\n\n\n\n \u2234We can say A \u00d7 C \u2282 B \u00d7 D<\/p>\n\n\n\n (ii)<\/strong> A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n\n\n\n Let us consider LHS A \u00d7 (B \u2229 C)<\/p>\n\n\n\n (B \u2229 C) = \u2205<\/p>\n\n\n\n A \u00d7 (B \u2229 C) = {1, 2} \u00d7 \u2205<\/p>\n\n\n\n = \u2205<\/p>\n\n\n\n Now, RHS<\/p>\n\n\n\n (A \u00d7 B) = {1, 2} \u00d7 {1, 2, 3, 4}<\/p>\n\n\n\n = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}<\/p>\n\n\n\n (A \u00d7 C) = {1, 2} \u00d7 {5, 6}<\/p>\n\n\n\n = {(1, 5), (1, 6), (2, 5), (2, 6)}<\/p>\n\n\n\n Since, there is no common element between A \u00d7 B and A \u00d7 C<\/p>\n\n\n\n (A \u00d7 B) \u2229 (A \u00d7 C) = \u2205<\/p>\n\n\n\n \u2234 A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n\n\n\n 5. If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, find Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}<\/p>\n\n\n\n (i)<\/strong> A \u00d7 (B \u2229 C)<\/p>\n\n\n\n (B \u2229 C) = {4}<\/p>\n\n\n\n A \u00d7 (B \u2229 C) = {1, 2, 3} \u00d7 {4}<\/p>\n\n\n\n = {(1, 4), (2, 4), (3, 4)}<\/p>\n\n\n\n (ii)<\/strong> (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n\n\n\n (A \u00d7 B) = {1, 2, 3} \u00d7 {3, 4}<\/p>\n\n\n\n = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}<\/p>\n\n\n\n (A \u00d7 C) = {1, 2, 3} \u00d7 {4, 5, 6}<\/p>\n\n\n\n = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n (A \u00d7 B) \u2229 (A \u00d7 C) = {(1, 4), (2, 4), (3, 4)}<\/p>\n\n\n\n (iii)<\/strong> A \u00d7 (B \u222a C)<\/p>\n\n\n\n (B \u222a C) = {3, 4, 5, 6}<\/p>\n\n\n\n A \u00d7 (B \u222a C) = {1, 2, 3} \u00d7 {3, 4, 5, 6}<\/p>\n\n\n\n = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n (iv)<\/strong> (A \u00d7 B) \u222a (A \u00d7 C)<\/p>\n\n\n\n (A \u00d7 B) = {1, 2, 3} \u00d7 {3, 4}<\/p>\n\n\n\n = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}<\/p>\n\n\n\n (A \u00d7 C) = {1, 2, 3} \u00d7 {4, 5, 6}<\/p>\n\n\n\n = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n (A \u00d7 B) \u222a (A \u00d7 C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n 6. Prove that: (ii) (A \u2229 B) x C = (A x C) \u2229 (B x C)<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) <\/strong>(A \u222a B) x C = (A x C) = (A x C) \u222a (B x C)<\/p>\n\n\n\n Let (x, y) be an arbitrary element of (A \u222a B) \u00d7 C<\/p>\n\n\n\n (x, y) \u2208 (A \u222a B) C<\/p>\n\n\n\n Since, (x, y) are elements of Cartesian product of (A \u222a B) \u00d7 C<\/p>\n\n\n\n x \u2208 (A \u222a B) and y \u2208 C<\/p>\n\n\n\n (x \u2208 A or x \u2208 B) and y \u2208 C<\/p>\n\n\n\n (x \u2208 A and y \u2208 C) or (x \u2208 Band y \u2208 C)<\/p>\n\n\n\n (x, y) \u2208 A \u00d7 C or (x, y) \u2208 B \u00d7 C<\/p>\n\n\n\n (x, y) \u2208 (A \u00d7 C) \u222a (B \u00d7 C) \u2026 (1)<\/p>\n\n\n\n Let (x, y) be an arbitrary element of (A \u00d7 C) \u222a (B \u00d7 C).<\/p>\n\n\n\n (x, y) \u2208 (A \u00d7 C) \u222a (B \u00d7 C)<\/p>\n\n\n\n (x, y) \u2208 (A \u00d7 C) or (x, y) \u2208 (B \u00d7 C)<\/p>\n\n\n\n (x \u2208 A and y \u2208 C) or (x \u2208 B and y \u2208 C)<\/p>\n\n\n\n (x \u2208 A or x \u2208 B) and y \u2208 C<\/p>\n\n\n\n x \u2208 (A \u222a B) and y \u2208 C<\/p>\n\n\n\n (x, y) \u2208 (A \u222a B) \u00d7 C \u2026 (2)<\/p>\n\n\n\n From 1 and 2, we get: (A \u222a B) \u00d7 C = (A \u00d7 C) \u222a (B \u00d7 C)<\/p>\n\n\n\n (ii)<\/strong> (A \u2229 B) x C = (A x C) \u2229 (B x C)<\/p>\n\n\n\n Let (x, y) be an arbitrary element of (A \u2229 B) \u00d7 C.<\/p>\n\n\n\n (x, y) \u2208 (A \u2229 B) \u00d7 C<\/p>\n\n\n\n Since, (x, y) are elements of Cartesian product of (A \u2229 B) \u00d7 C<\/p>\n\n\n\n x \u2208 (A \u2229 B) and y \u2208 C<\/p>\n\n\n\n (x \u2208 A and x \u2208 B) and y \u2208 C<\/p>\n\n\n\n (x \u2208 A and y \u2208 C) and (x \u2208 Band y \u2208 C)<\/p>\n\n\n\n (x, y) \u2208 A \u00d7 C and (x, y) \u2208 B \u00d7 C<\/p>\n\n\n\n (x, y) \u2208 (A \u00d7 C) \u2229 (B \u00d7 C) \u2026 (1)<\/p>\n\n\n\n Let (x, y) be an arbitrary element of (A \u00d7 C) \u2229 (B \u00d7 C).<\/p>\n\n\n\n (x, y) \u2208 (A \u00d7 C) \u2229 (B \u00d7 C)<\/p>\n\n\n\n (x, y) \u2208 (A \u00d7 C) and (x, y) \u2208 (B \u00d7 C)<\/p>\n\n\n\n (x \u2208A and y \u2208 C) and (x \u2208 Band y \u2208 C)<\/p>\n\n\n\n (x \u2208A and x \u2208 B) and y \u2208 C<\/p>\n\n\n\n x \u2208 (A \u2229 B) and y \u2208 C<\/p>\n\n\n\n (x, y) \u2208 (A \u2229 B) \u00d7 C \u2026 (2)<\/p>\n\n\n\n From 1 and 2, we get: (A \u2229 B) \u00d7 C = (A \u00d7 C) \u2229 (B \u00d7 C)<\/p>\n\n\n\n 7. If A x B \u2286 C x D and A \u2229 B \u2208 \u2205, Prove that A \u2286 C and B \u2286 D.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given:<\/p>\n\n\n\n A \u00d7 B \u2286<\/strong> C x D and A \u2229 B \u2208 \u2205<\/p>\n\n\n\n A \u00d7 B \u2286<\/strong> C x D denotes A \u00d7 B is subset of C \u00d7 D that is every element A \u00d7 B is in C \u00d7 D.<\/p>\n\n\n\n And A \u2229 B \u2208 \u2205 denotes A and B does not have any common element between them.<\/p>\n\n\n\n A \u00d7 B = {(a, b): a \u2208 A and b \u2208 B}<\/p>\n\n\n\n \u2234We can say (a, b) \u2286<\/strong> C \u00d7 D [Since, A \u00d7 B \u2286<\/strong> C x D is given]<\/p>\n\n\n\n a \u2208 C and b \u2208 D<\/p>\n\n\n\n a \u2208 A = a \u2208 C<\/p>\n\n\n\n A \u2286<\/strong> C<\/p>\n\n\n\n And<\/p>\n\n\n\n b \u2208 B = b \u2208 D<\/p>\n\n\n\n B \u2286<\/strong> D<\/p>\n\n\n\n Hence proved.<\/p>\n\n\n\n EXERCISE 2.3 PAGE NO: 2.20<\/p>\n\n\n\n 1. If A = {1, 2, 3}, B = {4, 5, 6}, which of the following are relations from A to B? (i) {(1, 6), (3, 4), (5, 2)}<\/strong><\/p>\n\n\n\n (ii) {(1, 5), (2, 6), (3, 4), (3, 6)}<\/strong><\/p>\n\n\n\n (iii) {(4, 2), (4, 3), (5, 1)}<\/strong><\/p>\n\n\n\n (iv) A \u00d7 B<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n A = {1, 2, 3}, B = {4, 5, 6}<\/p>\n\n\n\n A relation from A to B can be defined as:<\/p>\n\n\n\n A \u00d7 B = {1, 2, 3} \u00d7 {4, 5, 6}<\/p>\n\n\n\n = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}<\/p>\n\n\n\n (i)<\/strong> {(1, 6), (3, 4), (5, 2)}<\/p>\n\n\n\n No, it is not a relation from A to B. The given set is not a subset of A \u00d7 B as (5, 2) is not a part of the relation from A to B.<\/p>\n\n\n\n (ii)<\/strong> {(1, 5), (2, 6), (3, 4), (3, 6)}<\/p>\n\n\n\n Yes, it is a relation from A to B. The given set is a subset of A \u00d7 B.<\/p>\n\n\n\n (iii)<\/strong> {(4, 2), (4, 3), (5, 1)}<\/p>\n\n\n\n No, it is not a relation from A to B. The given set is not a subset of A \u00d7 B.<\/p>\n\n\n\n (iv)<\/strong> A \u00d7 B<\/p>\n\n\n\n A \u00d7 B is a relation from A to B and can be defined as:<\/p>\n\n\n\n {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6),(3, 4),(3, 5),(3, 6)}<\/p>\n\n\n\n 2. A relation R is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10} as follows: (x, y) R x is relatively prime to y. Express R as a set of ordered pairs and determine its domain and range.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Relatively prime numbers are also known as co-prime numbers. If there is no integer greater than one that divides both (that is, their greatest common divisor is one).<\/p>\n\n\n\n Given: (x, y) \u2208 R = x is relatively prime to y<\/p>\n\n\n\n Here,<\/p>\n\n\n\n 2 is co-prime to 3 and 7.<\/p>\n\n\n\n 3 is co-prime to 7 and 10.<\/p>\n\n\n\n 4 is co-prime to 3 and 7.<\/p>\n\n\n\n 5 is co-prime to 3, 6 and 7.<\/p>\n\n\n\n \u2234 R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}<\/p>\n\n\n\n Domain of relation R = {2, 3, 4, 5}<\/p>\n\n\n\n Range of relation R = {3, 6, 7, 10}<\/p>\n\n\n\n 3. Let A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x \u2264 y Solution:<\/strong><\/p>\n\n\n\n A is set of first five natural numbers.<\/p>\n\n\n\n So, A= {1, 2, 3, 4, 5}<\/p>\n\n\n\n Given: (x, y) R x \u2264 y<\/p>\n\n\n\n 1 is less than 2, 3, 4 and 5.<\/p>\n\n\n\n 2 is less than 3, 4 and 5.<\/p>\n\n\n\n 3 is less than 4 and 5.<\/p>\n\n\n\n 4 is less than 5.<\/p>\n\n\n\n 5 is not less than any number A<\/p>\n\n\n\n \u2234 R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}<\/p>\n\n\n\n \u201cAn inverse relation is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original relation. If the graph of a function contains a point (a, b), then the graph of the inverse relation of this function contains the point (b, a)\u201d.<\/p>\n\n\n\n \u2234 R-1<\/sup> = {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4) (5, 5)}<\/p>\n\n\n\n (i) Domain of R\u20111<\/sup> = {1, 2, 3, 4, 5}<\/p>\n\n\n\n (ii) Range of R = {1, 2, 3, 4, 5}<\/p>\n\n\n\n 4. Find the inverse relation R-1<\/sup> in each of the following cases: Solution:<\/strong><\/p>\n\n\n\n (i)<\/strong> Given:<\/p>\n\n\n\n R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}<\/p>\n\n\n\n So, R\u20111<\/sup> = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}<\/p>\n\n\n\n (ii)<\/strong> Given,<\/p>\n\n\n\n R= {(x, y): x, y \u2208 N; x + 2y = 8}<\/p>\n\n\n\n Here, x + 2y = 8<\/p>\n\n\n\n x = 8 \u2013 2y<\/p>\n\n\n\n As y \u2208 N, Put the values of y = 1, 2, 3,\u2026\u2026 till x \u2208 N<\/p>\n\n\n\n When, y = 1, x = 8 \u2013 2(1) = 8 \u2013 2 = 6<\/p>\n\n\n\n When, y = 2, x = 8 \u2013 2(2) = 8 \u2013 4 = 4<\/p>\n\n\n\n When, y = 3, x = 8 \u2013 2(3) = 8 \u2013 6 = 2<\/p>\n\n\n\n When, y = 4, x = 8 \u2013 2(4) = 8 \u2013 8 = 0<\/p>\n\n\n\n Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.<\/p>\n\n\n\n \u2234 R = {(2, 3), (4, 2), (6, 1)}<\/p>\n\n\n\n R\u20111<\/sup> = {(3, 2), (2, 4), (1, 6)}<\/p>\n\n\n\n (iii)<\/strong> Given,<\/p>\n\n\n\n R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x \u2013 3<\/p>\n\n\n\n Here,<\/p>\n\n\n\n x = {11, 12, 13} and y = (8, 10, 12}<\/p>\n\n\n\n y = x \u2013 3<\/p>\n\n\n\n When, x = 11, y = 11 \u2013 3 = 8 \u2208 (8, 10, 12}<\/p>\n\n\n\n When, x = 12, y = 12 \u2013 3 = 9 \u2209 (8, 10, 12}<\/p>\n\n\n\n When, x = 13, y = 13 \u2013 3 = 10 \u2208 (8, 10, 12}<\/p>\n\n\n\n \u2234 R = {(11, 8), (13, 10)}<\/p>\n\n\n\n R\u20111<\/sup> = {(8, 11), (10, 13)}<\/p>\n\n\n\n 5.<\/strong> Write the following relations as the sets of ordered pairs: (ii) A relation R on the set {1, 2, 3, 4, 5, 6, 7} defined by (x, y) <\/strong>\u2208 R \u21d4 x is relatively prime to y.<\/strong><\/p>\n\n\n\n (iii) A relation R on the set {0, 1, 2,\u2026,10} defined by 2x + 3y = 12.<\/strong><\/p>\n\n\n\n (iv) A relation R form a set A = {5, 6, 7, 8} to the set B = {10, 12, 15, 16, 18} defined by (x, y) R x divides y.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) <\/strong>A relation R from the set {2, 3, 4, 5, 6} to the set {1, 2, 3} defined by x = 2y.<\/p>\n\n\n\n Let A = {2, 3, 4, 5, 6} and B = {1, 2, 3}<\/p>\n\n\n\n Given, x = 2y where y = {1, 2, 3}<\/p>\n\n\n\n When, y = 1, x = 2(1) = 2<\/p>\n\n\n\n When, y = 2, x = 2(2) = 4<\/p>\n\n\n\n When, y = 3, x = 2(3) = 6<\/p>\n\n\n\n \u2234 R = {(2, 1), (4, 2), (6, 3)}<\/p>\n\n\n\n (ii)<\/strong> A relation R on the set {1, 2, 3, 4, 5, 6, 7} defined by (x, y) \u2208 R \u21d4 x is relatively prime to y.<\/p>\n\n\n\n Given:<\/p>\n\n\n\n (x, y) R x is relatively prime to y<\/p>\n\n\n\n Here,<\/p>\n\n\n\n 2 is co-prime to 3, 5 and 7.<\/p>\n\n\n\n 3 is co-prime to 2, 4, 5 and 7.<\/p>\n\n\n\n 4 is co-prime to 3, 5 and 7.<\/p>\n\n\n\n 5 is co-prime to 2, 3, 4, 6 and 7.<\/p>\n\n\n\n 6 is co-prime to 5 and 7.<\/p>\n\n\n\n 7 is co-prime to 2, 3, 4, 5 and 6.<\/p>\n\n\n\n \u2234 R ={(2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (6, 5), (6, 7), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7)}<\/p>\n\n\n\n (iii) <\/strong>A relation R on the set {0, 1, 2,\u2026, 10} defined by 2x + 3y = 12.<\/p>\n\n\n\n Given,<\/p>\n\n\n\n (x, y) R 2x + 3y = 12<\/p>\n\n\n\n Where x and y = {0, 1, 2,\u2026, 10}<\/p>\n\n\n\n 2x + 3y = 12<\/p>\n\n\n\n 2x = 12 \u2013 3y<\/p>\n\n\n\n x = (12-3y)\/2<\/p>\n\n\n\n When, y = 0, x = (12-3(0))\/2 = 12\/2 = 6<\/p>\n\n\n\n When, y = 2, x = (12-3(2))\/2 = (12-6)\/2 = 6\/2 = 3<\/p>\n\n\n\n When, y = 4, x = (12-3(4))\/2 = (12-12)\/2 = 0\/2 = 0<\/p>\n\n\n\n \u2234 R = {(0, 4), (3, 2), (6, 0)}<\/p>\n\n\n\n (iv) <\/strong>A relation R form a set A = {5, 6, 7, 8} to the set B = {10, 12, 15, 16, 18} defined by (x, y) \u2208 R \u21d4 x divides y.<\/p>\n\n\n\n Given,<\/p>\n\n\n\n (x, y) R x divides y<\/p>\n\n\n\n Where, x = {5, 6, 7, 8} and y = {10, 12, 15, 16, 18}<\/p>\n\n\n\n Here,<\/p>\n\n\n\n 5 divides 10 and 15.<\/p>\n\n\n\n 6 divides 12 and 18.<\/p>\n\n\n\n 7 divides none of the value of set B.<\/p>\n\n\n\n 8 divides 16.<\/p>\n\n\n\n \u2234 R = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}<\/p>\n\n\n\n 6. Let R be a relation in N defined by (x, y) <\/strong>\u2208 R <\/strong>\u21d4 x + 2y = 8. Express R and R-1<\/sup> as sets of ordered pairs.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n (x, y) R x + 2y = 8 where x \u2208 N and y \u2208 N<\/p>\n\n\n\n x + 2y= 8<\/p>\n\n\n\n x = 8 \u2013 2y<\/p>\n\n\n\n Putting the values y = 1, 2, 3,\u2026\u2026 till x \u2208 N<\/p>\n\n\n\n When, y = 1, x = 8 \u2013 2(1) = 8 \u2013 2 = 6<\/p>\n\n\n\n When, y = 2, x = 8 \u2013 2(2) = 8 \u2013 4 = 4<\/p>\n\n\n\n When, y = 3, x = 8 \u2013 2(3) = 8 \u2013 6 = 2<\/p>\n\n\n\n When, y = 4, x = 8 \u2013 2(4) = 8 \u2013 8 = 0<\/p>\n\n\n\n Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.<\/p>\n\n\n\n \u2234 R = {(2, 3), (4, 2), (6, 1)}<\/p>\n\n\n\n R\u20111<\/sup> = {(3, 2), (2, 4), (1, 6)}<\/p>\n\n\n\n 7. Let A = {3, 5} and B = {7, 11}. Let R = {(a, b): a <\/strong>\u2208 A, b <\/strong>\u2208 B, a-b is odd}. Show that R is an empty relation from A into B.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n A = {3, 5} and B = {7, 11}<\/p>\n\n\n\n R = {(a, b): a \u2208 A, b \u2208 B, a-b is odd}<\/p>\n\n\n\n On putting a = 3 and b = 7,<\/p>\n\n\n\n a \u2013 b = 3 \u2013 7 = -4 which is not odd<\/p>\n\n\n\n On putting a = 3 and b = 11,<\/p>\n\n\n\n a \u2013 b = 3 \u2013 11 = -8 which is not odd<\/p>\n\n\n\n On putting a = 5 and b = 7:<\/p>\n\n\n\n a \u2013 b = 5 \u2013 7 = -2 which is not odd<\/p>\n\n\n\n On putting a = 5 and b = 11:<\/p>\n\n\n\n a \u2013 b = 5 \u2013 11 = -6 which is not odd<\/p>\n\n\n\n \u2234 R = { } = \u03a6<\/p>\n\n\n\n R is an empty relation from A into B.<\/p>\n\n\n\n Hence proved.<\/p>\n\n\n\n 8. Let A = {1, 2} and B = {3, 4}. Find the total number of relations from A into B.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n Given,<\/p>\n\n\n\n A= {1, 2}, B= {3, 4}<\/p>\n\n\n\n n (A) = 2 (Number of elements in set A).<\/p>\n\n\n\n n (B) = 2 (Number of elements in set B).<\/p>\n\n\n\n We know,<\/p>\n\n\n\n n (A \u00d7 B) = n (A) \u00d7 n (B)<\/p>\n\n\n\n = 2 \u00d7 2<\/p>\n\n\n\n = 4 [since, n(x) = a, n(y) = b. total number of relations = 2ab<\/sup>]<\/p>\n\n\n\n \u2234 Number of relations from A to B are 24<\/sup> = 16.<\/p>\n\n\n\n 9. Determine the domain and range of the relation R defined by (ii) R= {(x, x3<\/sup>): x is a prime number less than 10}<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) <\/strong>R = {(x, x+5): x \u2208 {0, 1, 2, 3, 4, 5}<\/p>\n\n\n\n Given,<\/p>\n\n\n\n R = {(x, x+5): x \u2208 {0, 1, 2, 3, 4, 5}<\/p>\n\n\n\n \u2234 R = {(0, 0+5), (1, 1+5), (2, 2+5), (3, 3+5), (4, 4+5), (5, 5+5)}<\/p>\n\n\n\n R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}<\/p>\n\n\n\n So,<\/p>\n\n\n\n Domain of relation R = {0, 1, 2, 3, 4, 5}<\/p>\n\n\n\n Range of relation R = {5, 6, 7, 8, 9, 10}<\/p>\n\n\n\n (ii) <\/strong>R= {(x, x3<\/sup>): x is a prime number less than 10}<\/p>\n\n\n\n Given,<\/p>\n\n\n\n R = {(x, x3<\/sup>): x is a prime number less than 10}<\/p>\n\n\n\n Prime numbers less than 10 are 2, 3, 5 and 7<\/p>\n\n\n\n \u2234 R = {(2, 23<\/sup>), (3, 33<\/sup>), (5, 53<\/sup>), (7, 73<\/sup>)}<\/p>\n\n\n\n R = {(2, 8), (3, 27), (5, 125), (7, 343)}<\/p>\n\n\n\n So,<\/p>\n\n\n\n Domain of relation R = {2, 3, 5, 7}<\/p>\n\n\n\n Range of relation R = {8, 27, 125, 343}<\/p>\n\n\n\n 10. Determine the domain and range of the following relations: (ii) S= {a, b): b = |a-1|, a \u2208 Z and |a| \u2264 3}<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n (i) <\/strong>R= {a, b): a \u2208 N, a < 5, b = 4}<\/p>\n\n\n\n Given,<\/p>\n\n\n\n R= {a, b): a \u2208 N, a < 5, b = 4}<\/p>\n\n\n\n Natural numbers less than 5 are 1, 2, 3 and 4<\/p>\n\n\n\n a = {1, 2, 3, 4} and b = {4}<\/p>\n\n\n\n R = {(1, 4), (2, 4), (3, 4), (4, 4)}<\/p>\n\n\n\n So,<\/p>\n\n\n\n Domain of relation R = {1, 2, 3, 4}<\/p>\n\n\n\n Range of relation R = {4}<\/p>\n\n\n\n (ii) <\/strong>S= {a, b): b = |a-1|, a \u2208 Z and |a| \u2264 3}<\/p>\n\n\n\n Given,<\/p>\n\n\n\n S= {a, b): b = |a-1|, a \u2208 Z and |a| \u2264 3}<\/p>\n\n\n\n Z denotes integer which can be positive as well as negative<\/p>\n\n\n\n Now, |a| \u2264 3 and b = |a-1|<\/p>\n\n\n\n \u2234 a = {-3, -2, -1, 0, 1, 2, 3}<\/p>\n\n\n\n For, a = -3, -2, -1, 0, 1, 2, 3 we get,<\/p>\n\n\n\n S = {(-3, |-3 \u2013 1|), (-2, |-2 \u2013 1|), (-1, |-1 \u2013 1|), (0, |0 \u2013 1|), (1, |1 \u2013 1|), (2, |2 \u2013 1|), (3, |3 \u2013 1|)}<\/p>\n\n\n\n S = {(-3, |-4|), (-2, |-3|), (-1, |-2|), (0, |-1|), (1, |0|), (2, |1|), (3, |2|)}<\/p>\n\n\n\n S = {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}<\/p>\n\n\n\n b = 4, 3, 2, 1, 0, 1, 2<\/p>\n\n\n\n So,<\/p>\n\n\n\n Domain of relation S = {0, -1, -2, -3, 1, 2, 3}<\/p>\n\n\n\n Range of relation S = {0, 1, 2, 3, 4}<\/p>\n\n\n\n 11. Let A = {a, b}. List all relations on A and find their number.<\/strong><\/p>\n\n\n\n Solution:<\/strong><\/p>\n\n\n\n The total number of relations that can be defined from a set A to a set B is the number of possible subsets of A \u00d7 B. If n (A) = p and n (B) = q, then n (A \u00d7 B) = pq.<\/p>\n\n\n\n So, the total number of relations is 2pq<\/sup>.<\/p>\n\n\n\n Now,<\/p>\n\n\n\n A \u00d7 A = {(a, a), (a, b), (b, a), (b, b)}<\/p>\n\n\n\n Total number of relations are all possible subsets of A \u00d7 A:[{(a, a), (a, b), (b, a), (b, b)}, {(a, a), (a, b)}, {(a, a), (b, a)},{(a, a), (b, b)}, {(a, b), (b, a)}, {(a, b), (b, b)}, {(b, a), (b, b)}, {(a, a), (a, b), (b, a)}, {(a, b), (b, a), (b, b)}, {(a, a), (b, a), (b, b)}, {(a, a), (a, b), (b, b)}, {(a, a), (a, b), (b, a), (b, b)}]<\/p>\n\n\n\n n (A) = 2 \u21d2 n (A \u00d7 A) = 2 \u00d7 2 = 4<\/p>\n\n\n\n \u2234 Total number of relations = 24<\/sup> = 16<\/p>\n\n\n\n RD Sharma Solutions for Class 11 Maths Chapter 2\u2013Relations<\/p>\n\n\n\n Download PDF<\/strong>: RD Sharma Solutions for Class 11 Maths Chapter 2\u2013Relations PDF<\/a><\/p>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/09\/rd-sharma-solutions-for-class-11-maths-chapter-2.png\" "\\"RD")
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(i) A x (B <\/strong>\u222a C) = (A x B) <\/strong>\u222a (A x C)
(ii) A x (B \u2229 C) = (A x B) \u2229 (A x C)
(iii) A x (B \u2013 C) = (A x B) \u2013 (A x C)<\/strong><\/p>\n\n\n\n
(i) A x C \u2282 B x D
(ii) A x (B \u2229 C) = (A x B) \u2229 (A x C)<\/strong><\/p>\n\n\n\n
(i) A x (B \u2229 C)
(ii) (A x B) \u2229 (A x C)
(iii) A x (B \u222a C)
(iv) (A x B) \u222a (A x C)<\/strong><\/p>\n\n\n\n
(i) (A \u222a B) x C = (A x C) = (A x C) \u222a (B x C)<\/strong><\/p>\n\n\n\n
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Give reasons in support of your answer.<\/strong><\/p>\n\n\n\n
Express R and R-1<\/sup> as sets of ordered pairs. Determine also
(i) the domain of R\u20111<\/sup>
(ii) The Range of R.<\/strong><\/p>\n\n\n\n
(i) R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
(ii) R= {(x, y) : x, y <\/strong>\u2208 N; x + 2y = 8}
(iii) R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x \u2013 3<\/strong><\/p>\n\n\n\n
(i) A relation R from the set {2, 3, 4, 5, 6} to the set {1, 2, 3} defined by x = 2y.<\/strong><\/p>\n\n\n\n
(i) R = {(x, x+5): x <\/strong>\u2208 {0, 1, 2, 3, 4, 5}<\/strong><\/p>\n\n\n\n
(i) R= {a, b): a <\/strong>\u2208 N, a < 5, b = 4}<\/strong><\/p>\n\n\n\nRD Sharma Solutions for Class 11 Maths Chapter 2: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise RD Sharma Solutions for Class 11 Maths :<\/strong><\/h2>\n\n\n\n
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