 Hello and welcome to the session. In this session, we will discuss a question which says that if A and B belong to the set containing the elements 1, 2, 3, 4 and 5, then which of the following are functions in the given sets? First part is, F1 is a set containing the order pair AB such that A is equal to B minus 1 and second part is, F2 is a set containing the order pair AB such that A plus B is equal to 6. Now the first part in the solution of this question, we should know a result and that is string for a function. Now we can test whether a given relation is a function or not by applying the following tests. We have to examine whether the first set that is the domain and secondly we have to examine that the first members to use up, that means the elements 1, 2 and 3 of our group pairs that is 1, 2 and 3 are different. Therefore it is qualifying both these tests. Now in any case when the domain is not fully used up or the first members of all the ordered pairs are not different, then in that case the relation is not a function. Now this result will work out as a key idea to this question. And now we will start with the solution. We are given this and we will start with the first. Now for the first part given the order pair AB such that A is equal to B minus 1. B belongs to the set containing the elements 1, 2, 3, 4 and 5. We will express F1 as a set of ordered pairs B belongs to the set containing these elements. Now for A is equal to 1, A is equal to B minus 1 as it is given in the condition. Therefore A is equal to 1 minus 1 which is equal to 0. So the required ordered pair AB will be equal to 0, 1. But it is also given that the set containing these elements. But here A is equal to 0 which is not the element of the set. Therefore we will not consider this ordered pair equal to B minus 1 which is equal to 2 minus 1 which is 1. Therefore the ordered pair AB will be equal to 1, 2. Equal to B minus 1 which is equal to 3 minus 1 which is 2. Therefore the required ordered pair AB will be equal to 0, 3. Now for B is equal to 4, A is equal to B minus 1 which is equal to 3. Therefore the required ordered pair AB is equal to 3, 4. And lastly for B is equal to B minus 1 which is equal to 5 minus 1 which is 4. Therefore the required ordered pair AB is F1 is equal to set containing the ordered pairs 1, 2, 2, 3. Now F1 is a relation from the set to the set itself. So now we will draw an arrow diagram for this relation. Now F is that is the relation F1 from to the set itself. The first ordered pair is 1, 2 that means 1 is connected with 2. Second ordered pair is the third ordered pair is 3 and the last is 4, 5. Now using the results which are given on the key idea. Now in this case this is the domain and the domain are connected except this element. Therefore we can say that the domain is not is left associated that means the domain is not fully used up. Therefore the relation F1 is not a function. Now let us discuss the second part of the ordered ordered pair AB such that it is equal to 6. It is given that AB belong to the set containing the elements 1, 2, 3, 4 and 5 ordered pairs by using this condition. And also we will choose AB from this set. Then F2 is equal to the set containing the ordered pairs 1 as A is 1, B is 5 plus B that is 1 plus 5 is equal to. Now let us draw the pair we will do. Now here also A plus B is equal to 6 that means 2 plus 4 is equal to 6. Similarly there will be 3, 3. Then the next will be 4, 2. Then the next will be 5, 1. So in every other pair is equal to 6 itself. So now we will draw an L diagram for this relation. Now this relation is from this set to this set. In this relation the first ordered pair is 1, 5. That means 1 the next ordered pair is 3, 3. Now here it is containing the elements 1, 2, 3, 4 and 5. That means the domain is fully used up. The domain used up is the solution of the given question. And that's all for the session. Hope you all have enjoyed the session.