 Hello and welcome to the session. In this session, we will discuss a question which says that a set of white circular cones is such that each by the radius of 15 cm, the height of the cone varies, the maximum height being 28 cm. Let h represent the height and read the volume of the cones where r and h belong to the set of real numbers. Then first part define the relation between h and v as a set of ordered pairs. Second part states the domain and range of the relation. And third part is every relation a function. Now before starting the solution of this question, we should know a result. And that is testing for a function. Now you can test whether a given relation is a function or not by applying the following test. First is examine whether the first set that is the domain is fully used up. And secondly examine whether the first members of all ordered pairs are different. Now consider this relation from a to b. Now in this relation the domain which is a set containing the elements 1, 2 and 3 is fully used up. That means all the elements in the domain are connected with the unique elements of the set b. So the relations are from a to b as a set containing the ordered pairs 1, a, 2b and 3c. Now here you can see that the first members of all the ordered pairs are different. Now whenever the domain is fully used up the ordered pairs are different then the given relation is a function. Therefore the relation r is a function. Now in any case when the domain is more fully used up or the first members of all ordered pairs are not different then in that case the relation is not a function. Now this result will work out as a t-idea for solving out this question. And now we will start with the solution. Now here the radius of the cones is 15 cm and the height is represented by h and the volume of the cones is represented by v and also the set of real numbers. And in the first part we have to define the relation between h as a set of ordered pairs of radius 15 cm is given by is equal to 1 by 3 by r square h. And here 15 cm therefore v is equal to 1 by 3 into pi into 15 into 15 into h cm2. Now 3 into 5 is 15 so this is equal to 75 by h cm2. Now we have to define a relation and v as a set of ordered pairs. Now the relation and v belongs to the set of real numbers. And now we have to state the domain and range of the relation. Now taking the ordered pair h v such that v is equal to 75 by h where h and v belongs to the set of real numbers. Now it is given in the equation that the height of the cone varies and the maximum height being 28 cm. So given the maximum value 28 cm Now we know that domain is a set containing the ordered pairs in the relation. Now for the relation x the first components of all the ordered pairs and values of the height x. The maximum value of h is 28 cm. So where it will be equal to the set containing h such that 0 is equal to 28 where h belongs to set of real numbers. If your volume and height is, volume is equal to cm3. So for the maximum height which is equal to 28 cm the volume v will be equal to 75 into pi into 28 cm3. Which implies volume is equal to 75 into 22 by 7 into 28 cm3 which further gives volume is equal to, now here 7 into 4 is 28. So on solving this this will be 6600 cm3. Now we know that the values of all the ordered pairs in or for the relation of all the ordered pairs will be the different values of the volume v. Now here the maximum value of the volume is 6600 cm2 is equal to set containing element v 0 is less than v less than equal to 6600 where v belongs to set of real numbers. Now let us start with the third part. And for this we will use these tests which are given in the key I here. Now here the relation x is equal to the set containing the ordered pair hv 75 and v belongs to set of real numbers. Now for this relation for every different value values of h can have the same and secondly every value of h will associate different value say that the domain which is containing different values of h in this relation will have different values of the solution of the given question and that's all for this session. Hope you all have enjoyed the session.