 Hello and welcome to the session. In this session we will discuss the following question and the question says, which of the following relations are functions give reasons in case of a function determine its domain and range? Part a, r1 is equal to the set containing the ordered pair 2, 3, ordered pair 5, 10, ordered pair 3, 7, ordered pair minus 1, 4. Part b, r2 is equal to the set containing the ordered pair 1, 2, ordered pair 1, minus 3, ordered pair 2, 4, ordered pair 3, 6. Part c, r3 is equal to the set containing the ordered pair a1, b2, ordered pair a2, b1, ordered pair a3, b4. Now before we start solving the question let us first recall condition for a relation to be a function. A relation f which is a mapping from the set a to b is a function f1. All elements of the set a are used up to an element of set a is not paired with more than one element of set b. So this is our key idea for this question and using this key idea we will solve the question. Let's start the solution now. In part a we are given the relation r1 which is equal to the set containing the ordered pair 2, 3, ordered pair 5, 10, ordered pair 3, 7 and ordered pair minus 1, 4. We have to find if this relation is a function. First let's go back to the key idea. We have a relation f from a to b is a function if all elements of the set a are used up and an element of set a is not paired with more than one element of set b that is no two ordered pairs should have same first component. Now in r1 we can see that all the ordered pairs have unique first component. So we can say that since no two ordered pairs in r1 have same first component therefore r1 is a function. Now we will find out the range and domain of the function r1. We know that the domain of a function is the set of all first components of all the ordered pairs in the function. Now the first components of all the ordered pairs are 2, 5, 3, minus 1. So domain of r1 is equal to the set containing the elements 2, 5, 3, minus 1. Also we know that the range of a function is the set of all second components of all the ordered pairs in the function. Now the second components of all the ordered pairs are 3, 10, 7 and 4. So range of r1 is equal to the set containing the elements 3, 10, 7, 4. With this we complete part a. Now in part b we are given that relation r2 is equal to the set containing the ordered pair 1, 2, ordered pair 1, minus 3, ordered pair 2, 4 and ordered pair 3, 6. Now if we see the ordered pairs in the relation r2 we can see that the ordered pair 1, 2 and the ordered pair 1, minus 3. We can see that the ordered pair 1, 2 and the ordered pair 1, minus 3 have same first component so the element 1 of domain is paired with more than one element of range. So we can say that since ordered pair 1, 2 and ordered pair 1, minus 3 have the same first components therefore r2 is not a function so this completes part b. Now in part c we are given the relation r3 which is equal to the set containing the ordered pair a1, b2, ordered pair a2, b1 and the ordered pair a3, b4. We can see that all the ordered pairs have unique first components so we can say that since no two ordered pairs have same first components therefore r3 is a function now we will find the range and domain of r3. All the first components of the ordered pairs in r3 are a1, a2 and a3. So domain of r3 is equal to the set containing the elements a1, a2, a3. Also all the second components of the ordered pairs in r3 are b2, b1, b4. So the range of r3 is equal to the set containing the elements b2, b1, b4. This completes part c and with this we end our session hope you enjoyed the session.