Let $ A = \left\{ {1,2,3,...45} \right\} $ and R be the relation ‘Is square of’ in A. Which of the following is false?
A. $ R = \left\{ {(1,1),(4,2),(9,3),(16,4),(25,5),(36,6)} \right\} $
B.Domain of $ R = \left\{ {1,4,9,16,25,36} \right\} $
C.Range of $ R = \left\{ {1,2,3,4,5,6} \right\} $
D.Range of $ R = \left\{ {3,4,5,6} \right\} $

Hint: According to given in the question we have a se i.e. $ A = \left\{ {1,2,3,...45} \right\} $ . We have to find the relation and domain. Let us first know what is the definition of Domain. The domain of the function is the complete set of possible values of the independent variables. So we have to find all the possible values i.e. the square of all the numbers in the function which is the relation of A .

Complete step-by-step answer:
Here we have $ A = \left\{ {1,2,3,...45} \right\} $ and R be the relation ‘Is square of’ in A.
To find the R we can write $ {(1)^2} = 1,{(2)^2} = 4,{(3)^2} = 9,{(4)^2} = 16,{(5)^2} = 25,{(6)^2} = 36 $ .
Therefore we have $ R = \left\{ {(1,1),(4,2),(9,3),(16,4),(25,5),(36,6)} \right\} $ .
From the definition we know that Domain of $ R = \left\{ {x;(x,y) \in R} \right\} = \left\{ {1,4,9,16,25,36} \right\} $
So we can write a range of $ $ $ R = \left\{ {y;(x,y) \in R} \right\} = \left\{ {1,2,3,4,5,6} \right\} $ .
We can see that all the options given in the question match the solution except (d).
Hence we can say that option (d) is false.
So, the correct answer is “Option d”.

Note: We should note that the definition of Range says that the range of a function is the complete set of all possible values of the dependent variable i.e. $ y $ after we have substituted the domain. So in the above solution we have first substituted the domain of $ R $ and then the range of $ R $ . We have here $ x = 1,4,9,16,25,36 $ and for variable $ y = $ $ 1,2,3,4,5,6 $ .