Is the given relation a function? Give reasons for your answer.
(i) h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}
(ii) f = {(x, x) | x is a real number}
(iii) g = n, (1/n) |n is a positive integer
(iv) s = {(n, n2) | n is a positive integer}
(v) t = {(x, 3) | x is a real number.
(i) According to the question,
h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}
Therefore, element 3 has two images, namely, 9 and 11.
A relation is said to be function if every element of one set has one and only one image in other set.
Hence, h is not a function.
(ii) According to the question,
f = {(x, x) | x is a real number}
This means the relation f has elements which are real number.
Therefore, for every x ∈ R there will be unique image.
A relation is said to be function if every element of one set has one and only one image in other set.
Hence, f is a function.
(iii) According to the question,
g = n, (1/n) |n is a positive integer
Therefore, the element n is a positive integer and the corresponding 1/n will be a unique and distinct number.
Therefore, every element in the domain has unique image.
A relation is said to be function if every element of one set has one and only one image in other set.
Hence, g is a function.
(iv) According to the question,
s = {(n, n2) | n is a positive integer}
Therefore, element n is a positive integer and the corresponding n2 will be a unique and distinct number, as square of any positive integer is unique.
Therefore, every element in the domain has unique image.
A relation is said to be function if every element of one set has one and only one image in other set.
Hence, s is a function.
(v) According to the question,
t = {(x, 3) | x is a real number.
Therefore, the domain element x is a real number.
Also, range has one number i.e., 3 in it.
Therefore, for every element in the domain has the image 3, it is a constant function.
A relation is said to be function if every element of one set has one and only one image in other set.
Hence, t is a function.