**Question:**

** 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.**

**Solution:**

(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.

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.

Hence, t is a function.

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