1.

Examine which of the following is/are functions:(i) {(1, 2), (2, 3), (3, 4), (2, 1)}(ii) {(a, 0), (b, 0), (c, 1), (d, 1)}(iii) {(1, a), (2, b), (1, b), (2, a)}(iv) {(a, a), (b, b), (c, c)}(v) {(a, b)}(vi) {(4, 1), (4, 2), (4, 3), (4, 4)}(vii) {(1, 4), (2, 4), (3, 4), (4, 4)}(viii) {(x, y) | x, y ∈ R ∧ y2 = x}(ix) {(x, y) | x, y ∈ R ∧ x2 = y}(x) {(x, y) | x, y ∈ R ∧ x = y3}(xi) {(x, y) | x, y ∈ R ∧ y = x3}

Answer»

(i) {(1, 2), (2, 3), (3, 4), (2, 1)}

It is not a function because element 2 corresponds to two elements 3 and 1.

(ii) {(a, 0), (b, 0), (c, 1), (d, 1)}

It is a function because under this each element corresponds to one and only one element.

(iii) {(1, a), (2, b), (1, b), (2, a)}

It is not a function because element 1 corresponds to two elements a and b.

(iv) {(a, a), (b, b), (c, c)}

It is a function because first element of ordered pair set is not same.

(v) {a, b}

It is a function because a corresponds to b

(vi) {(4, 1), (4, 2), (4, 3), (4, 4)}

It is not a function because first element of ordered pair set is same.

(vii) {(1, 4), (2, 4), (3, 4), (4, 4)}

It is a function because first element of ordered pair set is unequal.

(viii) {(x, y) : x, y ∈ R, y2 = x}

Here y2 = x ⇒ y = ±√x and if x = 4 then y = ±2

Hence, element of y is related with 2 and -2 so, it is not a function.

(ix) {(x, y) : x, y ∈ R, x2 = y}

It is a function because for y = x2, each real value of x there is a unique image in R for each element of R

(x) {(x, y) : x, y ∈ R, x = y3}

It is a function because y = x1/3, ∀ x ∈ R unique image is the set B.

(xi) {(x, y) : x, y ∈ R, y = x3}

It is also a function because for y = x3 ∀ x ∈ R unique image is in set B.



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