| 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. |
|