Question -
Answer -
(i) Consider f1 ={(1, 3), (2, 5), (3, 7)}; A = {1, 2, 3}, B = {3, 5, 7}
Injectivity:
f1 (1) = 3
f1 (2) = 5
f1 (3) = 7
⇒ Every elementof A has different images in B.
So, f1 is one-one.
Surjectivity:
Co-domain of f1 = {3, 5, 7}
Range of f1 =set of images = {3, 5, 7}
⇒ Co-domain = range
So, f1 is onto.
(ii) Consider f2 ={(2, a), (3, b), (4, c)}; A = {2, 3, 4}, B ={a, b, c}
Injectivity:
f2 (2) = a
f2 (3) = b
f2 (4) = c
⇒ Every elementof A has different images in B.
So, f2 is one-one.
Surjectivity:
Co-domain of f2 = {a, b, c}
Range of f2 =set of images = {a, b, c}
⇒ Co-domain = range
So, f2 isonto.
(iii) Consider f3 ={(a, x), (b, x), (c, z), (d, z)} ; A ={a, b, c, d,}, B = {x, y, z}
Injectivity:
f3 (a) = x
f3 (b) = x
f3 (c) = z
f3 (d) = z
⇒a and b have the same image x.
Also c and d havethe same image z
So, f3 is not one-one.
Surjectivity:
Co-domain of f3 ={x, y, z}
Range of f3 =set of images = {x, z}
So, the co-domain is not same as the range.
So, f3 is not onto.