3 years ago
Identify the correct statements for the given functions from the set {a, b, c, d} to itself.
a) f(a) = b, f(b) = a, f(c) = c, f(d) = d
b) f(a) = b, f(b) = a, f(c) = c, f(d) = d
c) f(a) = b, f(b) = b, f(c) = d, f(d) = c
d) f(a) = b, f(b) = b, f(c) = d, f(d) = c
e) f(a) = d, f(b) = b, f(c) = c, f(d) = d
f) f(a) = d, f(b) = b, f(c) = c, f(d) = d
June Staff answered 3 years ago
Step 1
One to one: Every image has exactly one unique pre-image in domain.
a) f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Srep 2
Therefore it is one to one.
b) f(a) = b, f(b)=a, f(c)=c, f(d)=d
b has a pre-image in domain i.e a
a has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Step 3
Therefore it is one to one.
c) f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Step 4
Therefore it is not a one to one mapping.
d) f(a) = b, f(b)=b, f(c)=d, f(d)=c
b has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
b has two pre image.
Here all elements have not a unique pre- image in domain.
Step 5
Therefore it is not a one to one mapping.
e) f(a) = d, f(b)=b, f(c)=d, f(d)=c
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
d has a pre-image in domain i.e c
c has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Step 6
Therefore it is a one to one mapping.
f) f(a)=d, f(b)=b, f(c)=c, f(d)=d
d has a pre-image in domain i.e a
b has a pre-image in domain i.e b
c has a pre-image in domain i.e c
d has a pre-image in domain i.e d
Here all elements have a unique pre- image in domain.
Therefore it is a one to one mapping.
courtneybarnettticketsvectorarena Staff answered 3 years ago
Step 1
Definition f from A to B has the property that each element of A has been assigned to exactly one element of B.
The function f is onto if and only if for every element bin B there exist an element ain A such that f(a)=b
Step 2
Solution
A={a, b, c, d}
B={a, b, c, d}
a) Given: f(a)=b, f(b)=a, f(c)=c, f(d)=d
The function f is onto, because every element of B={a, b, c, d} is the image of an element.
Step 3
b) Given: f(a)=b, f(b)=b, f(c)=d, f(d)=c
The function is not onto, because the element a is not image of any element.
Step 4
c) Given: f(a)=d, f(b)=b, f(c)=c, f(d)=d
The function is not onto, because the element a is not the image of any element.
Ideknmn Staff answered 3 years ago
Step 1
In order for a function to be a one-to-one function, for each y-value, there is exactly one corresponding x-value. Therefore,
a) Yes since for each y-value, there is exactly one corresponding x-value.
b) No since for the y-value b, there are two corresponding x-value: a and b.
c) No since for the y-value d, there are two corresponding x-value: a and d.