Test: Types of Functions - SAT MCQ

The function f = {(5, 3), (5, 4), (6, 4), (8, 9)} is neither one-one nor onto.
The function is not one – one 8 does not have an image in the codomain N and we know that a function can only be one – one if every element in the set M has an image in the codomain N.
A function can be onto only if each element in the co-domain has a pre-image in the domain X. In the function f = {(5, 3), (5, 4), (6, 4), (8, 9)}, 10 in the co-domain N does not have a pre-image in the domain X.
f = {(5, 3), (6, 4), (7, 9), (8, 10)} is both one-one and onto.
f = {(5, 4), (5, 9), (6, 3), (7, 10), (8, 10)} and f = {(6, 4), (7, 3),(7, 9), (8, 10)} are many – one onto.

f = {(1, 4), (2, 5), (3, 6)} is a bijective function.
One-one: It is a one-one function because every element in set A = {1, 2, 3} has a distinct image in set B = {4, 5, 6}.
Onto: It is an onto function as every element in set B = {4, 5, 6} is the image of some element in set A = {1, 2, 3}.
f = {(2, 4), (2, 5), (2, 6)} and f = {(1, 4), (1, 5), (1, 6)} are many-one onto.
f = {(1, 5), (2, 4), (3, 4)} is neither one – one nor onto.

The given function is bijective i.e. both one-one and onto.
one – one : Every element in the domain X has a distinct image in the codomain Y. Thus, the given function is one- one.
onto: Every element in the co- domain Y has a pre- image in the domain X. Thus, the given function is onto.

The above function is onto or surjective. A function f: X → Y is said to be surjective or onto if, every element of Y is the image of some elements in X.
The condition for a surjective function is for every y ∈ Y, there is an element in X such that f(x) = y.

The function f = {(10, 5), (20, 10), (30, 15)} is one-one and not onto. The function is one-one because element is set P = {10, 20, 30} has a distinct image in set Q = {5, 10, 15, 20}. The function is not onto because every element in set Q = {5, 10, 15, 20} does not have a pre-image in set P = {10, 20, 30} (20 does not have a pre-image in set P).
f = {(10, 5), (10, 10), (10, 15), (10, 20)} and f = {(10, 5), (10, 10), (20, 15), (30, 20)} are many – one onto.
f = {(20, 5), (20, 10), (30, 10)} is neither one – one nor onto.

The above statement is false. f is neither one-one nor onto.
For one-one: Consider f(x₁) = f(x₂)
∴ 5x₁⁴ + 2 = 5x₂⁴ + 2
⇒ x₁ = ± x₂.
Hence, the function is not one – one.
For onto: Consider the real number 1 which lies in co-domain R, and let 
Clearly, there is no real value of x which lies in the domain R such that f(x) = y.
Therefore, f is not onto as every element lying in the codomain must have a pre-image in the domain.

The given statement is true. f is both one-one and onto.
For one-one: Consider f(x₁) = f(x₂)
∴7x₁ + 4 = 7x₂ + 4
⇒ x₁ = x₂.
Thus, f is one – one.
For onto: Now for any real number y which lies in the co- domain R, there exists an element x=(y-4)/7
such that  Therefore, the function is onto.

The above function is an injective or one-one function.
Consider f(x₁) = f(x₂)
∴ x₁² + 12 = x₂² + 12
⇒ x₁ = x₂
Hence, it is an injective function.

The above function is one – one. A function f: X → Y is said to be one – one if each of the elements in X has a distinct image in Y.
The condition for a one-one function is for every x₁, x₂ ∈ X, f(x₁) = f(x₂) ⇒ x₁ = x₂.