Test: Types of Functions - Grade 12 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.

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.

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 is a many -one function.
Consider f(x₁)=f(x₂)
∴5x₁³-8=5x₂³-8
5x₁³=5x₂³
⇒x₁ = ±x₂. Hence, the function is many – one.

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 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 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.

Given, set A = {1, 2, 3}
The equivalence relations for the given set are:
R₁ = {(1, 1), (2, 2), (3, 3)}
R₂ = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R₃ = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}
R₄ = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R₅ = {(1, 2, 3) ⇔ A x A = A²}
Therefore, the maximum number of an equivalence relation is ‘5’.

Given,
A = {1, 2, 3}
R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.
Let us write the combination of elements to check whether the given relation is reflexive, symmetric, and transitive.
R is reflexive because (1, 1),(2, 2),(3, 3) ∈ R.
R is not symmetric because (1, 2), (2, 3), (1, 3) ∈ R but (2, 1), (3, 2), (3, 1) ∉ R.
R is transitive because (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R
Therefore, R is reflexive, transitive, but not symmetric.