Test: Types of Functions - Class 10 MCQ

# Test: Types of Functions - Class 10 MCQ

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## 10 Questions MCQ Test The Complete SAT Course - Test: Types of Functions

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Test: Types of Functions - Question 1

### Let M = {5, 6, 7, 8} and N = {3, 4, 9, 10}. Which one of the following functions is neither one-one nor onto?

Detailed Solution for Test: Types of Functions - Question 1

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.

Test: Types of Functions - Question 2

### Let A = {1, 2, 3} and B = {4, 5, 6}. Which one of the following functions is bijective?

Detailed Solution for Test: Types of Functions - Question 2

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.

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Test: Types of Functions - Question 3

### The following figure depicts which type of function?

Detailed Solution for Test: Types of Functions - Question 3

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.

Test: Types of Functions - Question 4

A function f: R → R is defined by f(x) = 5x- 8. The type of function is _________

Detailed Solution for Test: Types of Functions - Question 4

Test: Types of Functions - Question 5

The following figure represents which type of function?

Detailed Solution for Test: Types of Functions - Question 5

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.

Test: Types of Functions - Question 6

Let P = {10, 20, 30} and Q = {5, 10, 15, 20}. Which one of the following functions is one – one and not onto?

Detailed Solution for Test: Types of Functions - Question 6

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.

Test: Types of Functions - Question 7

A function f: R → R defined by f(x) = 5x+ 2 is one – one but not onto.

Detailed Solution for Test: Types of Functions - Question 7

The above statement is false. f is neither one-one nor onto.
For one-one: Consider f(x1) = f(x2)
∴ 5x1+ 2 = 5x2+ 2
⇒ x= ± 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.

Test: Types of Functions - Question 8

The function f: R → R defined as f(x) = 7x + 4 is both one-one and onto.

Detailed Solution for Test: Types of Functions - Question 8

The given statement is true. f is both one-one and onto.
For one-one: Consider f(x1) = f(x2)
∴7x+ 4 = 7x+ 4
⇒ x= 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.

Test: Types of Functions - Question 9

A function f∶ N → N is defined by f(x) = x+ 12. What is the type of function here?

Detailed Solution for Test: Types of Functions - Question 9

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

Test: Types of Functions - Question 10

The following figure depicts which type of function?

Detailed Solution for Test: Types of Functions - Question 10

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) ⇒ x= x2.

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