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Easily improve your preparation level with the help of or Functions Worksheet. Check out the step-by-step solution and get complete knowledge of the concept. Mainly domain, co-domain, and range of functions are covered in the Worksheet on Mapping or Functions.
Do Read: Worksheet on Math Relation
Functions or Mapping Worksheet with Solutions
1. Which of the following represents a mapping?
(a) {(5, 3); (6, 4); (8, 6); (10, 8)}
(b) {(3, 9); (4, 13); (5, 17)}
(c) {(4, 8); (4, 12); (5, 10); (6, 12)}
(d) {(2, 3); (3, 4); (4, 5); (5, 6)}
(e) {(3, 2); (4, 2); (6, 2); (8, 2)}
(f) {(2, 4); (2, 6); (3, 6)}
Solution:
(a) Given that {(5, 3); (6, 4); (8, 6); (10, 8)}
Let the two sets are P and Q.
The required diagram is
Different elements of P can have the same image in Q. Adjoining figure represents a mapping.
(b) Given that {(3, 9); (4, 13); (5, 17)}
Let the two sets are P and Q.
The required diagram is
Different elements of P can have the same image in Q. Adjoining figure represents a mapping.
(c) Given that {(4, 8); (4, 12); (5, 10); (6, 12)}
Let the two sets are P and Q.
The required diagram is
No element of P must have more than one image. The adjoining figure does not represent a mapping since element 4 in set P is associated with two elements 8, 12 of set Q.
(d) Given that {(2, 3); (3, 4); (4, 5); (5, 6)}
Let the two sets are P and Q.
The required diagram is
Different elements of P can have the same image in Q. Adjoining figure represents a mapping.
(e) Given that {(3, 2); (4, 2); (6, 2); (8, 2)}
Let the two sets are P and Q.
The required diagram is
Different elements of P can have the same image in Q. Adjoining figure represents a mapping.
(f) Given that {(2, 4); (2, 6); (3, 6)}
Let the two sets are P and Q.
The required diagram is
No element of P must have more than one image. The adjoining figure does not represent a mapping since element 2 in set P is associated with two elements 4, 6 of set Q.
Therefore, (a), (b), (d), (e) represents a mapping.
2. Which of the following arrow diagrams represents a mapping?
(a)
(b)
(c)
(d)
Solution:
(a) Different elements of P can have the same image in Q. Adjoining figure represents a mapping.
(b) Every element of P must have an image in Q. Adjoining figure does not represent a mapping since the element z in set P is not associated with any element of set Q.
No element of P must have more than one image. The adjoining figure does not represent a mapping since element x in set P is associated with two elements m, n of set Q.
(c) Every element of P must have an image in Q. Adjoining figure does not represent a mapping since the element y in set P is not associated with any element of set Q.
(d) No element of P must have more than one image. The adjoining figure does not represent a mapping since element x in set P is associated with two elements m, o of set Q.
Therefore, (a) represents a mapping.
3. A function f is defined by f(x) = 3x – 5. Write the values of
(a) f(0)
(b) f(-3)
(c) f(4)
(d) f(-2)
Solution:
Given that a function f is defined by f(x) = 3x – 5.
(a) f(0)
To find the f(0), substitute 0 in the place of the x.
The given equation is f(x) = 3x – 5
f(0) = 3 (0) – 5 = 0 – 5 = -5
f(0) = -5
Therefore, the answer is f(0) = -5
(b) f(-3)
To find the f(-3), substitute -3 in the place of the x.
The given equation is f(x) = 3x – 5
f(-3) = 3 (-3) – 5 = -9 – 5 = -14
f(-3) = -14
Therefore, the answer is f(-3) = -14
(c) f(4)
To find the f(4), substitute 4 in the place of the x.
The given equation is f(x) = 3x – 5
f(4) = 3 (4) – 5 = 12 – 5 = 7
f(4) = 7
Therefore, the answer is f(4) = 7
(d) f(-2)
To find the f(-2), substitute -2 in the place of the x.
The given equation is f(x) = 3x – 5
f(-2) = 3 (-2) – 5 = -6 – 5 = -11
f(-2) = -11
Therefore, f(0) = -5; f(-3) = -14; f(4) = 7; f(-2) = -11
4. Find the range of each of the following functions.
(a) f(x) = 7 – x, x ∈ N, x > 0
(b) f(x) = x² + 4, x ∈ R
Solution:
(a) Given that f(x) = 7 – x, x ∈ N, x > 0
x > 0
Multiply -1 on both sides.
– x < 0
Add 7 on both sides
7 – x < 0 + 7
7 – x < 7
We know that f(x) = 7 – x
f(x) < 7
We know that value of f(x) is less than 7
Hence, Range = (-∞, 7)
(b) Given that f(x) = x² + 4, x ∈ R
x² ≥ 0
Add 4 on both sides.
x² + 4 ≥ 0 + 4
x² + 4 ≥ 4
We know that f(x) = x² + 4
f(x) ≥ 4
Hence, Range of f(x) = (4, ∞)
5. Let M = {2, 4, 8, 7} and N = {5, 9, 17, 18, 19}
Consider the rule f(x) = x + 3, where x ∈ A.
Represent the mapping in the roster form.
Also, find the domain and range of the mapping.
Solution:
Given that M = {2, 4, 7, 8} and N = {5, 9, 17, 18, 19}
Consider the rule f(x) = x + 3, where x ∈ A.
For the required Relation we have to find the ordered pairs where x co-ordinate is from set M which is mapped to the y co-ordinate from set N.
If f(2) = 2 + 3 = 5
f(4) = 4 + 3 = 7
f(7) = 7 + 3 = 10
f(8) = 8 + 3 = 11
A set of ordered pairs r = {(2, 5); (4, 7); (7, 10); (8, 11)}
Domain = Set of all first elements in a relation = {2, 4, 7, 8}
Range = Set of all second elements in a relation = {5, 7, 10, 11}
6. Let A = {2, 3, 6} B = {3, 4, 5, 6, 8, 12}
Draw the arrow diagram to represent the rule f(x) = 2x from A to B.
Solution:
Given that A = {2, 3, 6} B = {3, 4, 5, 6, 8, 12}
Consider the rule f(x) = 2x from A to B
For the required Relation we have to find the ordered pairs where x co-ordinate is from set A which is mapped to the y co-ordinate from set B.
If f(2) = 2 (2) = 4
f(3) = 2 (3) = 6
f(6) = 2 (6) = 12
A set of ordered pairs r = {(2, 4); (3, 6); (6, 12)}
7. Let A = {4, 9, 12} and B = {2, 3, 4}
(a) Show that the relation R = {(4, 2), (9, 3)} is not a mapping from A to B.
(b) Show that the relation R = {(4, 2); (4, 4); (9, 3); (12, 2); (12, 4)} from A to B is not a mapping from A to B.
Solution:
Given that A = {4, 9, 12} and B = {2, 3, 4}
(a) R = {(4, 2), (9, 3)} is not a mapping from A to B.
Every element of A must have an image in B. From the given information, set A element 12 is not associated with any element of set B.
Therefore, R = {(4, 2), (9, 3)} is not a mapping from A to B.
(b) R = {(4, 2); (4, 4); (9, 3); (12, 2); (12, 4)} from A to B
No element of A must have more than one image. From the given information, set A elements 4 and 12 are associated with two elements of set B. Element 4 of Set A associated with 2 and 4. Also, the 12 of Set A associated with 2 and 4.
Therefore, R = {(4, 2); (4, 4); (9, 3); (12, 2); (12, 4)} from A to B is not a mapping from A to B.
8. Let A = {3, 4, 5} and B = {12, 17, 22}
Consider the rule f(x) = 5x – 3, where x ∈ A
(a) Show that f is a mapping from A to B.
(b) Find the domain and range of the mapping.
(c) Represent the mapping in the roster form.
(d) Draw the arrow diagram to represent the mapping.
Solution:
Given that A = {3, 4, 5} and B = {12, 17, 22}
(a) If f(x) = 5x – 3, where x ∈ A
Substitute elements of A in f(x)
f(3) = 5 (3) – 3 = 12
f(4) = 5 (4) – 3 = 17
f(5) = 5 (5) – 3 = 22
Different elements of A can have the same image in B. Adjoining figure represents a mapping.
Therefore, f is a mapping from A to B.
(b) Domain = Set of all first elements in a relation = {3, 4, 5}
Range = Set of all second elements in a relation = {12, 17, 22}
(c) The mapping in the roster form r = {(3, 12); (4, 17); (5, 22)}
(d)