Exercise 2.1
Question1. Iffind the values of x and y
Solution :
Here
x/3 + 1 = 5/3 and y – 2/3 = 1/3
x/3 + 1 = 5/3 and 3y -2 = 1
x = 2 and 3y = 3
x = 2 and y =1and
Question2.If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A×B).
Solution :
Number of elements in set A = 3 and Number of elements in set B = 3
∴ Number of elements in A × B = 3 × 3 = 9
Question3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Solution :
Given: G = {7, 8} and H = {5, 4, 2}
∴ G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
And H × G = {(5, 7), (4, 7), (2, 7), (5, 8), (4, 8), (2, 8)}
Question4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = { n, m}, then P × Q = {(m, n),(n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ ϕ) = ϕ
Solution :
(i) Here P = {m, n} and Q = { n, m}
Number of elements in set P = 2 and Number of elements in set Q = 2
∵ Number of elements in P × Q = 2 × 2 = 4
But P × Q = {(m,n),(),}and here number of elements in P × Q = 2
Therefore, statement is false.
(ii) True
(iii) True
Question5. If A = {–1, 1}, find A × A × A.
Solution :
It is known that for any non-empty set A, A × A × A is defined as
A × A × A = {(a, b, c): a, b, c ∈ A}
It is given that A = {–1, 1}
∴ A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}
Question6. If A × B = {(a, x),(a , y), (b, x), (b, y)}. Find A and B
Solution :
It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}
We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
∴ A is the set of all first elements and B is the set of all second elements.
Thus, A = {a, b} and B = {x, y}
Question7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}.
Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C).
(ii) A × C is a subset of B × D.
Solution :
Given: A = {1, 2}, B = {1, 2, 3, 4}, C
= {5, 6} and D = {5, 6, 7, 8}
Question8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?
List them.
Solution :
A = {1, 2} and B = {3, 4}
∴A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
⇒ n(A × B) = 4
We know that if C is a set with n(C) = m, then n[P(C)] = 2m.
Therefore, the set A × B has 24
= 16 subsets. These are ϕ
, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)},
{(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)},
{(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}
Question9: Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.
Solution :We are provided with the fact that n(A)=3 and n(B)=2 ; and(x,1),(y,2),(z,1)
are in A×B
.
We also know that, A
is the set of all the first elements and B
is the set of all the second elements.
So, we can conclude, A
having elements x,y,z and B having elements 1,2
.
Thus, we get, n(A)=3,n(B)=2
.
So, A={x,y,z},B={1,2}
Question10: The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0,1). Find the set A and the remaining elements of A × A.
Solution :.We are provided with, n(A×A)=9
.
We also know that, if n(A)=a,n(B)=b
, then n(A×B)=ab
As it is given that, n(A×A)=9
It can be written as,
n(A)×n(A)=9
⇒n(A)=3
And it is also given that (−1,0),(0,1)
are the two elements of A×A
.
Again, the fact is also known that, A×A={(a,a):a∈A}
. And also −1,0,1 are the elements of A
.
Also, n(A)=3
, implies A={−1,0,1}
.
So, (−1,−1),(−1,1),(0,−1),(0,0),(1,−1),(1,0),(1,1)
are the remaining elements of A×A .
Exercise 2.2
Question1. Let A = {1, 2, 3,…,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.
Solution :
Given: A = {1, 2, 3, ……….., 14}
The ordered pairs which satisfy 3x – y=0 are (1, 3), (2, 6), (3, 9) and (4, 12).
∴ R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Domain = {1, 2, 3, 4}
Range = {3, 6, 9, 12}
Co-domain = {1, 2, 3, ……….., 14}
Question2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈N}. Depict this relationship using roster form. Write down the domain and the range.
Solution :
R = {(x, y): y = x + 5, x is a natural number less than 4, x, y∈ N}
The natural numbers less than 4 are 1, 2, and 3.
∴R = {(1, 6), (2, 7), (3, 8)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
∴ Range of R = {6, 7, 8}
Question3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Solution :
A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
∴R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
Question4. Figure shows a relationship between the sets P and Q. Write this relation:
(i) in set-builder form
(ii) roster form
What is its domain and range?
Solution :
According to the given figure, P = {5, 6, 7}, Q = {3, 4, 5}
(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}
Question5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a , b ∈A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Solution :
A = {1, 2, 3, 4, 6}, R = {(a, b): a, b ∈ A, b is exactly divisible by a}
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) Domain of R = {1, 2, 3, 4, 6}
(iii) Range of R = {1, 2, 3, 4, 6}
Question6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
Solution :
R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
∴ R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
∴Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {5, 6, 7, 8, 9, 10}
Question7: Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Solution :
R = {(x, x3) : x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
∴R = {(2, 8), (3, 27), (5, 125), (7, 343)}
Question8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution :
It is given that A = {x, y, z} and B = {1, 2}.
∴ A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
Since n(A × B) = 6, the number of subsets of A × B is 26.
Therefore, the number of relations from A to B is 26.
Question9. Let R be the relation on Z defined by R = {(a,b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution :
R = {(a, b): a, b ∈ Z, a – b is an integer}
It is known that the difference between any two integers is always an integer.
∴Domain of R = Z
Range of R = Z
Exercise 2.3
Question1. Which of the following are functions? Give reasons. If it is a function determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}
Solution :We have the given relation as, {(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)}
.
Thus, we can see, the domain of the relation consists of {2,5,8,11,14,17}
and range is {1}
.
And we also have, every element of the domain is having their unique images, then it is a function.
(ii) {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)}
Ans: We have our given relation, {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)}
.
Thus, we have our domain as, {2,4,6,8,10,12,14}
and range as, {1,2,3,4,5,6,7}
.
Every element of the domain is having their unique images, so this is a function.
(iii) {(1,3),(1,5),(2,5)}
Ans: Our given relation is, {(1,3),(1,5),(2,5)}
.
From the domain of the relation the element 1
is having two different images 3,5
.
So,every element of the domain is not having their unique images. So, this is not a function.
Question2. Find the domain and range of the following real functions:
(i) f(x) = -|x|
(ii) f(x) = √(9 – x²)
Solution :
Question3. A function f is defined by f(x) = 2x –5. Write down the values of
(i) f (0), (ii) f (7), (iii) f (–3).
Solution :
Given: f(x) 2x – 5
(i) Putting We have the given function as, f(x)=2x−5
So, the value of,
f(0)=2×0−5=−5
(ii) Putting
We have the given function as, f(x)=2x−5
So, the value of,
f(7)=2×7−5=14−5=9
(iii) Putting
We have the given function as, f(x)=2x−5
So, the value of,
f(−3)=2×(−3)−5=−6−5=−11
Question4. The function t which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by Find:
(i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212.
Solution :
Question5. Find the range of each of the following functions.
(i) f (x) = 2 – 3x, x ∈ R , x > 0.
(ii) f (x) = x 2 + 2, x is a real number.
(iii) f (x) = x, x is a real number.
Solution :
(i) f(x) = 2 – 3x, x ∈ R, x > 0
The values of f(x) for various values of real numbers x > 0 can be written in the tabular form as
| x | 0.01 | 0.1 | 0.9 | 1 | 2 | 2.5 | 4 | 5 | … |
| f(x) | 1.97 | 1.7 | –0.7 | –1 | –4 | –5.5 | –10 | –13 | … |
Thus, it can be clearly observed that the range of f is the set of all real numbers less than 2.
i.e., range of f = (–∞, 2)
Alter:
Let x > 0
⇒ 3x > 0
⇒ 2 –3x < 2
⇒ f(x) < 2
∴Range of f = (–∞, 2)
(ii) f(x) = x2 + 2, x, is a real number
The values of f(x) for various values of real numbers x can be written in the tabular form as
| x | 0 | ±0.3 | ±0.8 | ±1 | ±2 | ±3 | … | |
| f(x) | 2 | 2.09 | 2.64 | 3 | 6 | 11 | ….. |
Thus, it can be clearly observed that the range of f is the set of all real numbers greater than 2.
i.e., range of f = [2, ∞)
Alter:
Let x be any real number.
Accordingly,
x2 ≥ 0
⇒ x2 + 2 ≥ 0 + 2
⇒ x2 + 2 ≥ 2
⇒ f(x) ≥ 2
∴ Range of f = [2,∞)
(iii) f(x) = x, x is a real number
It is clear that the range of f is the set of all real numbers.
∴ Range of f = R
Miscellaneous Exercise
Question1. The relation f is defined by.
The relation g is defined by. Show that f is a function and g is not a function.
Solution :
Given: f and
Question2. If f(x) = x2 find
Solution :
Given: f(x) = x2
At
Question3. Find the domain of the function
Solution :
Question4. Find the domain and range of the real function f defined by f(x) = √|x – 1|
Solution :
The given real function is f(x)=√|x−1
It can be seen that√| x−1
is defined for (x−1)≥0
i.e., f(x)= √|(x−1)
is defined for x≥1
Therefore the domain of f is the set of all real numbers greater than or equal to 1 i.e.,
the domain of f =[1,∞)
As x≥1
⇒(x−1)≥0
⇒√|x−1≥0
f(x)≥0
Therefore the range of f is the set of all real numbers greater than or equal to 0
i.e., the range of f =[0,∞)
Question5. Find the domain and the range of the real function f defined by f (x) = |x – 1|.
Question6. Letbe a function from R into R. Determine the range of f
Solution :
Here
Put
substituting values and Determing the image , we have
Question7. Let R → R be defined respectively by Find f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.
Solution :
Given: We have the functions defined as, f,g:R→Ris defined as, f(x)=x+1,g(x)=2x−3
Thus, the function
(f+g)(x)=f(x)+g(x)
=(x+1)+(2x−3)
=3x−2
So, the function (f+g)(x)=3x−2
.
Again, the function,
(f−g)(x)=f(x)−g(x)
=(x+1)−(2x−3)
=−x+4
So, the function (f−g)(x)=−x+4
.
Similarly,
(fg)(x)=f(x)g(x)
where g(x)≠0 and also x∈R
.
Now, putting the values,
(fg)(x)=x+12x−3
where,
2x−3≠0
⇒x≠32
Question 8. Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Solution :
f = {(1, 1), (2, 3), (0, –1), (–1, –3)}
f(x) = ax + b
(1, 1) ∈ f
⇒ f(1) = 1
⇒ a × 1 + b = 1
⇒ a + b = 1
(0, –1) ∈ f
⇒ f(0) = –1
⇒ a × 0 + b = –1
⇒ b = –1
On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2.
Thus, the respective values of a and b are 2 and –1.
Question9. Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b2}. Are the following true?
(i) (a, a) ∈ R, for all a ∈ N
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
Justify your answer in each case.
Solution :
R = {(a, b): a, b ∈ N and a = b2}
(i) It can be seen that 2 ∈ N;however, 2 ≠ 22 = 4.
Therefore, the statement “(a, a) ∈ R, for all a ∈ N” is not true.
(ii) It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.
Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N
Therefore, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.
(iii) It can be seen that (16, 4) ∈ R, (4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 42 and 4 = 22.
Now, 16 ≠ 22 = 4; therefore, (16, 2) ∉ N
Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.
Question10. Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}. Are the following true?
(i) f is a relation from A to B (ii) f is a function from A to B. Justify your answer in each case.
Solution :
A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}
∴A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.
It is observed that f is a subset of A × B.
Thus, f is a relation from A to B.
(ii) Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.
Question11. Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z: justify your answer.
Solution :
The relation f is defined as f = {(ab, a + b): a, b ∈ Z}
We know that a relation f from a set A to a set B is said to be a function if every element of set A has unique images in set B.
Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f
i.e., (12, 8), (12, –8) ∈ f
It can be seen that the same first element i.e., 12 corresponds to two different images i.e., 8 and –8. Thus, relation f is not a function.
Question12. Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
Solution :f(n)= the highest prime factor of n
.
The prime factor of 9 is 3
.
The prime factors of 10 is 2,5
.
The prime factor of 11 is 11
.
The prime factor of 12 is 2,3
.
The prime factor of 13 is 13
.
Thus, it can be said,
f(9) = the highest prime factor of 9=3
f(10) = the highest prime factor of 10=5
f(11) = the highest prime factor of 11=11
f(12) = the highest prime factor of 12=3
f(13)= the highest prime factor of 13=13
.Now, the range of the function will be, {3,5,11,13}