NCERT Solutions Class 11 Maths Relations and Functions||Relations and Functions class 11 solution||class 11 maths solution
Exercise 2.1 Page No: 33
1. If 
, find the values of x and y.
Solution:
Given,
As the ordered pairs are equal, the corresponding elements should also be equal.
Thus,
x/3 + 1 = 5/3 and y – 2/3 = 1/3
Solving, we get
x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M and adding]
x = 2 and 3y = 3
Therefore,
x = 2 and y = 1
2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?
Solution:
Given, set A has 3 elements and the elements of set B are {3, 4, and 5}.
So, the number of elements in set B = 3
Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)
= 3 × 3 = 9
Therefore, the number of elements in (A × B) will be 9.
3. 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}
We know that,
The Cartesian product of two non-empty sets P and Q is given as
P × Q = {(p, q): p ∈ P, q ∈ Q}
So,
G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
4. 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) The statement is False. The correct statement is:
If P = {m, n} and Q = {n, m}, then
P × Q = {(m, m), (m, n), (n, m), (n, n)}
(ii) True
(iii) True
5. If A = {–1, 1}, find A × A × A.
Solution:
The A × A × A for a non-empty set A is given by
A × A × A = {(a, b, c): a, b, c ∈ A}
Here, It is given A = {–1, 1}
So,
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)}
6. If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution:
Given,
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 given by:
P × Q = {(p, q): p ∈ P, q ∈ Q}
Hence, A is the set of all first elements and B is the set of all second elements.
Therefore, A = {a, b} and B = {x, y}
7. 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}
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
Thus,
L.H.S. = A × (B ∩ C) = A × Î¦ = Φ
Next,
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
Thus,
R.H.S. = (A × B) ∩ (A × C) = Φ
Therefore, L.H.S. = R.H.S
– Hence verified
(ii) To verify: A × C is a subset of B × D
First,
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
And,
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.
Thus, A × C is a subset of B × D.
– Hence verified
8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Solution:
Given,
A = {1, 2} and B = {3, 4}
So,
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Number of elements in A × B is n(A × B) = 4
We know that,
If C is a set with n(C) = m, then n[P(C)] = 2m.
Thus, the set A × B has 24 = 16 subsets.
And, these subsets are as below:
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)}
9. 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:
Given,
n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.
We know that,
A = Set of first elements of the ordered pair elements of A × B
B = Set of second elements of the ordered pair elements of A × B.
So, clearly x, y, and z are the elements of A; and
1 and 2 are the elements of B.
As n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set B = {1, 2}.
10. 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 know that,
If n(A) = p and n(B) = q, then n(A × B) = pq.
Also, n(A × A) = n(A) × n(A)
Given,
n(A × A) = 9
So, n(A) × n(A) = 9
Thus, n(A) = 3
Also given that, the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.
And, we know in A × A = {(a, a): a ∈ A}.
Thus, –1, 0, and 1 has to be the elements of A.
As n(A) = 3, clearly A = {–1, 0, 1}.
Hence, the remaining elements of set A × A are as follows:
(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)
Exercise 2.2 Page No: 35
1. 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:
The relation R from A to A is given as:
R = {(x, y): 3x – y = 0, where x, y ∈ A}
= {(x, y): 3x = y, where x, y ∈ A}
So,
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Now,
The domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
Hence, Codomain of R = A = {1, 2, 3, …, 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
Hence, Range of R = {3, 6, 9, 12}
2. 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:
The relation R is given by:
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.
So,
R = {(1, 6), (2, 7), (3, 8)}
Now,
The domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
Hence, Range of R = {6, 7, 8}
3. 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:
Given,
A = {1, 2, 3, 5} and B = {4, 6, 9}
The relation from A to B is given as:
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
Thus,
R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
4. The figure shows a relationship between the sets P and Q. write this relation
(i) in set-builder form (ii) in roster form.
What is its domain and range?
Solution:
From the given figure, it’s seen that
P = {5, 6, 7}, Q = {3, 4, 5}
The relation between P and Q:
Set-builder form
(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}
Roster form
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}
5. 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:
Given,
A = {1, 2, 3, 4, 6} and relation R = {(a, b): a, b ∈ A, b is exactly divisible by a}
Hence,
(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}
6. Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution:
Given,
Relation R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
Thus,
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
So,
Domain of R = {0, 1, 2, 3, 4, 5} and,
Range of R = {5, 6, 7, 8, 9, 10}
7. Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Solution:
Given,
Relation R = {(x, x3): x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
Therefore,
R = {(2, 8), (3, 27), (5, 125), (7, 343)}
8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution:
Given, A = {x, y, z} and B = {1, 2}.
Now,
A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
As n(A × B) = 6, the number of subsets of A × B will be 26.
Thus, the number of relations from A to B is 26.
9. 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:
Given,
Relation R = {(a, b): a, b ∈ Z, a – b is an integer}
We know that the difference between any two integers is always an integer.
Therefore,
Domain of R = Z and Range of R = Z
Exercise 2.3 Page No: 44
1. Which of the following relations 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:
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called as a function.
Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called as a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
It’s seen that the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation cannot be called as a function.
2. Find the domain and range of the following real function:
(i) f(x) = –|x| (ii) f(x) = √(9 – x2)
Solution:
(i) Given,
f(x) = –|x|, x ∈ R
We know that,
As f(x) is defined for x ∈ R, the domain of f is R.
It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.
Therefore, the range of f is given by (–∞, 0].
(ii) f(x) = √(9 – x2)
As √(9 – x2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x2 ≥ 0.
So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].
Now,
For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.
Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].
3. 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,
Function, f(x) = 2x – 5.
Therefore,
(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5
(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9
(iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11
4. 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:
5. Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x > 0.
(ii) f(x) = x2 + 2, x is a real number.
(iii) f(x) = x, x is a real number.
Solution:
(i) Given,
f(x) = 2 – 3x, x ∈ R, x > 0.
We have,
x > 0
So,
3x > 0
-3x < 0 [Multiplying by -1 both the sides, the inequality sign changes]
2 – 3x < 2
Therefore, the value of 2 – 3x is less than 2.
Hence, Range = (–∞, 2)
(ii) Given,
f(x) = x2 + 2, x is a real number
We know that,
x2 ≥ 0
So,
x2 + 2 ≥ 2 [Adding 2 both the sides]
Therefore, the value of x2 + 2 is always greater or equal to 2 for x is a real number.
Hence, Range = [2, ∞)
(iii) Given,
f(x) = x, x is a real number
Clearly, the range of f is the set of all real numbers.
Thus,
Range of f = R
Miscellaneous Exercise Page No: 46
1. 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:
The given relation f is defined as:
It is seen that, for 0 ≤ x < 3,
f(x) = x2 and for 3 < x ≤ 10,
f(x) = 3x
Also, at x = 3
f(x) = 32 = 9 or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9 [Single image]
Hence, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Therefore, the given relation is a function.
Now,
In the given relation g is defined as
It is seen that, for x = 2
g(x) = 22 = 4 and g(x) = 3 × 2 = 6
Thus, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6.
Therefore, this relation is not a function.
2. If f(x) = x2, find
Solution:
Given,
f(x) = x2
Hence,
3. Find the domain of the function
Solution:
Given function,
.
It clearly seen that, the function f is defined for all real numbers except at x = 6 and x = 2 as the denominator becomes zero otherwise.
Therefore, the domain of f is R – {2, 6}.
4. Find the domain and the range of the real function f defined by f(x) = √(x – 1).
Solution:
Given real function,
f(x) = √(x – 1)
Clearly, √(x – 1) is defined for (x – 1) ≥ 0.
So, the function f(x) = √(x – 1) is defined for x ≥ 1.
Thus, the domain of f is the set of all real numbers greater than or equal to 1.
Domain of f = [1, ∞).
Now,
As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √(x – 1) ≥ 0
Thus, the range of f is the set of all real numbers greater than or equal to 0.
Range of f = [0, ∞).
5. Find the domain and the range of the real function f defined by f (x) = |x – 1|.
Solution:
Given real function,
f (x) = |x – 1|
Clearly, the function |x – 1| is defined for all real numbers.
Hence,
Domain of f = R
Also, for x ∈ R, |x – 1| assumes all real numbers.
Therefore, the range of f is the set of all non-negative real numbers.
6. Let 
be a function from R into R. Determine the range of f.
Solution:
Given function,
Substituting values and determining the images, we have
The range of f is the set of all second elements. It can be observed that all these elements are greater than or equal to 0 but less than 1.
[As the denominator is greater than the numerator.]Or,
We know that, for x ∈ R,
x2 ≥ 0
Then,
x2 + 1 ≥ x2
1 ≥ x2 / (x2 + 1)
Therefore, the range of f = [0, 1)
7. Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.
Solution:
Given, the functions f, g: R → R is defined as
f(x) = x + 1, g(x) = 2x – 3
Now,
(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
Thus, (f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4
Thus, (f – g) (x) = –x + 4
f/g(x) = f(x)/g(x), g(x) ≠ 0, x ∈ R
f/g(x) = x + 1/ 2x – 3, 2x – 3 ≠ 0
Thus, f/g(x) = x + 1/ 2x – 3, x ≠ 3/2
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:
Given, f = {(1, 1), (2, 3), (0, –1), (–1, –3)}
And the function defined as, f(x) = ax + b
For (1, 1) ∈ f
We have, f(1) = 1
So, a × 1 + b = 1
a + b = 1 …. (i)
And for (0, –1) ∈ f
We have f(0) = –1
a × 0 + b = –1
b = –1
On substituting b = –1 in (i), we get
a + (–1) = 1 ⇒ a = 1 + 1 = 2.
Therefore, the values of a and b are 2 and –1, respectively.
9. 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:
Given relation R = {(a, b): a, b ∈ N and a = b2}
(i) It can be seen that 2 ∈ N; however, 2 ≠ 22 = 4.
Thus, the statement “(a, a) ∈ R, for all a ∈ N” is not true.
(ii) Its clearly seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 32.
Now, 3 ≠ 92 = 81; therefore, (3, 9) ∉ N
Thus, the statement “(a, b) ∈ R, implies (b, a) ∈ R” is not true.
(iii) Its clearly 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
Thus, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.
10. 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:
Given,
A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16}
So,
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)}
Also 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’s clearly seen that f is a subset of A × B.
Therefore, f is a relation from A to B.
(ii) As the same first element i.e., 2 corresponds to two different images (9 and 11), relation f is not a function.
11. 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:
Given 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.
As 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), (–2 × –6, –2 + (–6)) ∈ f
i.e., (12, 8), (12, –8) ∈ f
It’s clearly seen that, the same first element, 12 corresponds to two different images (8 and –8).
Therefore, the relation f is not a function.
12. 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:
Given,
A = {9, 10, 11, 12, 13}
Now, f: A → N is defined as
f(n) = The highest prime factor of n
So,
Prime factor of 9 = 3
Prime factors of 10 = 2, 5
Prime factor of 11 = 11
Prime factors of 12 = 2, 3
Prime factor of 13 = 13
Thus, it can be expressed as
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
The range of f is the set of all f(n), where n ∈ A.
Therefore,
Range of f = {3, 5, 11, 13}
By-Yatendra Kumar
