NCERT Solutions for Class 11 Humanities Maths Chapter 2 Relations And Functions are provided here with simple stepbystep explanations. These solutions for Relations And Functions are extremely popular among class 11 Humanities students for Maths Relations And Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 11 Humanities Maths Chapter 2 are provided here for you for free. You will also love the adfree experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class 11 Humanities Maths are prepared by experts and are 100% accurate.
Page No 33:
Question 1:
If, find the values of x and y.
Answer:
It is given that.
Since the ordered pairs are equal, the corresponding elements will also be equal.
Therefore, and.
∴ x = 2 and y = 1
Page No 33:
Question 2:
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?
Answer:
It is given that set A has 3 elements and the elements of set B are 3, 4, and 5.
⇒ Number of elements in set B = 3
Number of elements in (A × B)
= (Number of elements in A) × (Number of elements in B)
= 3 × 3 = 9
Thus, the number of elements in (A × B) is 9.
Page No 33:
Question 3:
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Answer:
G = {7, 8} and H = {5, 4, 2}
We know that the Cartesian product P × Q of two nonempty sets P and Q is defined as
P × Q = {(p, q): p∈ P, q ∈ Q}
∴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)}
Page No 33:
Question 4:
State whether each of the following statement 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 nonempty sets, then A × B is a nonempty 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.
Answer:
(i) False
If P = {m, n} and Q = {n, m}, then
P × Q = {(m, m), (m, n), (n, m), (n, n)}
(ii) True
(iii) True
Page No 33:
Question 5:
If A = {–1, 1}, find A × A × A.
Answer:
It is known that for any nonempty 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)}
Page No 33:
Question 6:
If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Answer:
It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}
We know that the Cartesian product of two nonempty 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}
Page No 33:
Question 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
Answer:
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
∴L.H.S. = A × (B ∩ C) = A × Φ = Φ
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)}
∴ R.H.S. = (A × B) ∩ (A × C) = Φ
∴L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) To verify: A × C is a subset of B × D
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
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)}
We can observe that all the elements of set A × C are the elements of set B × D.
Therefore, A × C is a subset of B × D.
Page No 33:
Question 8:
Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Answer:
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)] = 2^{m}.
Therefore, the set A × B has 2^{4} = 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)}
Page No 33:
Question 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.
Answer:
It is given that 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.
∴ x, y, and z are the elements of A; and 1 and 2 are the elements of B.
Since n(A) = 3 and n(B) = 2, it is clear that A = {x, y, z} and B = {1, 2}.
Page No 34:
Question 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.
Answer:
We know that if n(A) = p and n(B) = q, then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is given that n(A × A) = 9
∴ n(A) × n(A) = 9
⇒ n(A) = 3
The ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.
We know that A × A = {(a, a): a ∈ A}. Therefore, –1, 0, and 1 are elements of A.
Since n(A) = 3, it is clear that A = {–1, 0, 1}.
The remaining elements of set A × A are (–1, –1), (–1, 1), (0, –1), (0, 0),
(1, –1), (1, 0), and (1, 1)
Page No 35:
Question 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.
Answer:
The relation R from A to A is given as
R = {(x, y): 3x – y = 0, where x, y ∈ A}
i.e., R = {(x, y): 3x = y, where x, y ∈ A}
∴R = {(1, 3), (2, 6), (3, 9), (4, 12)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
∴Domain of R = {1, 2, 3, 4}
The whole set A is the codomainof the relation R.
∴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.
∴Range of R = {3, 6, 9, 12}
Page No 36:
Question 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.
Answer:
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}
Page No 36:
Question 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.
Answer:
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)}
Page No 36:
Question 4:
The given figure shows a relationship between the sets P and Q. write this relation
(i) in setbuilder form (ii) in roster form.
What is its domain and range?
Answer:
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}
Page No 36:
Question 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.
Answer:
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}
Page No 36:
Question 6:
Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Answer:
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}
Page No 36:
Question 7:
Write the relation R = {(x, x^{3}): x is a prime number less than 10} in roster form.
Answer:
R = {(x, x^{3}): 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)}
Page No 36:
Question 8:
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Answer:
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 2^{6}.
Therefore, the number of relations from A to B is 2^{6}.
Page No 36:
Question 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.
Answer:
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
Page No 44:
Question 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)}
Answer:
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
Since 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation is 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)}
Since 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation is 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)}
Since the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation is not a function.
Page No 44:
Question 2:
Find the domain and range of the following real function:
(i) f(x) = –x (ii)
Answer:
(i) f(x) = –x, x ∈ R
We know that x =
Since f(x) is defined for x ∈ R, the domain of f is R.
It can be observed that the range of f(x) = –x is all real numbers except positive real numbers.
∴The range of f is (–, 0].
(ii)
Sinceis defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, the domain of f(x) is {x : –3 ≤ x ≤ 3} or [–3, 3].
For any value of x such that –3 ≤ x ≤ 3, the value of f(x) will lie between 0 and 3.
∴The range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].
Page No 44:
Question 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)
Answer:
The given function is 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
Page No 44:
Question 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
Answer:
The given function is.
Therefore,
(i)
(ii)
(iii)
(iv) It is given that t(C) = 212
Thus, the value of t, when t(C) = 212, is 100.
Page No 44:
Question 5:
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
Answer:
(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) = x^{2} + 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,
x^{2}≥ 0
⇒ x^{2} + 2 ≥ 0 + 2
⇒ x^{2} + 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
Page No 46:
Question 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.
Answer:
The relation f is defined as
It is observed that for
0 ≤ x < 3, f(x) = x^{2}
3 < x ≤ 10, f(x) = 3x
Also, at x = 3, f(x) = 3^{2} = 9 or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9
Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Thus, the given relation is a function.
The relation g is defined as
It can be observed that for x = 2, g(x) = 2^{2} = 4 and g(x) = 3 × 2 = 6
Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.
Page No 46:
Question 2:
If f(x) = x^{2}, find.
Answer:
Page No 46:
Question 3:
Find the domain of the function
Answer:
The given function is.
It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.
Hence, the domain of f is R – {2, 6}.
Page No 46:
Question 4:
Find the domain and the range of the real function f defined by.
Answer:
The given real function is.
It can be seen that is defined for (x – 1) ≥ 0.
i.e., 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 ⇒
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,).
Page No 46:
Question 5:
Find the domain and the range of the real function f defined by f (x) = x – 1.
Answer:
The given real function is f (x) = x – 1.
It is clear that x – 1 is defined for all real numbers.
∴Domain of f = R
Also, for x ∈ R, x – 1 assumes all real numbers.
Hence, the range of f is the set of all nonnegative real numbers.
Page No 46:
Question 6:
Letbe a function from R into R. Determine the range of f.
Answer:
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.
[Denominator is greater numerator]
Thus, range of f = [0, 1)
Page No 46:
Question 7:
Let f, g: R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and.
Answer:
f, g: R → R is defined as f(x) = x + 1, g(x) = 2x – 3
(f + g) (x) = f(x) + g(x) = (x + 1) + (2x – 3) = 3x – 2
∴(f + g) (x) = 3x – 2
(f – g) (x) = f(x) – g(x) = (x + 1) – (2x – 3) = x + 1 – 2x + 3 = – x + 4
∴ (f – g) (x) = –x + 4
Page No 46:
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.
Answer:
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.
Page No 46:
Question 9:
Let R be a relation from N to N defined by R = {(a, b): a, b ∈ N and a = b^{2}}. 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.
Answer:
R = {(a, b): a, b ∈ N and a = b^{2}}
(i) It can be seen that 2 ∈ N;however, 2 ≠ 2^{2} = 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 = 3^{2}.
Now, 3 ≠ 9^{2} = 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 = 4^{2} and 4 = 2^{2}.
Now, 16 ≠ 2^{2} = 4; therefore, (16, 2) ∉ N
Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R” is not true.
Page No 46:
Question 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.
Answer:
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 nonempty set A to a nonempty 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.
Page No 47:
Question 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.
Answer:
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.
Page No 47:
Question 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.
Answer:
A = {9, 10, 11, 12, 13}
f: A → N is defined as
f(n) = The highest prime factor of n
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
∴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.
∴Range of f = {3, 5, 11, 13}
View NCERT Solutions for all chapters of Class 14