NCERT Solutions Class 11 Mathematics Chapter 2 Relations and Functions

Exercise 2.1

Question 1. If (x/3 + 1, y - 2/3) = (5/3 1/3) , find the values of x and y. 

Answer :

It is given that (x/3 + 1, y - 2/3) = (5/3, 1/3). 

Since the ordered pairs are equal, the corresponding elements will also be equal. 

Therefore, x/3 + 1= 5/3 and y - 2/3 = 1/3. 

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.

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 non-empty 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)}

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 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.

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

Question 5. If A = {–1, 1}, find A × A × A.

Answer :

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)}

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 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}

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.

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 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)}

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}.

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)

Exercise 2.2

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 :

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}

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} 

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)} 

Question 4. 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?

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} 

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} 

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} 

Question 7: Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.

Answer :

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)} 

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 26.

Therefore, the number of relations from A to B is 26.

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

Exercise 2.3

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.

Question 2. Find the domain and range of the following real function:
(i) f(x) = –|x|
(ii) f(x) = √(9 – x²)

Answer :

(i) f(x) = - |x| , x ∈ R 

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) f(x) = √(9 – x²)

Since √(9 – x²) is 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].

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

Question 4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by t(C)=9C/5 + 32
Find
(i) t(0)
(ii) t(28)
(iii) t(–10)
(iv) The value of C, when t(C) = 212. 

Answer :

The given function is t(C) = 9C/5 + 32 . 

Therefore, 

Question 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

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

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

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

Question 1. The relation f 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) = x²

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

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) = 2² = 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.

Question 2. If f(x) = x², find [f(1.1) - f(1)]/(1.1 - 1)

Answer :

Question 3. Find the domain of the function f(x) = (x² + 2x + 1)/(x² - 8x + 12) 

Answer :

The given function is f(x) = (x² + 2x + 1)/(x² - 8x + 12) 

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}.

Question 4. 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 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

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,∞).

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 non-negative real numbers.

Question 6. Let f = [{x , x² /(1 + x² )} : x ∈ R] be 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)

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 f/g. 

Answer :

f, g: R → Ris 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

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.

Question 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.

Answer :

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.

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 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.

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.

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}