NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Maths
Chapter : 2
Chapter Name : Relations And Functions

Exercise 2.1

Q1 If ( , nd the values of x and y.
x 2 5 1
+ 1, y − ) = ( , )
3 3 3 3

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, + 1 = x

3
and
5
y −
3
= . 2

3
1

3
x 5
+ 1 =
3 3

,y−
x 5 2 1
⇒ = − 1 =
3 3 3 3
x 2 1 2
⇒ = ⇒ y = +
3 3 3 3

⇒ x = 2 ⇒ y = 1

∴ x = 2 and y = 1

Page : 33 , Block Name : Exercise 2.1

Q2 If the set A has 3 elements and the set B = {3, 4, 5}, then nd 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 x B)
= (Number of elements in A) x (Number of elements in B)
= 3x3 =9
Thus, the number of elements in (A x B) is 9.

Page : 33 , Block Name : Exercise 2.1

Q3 If G = {7, 8} and H = {5, 4, 2}, nd 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 de ned 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 : 33 , Block Name : Exercise 2.1

Q4 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

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 : 33 , Block Name : Exercise 2.1

Q5 If A = {–1, 1}, nd A × A × A.

Answer. It is known that for any non-empty set A, A × A × A is de ned 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 : 33 , Block Name : Exercise 2.1

Q6 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 de ned as P × Q = {(p, q):
p ∈ P, q ∈ Q}
∴ A is the set of all rst elements and B is the set of all second elements.
Thus, A = {a, b} and B = {x, y}

Page : 33 , Block Name : Exercise 2.1

Q7 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 : 33 , Block Name : Exercise 2.1

Q8 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 = 16 subsets. These are ϕ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3),
4

(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 : 33 , Block Name : Exercise 2.1

Q9 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, nd
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 rst 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 : 33 , Block Name : Exercise 2.1

Q10 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 : 34 , Block Name : Exercise 2.1

Exercise 2.2

Q1 Let A = {1, 2, 3,...,14}. De ne 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 rst elements of the ordered pairs in the relation.
∴Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of 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 : 35 , Block Name : Exercise 2.2

Q2 De ne 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
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 rst 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 : 36 , Block Name : Exercise 2.2

Q3 A = {1, 2, 3, 5} and B = {4, 6, 9}. De ne 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 : 36 , Block Name : Exercise 2.2

Q4 The Fig 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 gure, 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 : 36 , Block Name : Exercise 2.2

Q5 Let A = {1, 2, 3, 4, 6}. Let R be the relation on A de ned 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 : 36 , Block Name : Exercise 2.2

Q6 Determine the domain and range of the relation R de ned 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 : 36 , Block Name : Exercise 2.2

Q7 Write the relation R = {(x, x ) : x is a prime number less than 10} in roster form.
3

Answer. R = {(x, x ) : x is a prime number less than 10}
3

The prime numbers less than 10 are 2, 3, 5, and 7.
∴R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Page : 36 , Block Name : Exercise 2.2

Q8 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 : 36 , Block Name : Exercise 2.2

Q9 Let R be the relation on Z de ned 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 : 36 , Block Name : Exercise 2.2

Exercise 2.3

Q1 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 rst element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation
is not a function.

Page : 44 , Block Name : Exercise 2.3

Q2 Find the domain and range of the following real functions:
(i) f(x) = - |x|
(ii) f(x) = √9 − x 2

Answer. (i) f(x) = - |x| , x ∈ R
x, x ≥ 0
We know that |x| = {
−x, x < 0

−x, x ≥ 0
∴ f (x) = −|x| = {
x, x < 0

Since f(x) is de ned 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(-\infty, 0] .
(ii) f(x) = √9 − x 2

Since √9 − x is de ned for all real numbers that are greater than or equal to –3 and less than
2

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 : 44 , Block Name : Exercise 2.3

Q3 A function f is de ned 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, 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 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,x, is a real number
2

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,
2
x ≥ 0

2
⇒ x + 2 ≥ 0 + 2

2
⇒ x + 2 ≥ 2

⇒ f (x) ≥ 2

Therefore, 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.
Therefore, Range of f = R

Page : 44 , Block Name : Exercise 2.3

Miscellaneous Exercise

2
x ,0 ≤ x ≤ 3
Q1 The relation f is de ned by f (x) = {
3x, 3 ≤ x ≤ 10
2
x , 0 ≤ x ≤ 2
The relation g is de ned by g(x) = {
3x, 2 ≤ x ≤ 10

Show that f is a function and g is not a function.

2
x ,0 ≤ x ≤ 3
Answer. The relation f is de ned by f (x) = {
3x, 3 ≤ x ≤ 10

It is observed that for
2
0 ≤ x < 3, f (x) = x

3 < x ≤ 10, f (x) = 3x

Also, at x=3, f(x) =3 = 9 or f(x) = 3 x 3 = 9
2

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.
2
x , 0 ≤ x ≤ 2
The relation g is de ned by g(x) = {
3x, 2 ≤ x ≤ 10

It can be observed that for x = 2, g(x) = 2 = 4 and g(x) = 3 x 2 = 6
2

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 : 46 , Block Name : Miscellaneous Exercise

f (1.1)−f (1)
Q2 If f(x) = x , nd
2

(1.1−1)
.

Answer. f(x) = x 2

2 2
f (1.1)−f (1) (1.1) −(1)
Therefore, =
1.21−1 0.21
= = = 2.1
(1.1−1) (1.1−1) 0.1 0.1

Page : 46 , Block Name : Miscellaneous Exercise

2

Q3 Find the domain of the function f (x) = .
x +2x+1

2
x −8x+12

2

Answer. The given function is f (x) =
x +2x+1

2
x −8x+12
2 2

f (x) =
x +2x+1

x2 −8x+12
= x +2x+1

(x−6)(x−2)

It can be seen that function f is de ned for all real numbers except at x = 6 and x = 2.
Hence, the domain of f is R - {2,6}.

Page : 46 , Block Name : Miscellaneous Exercise

Q4 Find the domain and the range of the real function f de ned by f (x) = √(x − 1).

Answer. He given real function is f (x) = √(x − 1)
It can be seen that √(x − 1) is de ned for (x-1) ≥ 0.
i.e.,
f (x) = √(x − 1) is de ned for x ≥ 1.
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 : 46 , Block Name : Miscellaneous Exercise

Q5 Find the domain and the range of the real function f de ned by f (x) = |x − 1|.

Answer. The given real function is f(x) = |x − 1|.
It is clear that |x − 1| is de ned for all real numbers.
Therefore, 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.

Page : 46 , Block Name : Miscellaneous Exercise

2

Q6 Let f = {(x, x

1+x
2
) : x ∈ R} be a function from R into R. Determine the range of f.

Answer.
2
x
f = {(x, 2
) : x ∈ R}
1+x

1 1 9 4 9 16
= {(0, 0), (±0.5, ) , (±1, ) , (±1.5, ) , (±2, ) , (3, ) , (4, ),…
5 2 13 5 10 17

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 : 46 , Block Name : Miscellaneous Exercise

Q7 Let f, g : R → R be de ned, respectively by f(x) = x + 1, g(x) = 2x - 3. Find f + g , f - g and .
f

g

Answer. f, g : R → R is de ned 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

f f (x)
( ) (x) = , g(x) ≠ 0, x ∈ R
g g(x)

f x+1
∴ ( ) (x) = , 2x − 3 ≠ 0 or 2x ≠ 3
g 2x−3

f x+1 3
∴ ( ) (x) = ,x ≠
g 2x−3 2

Page : 46 , Block Name : Miscellaneous Exercise

Q8 Let f = {(1,1), (2,3), (0,-1), (-1, -3)} be a function from Z to Z de ned 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 : 46 , Block Name : Miscellaneous Exercise

Q9 Let R be a relation from N to N de ned by R = {(a, b) : a, b ∈N
N and a = b }. Are the following
2

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 ∈NN and a = b } 2

(i) It can be seen that 2 ∈ N; however, 2 ≠ 2 = 4 2

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 = 81; therefore, (3, 9) ∉ N
2

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 and 4 = 2 .
2 2

Now, 16 ≠ 2 = 4; therefore, (16, 2) ∉ N
2

Therefore, the statement “(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R”is not true.

Page : 46 , Block Name : Miscellaneous Exercise

Q10 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
x B.
It is observed that f is a subset of A x B.
Thus, f is a relation from A to B.
(ii) Since the same rst element i.e., 2 corresponds to two different images i.e., 9 and 11, relation
f is not a function.

Page : 46 , Block Name : Miscellaneous Exercise

Q11 Let f be the subset of Z × Z de ned by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z?
Justify your answer.

Answer. The relation f is de ned 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 rst element i.e., 12 corresponds to two different images i.e., 8 and
–8. Thus, relation f is not a function.

Page : 47 , Block Name : Miscellaneous Exercise

Q12 Let A = {9,10,11,12,13} and let f : A → N be de ned 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 de ned 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}

Page : 47 , Block Name : Miscellaneous Exercise