Class 11 Maths Chapter 1 Sets Exercise 1.2 Solutions in English Medium
Free NCERT Solutions for Class 11 Math’s Chapter 1 Sets Exercise 1.1, Exercise 1.2, Exercise 1.3, Exercise 1.4, Exercise 1.5, Exercise 1.6 and Miscellaneous Exercise in English Medium for CBSE.
NCERT Maths Class 11 Sets. Just click on the Exercise wise links given below to practice the Maths solutions for the respective exercise.
| SETS | Solutions Link |
|---|---|
| Exercise 1.1 | Click here |
| Exercise 1.2 | Click here |
| Exercise 1.3 | Click here |
| Exercise 1.4 | Click here |
| Exercise 1.5 | Click here |
| Exercise 1.6 | Click here |
EXERCISE : 1.2
1. Which of the following
are examples of the null set
(i)
Set of odd
natural numbers divisible by 2
(ii)
Set of
even prime numbers
(iii)
{ x : x is
a natural numbers, x < 5 and X > 7 }
(iv)
{ y : y is
a point common to any two parallel lines }
Solution:-
(i)
A set of
odd natural numbers divisible by 2 is a
null set because no odd number is divisible
by 2 .
(ii)
{ 2 } is a
Set of even prime numbers which is not a
null set .
(iii)
{ x : x is a natural numbers, x < 5 and X
> 7 } is a null set as a number cannot both less than 5 and greater than 7.
(iv)
{ y : y is
a point common to any two parallel lines }is a null set because parallel lines
do not interest. Hence, they have no common point .
2. Which of the following
sets are finite and infinite
(i)
The set of
months of a year
(ii)
{ 1, 2, 3, . . . . . }
(iii)
{ 1, 2, 3,
. . . 99, 100 }
(iv)
The set of
positive integers greater than 100
(v)
The set of
prime numbers less than 99
Solution:-
(i)
The set of
months of a year is a finite set because it has 12 elements.
(ii)
{ 1, 2, 3,
. . . . . } is an infinite set because there are infinite element in the set.
(iii)
{ 1, 2, 3,
. . . 99, 100 } is a finite set because the set contains finite number of the
elements.
(iv)
The set of positive integers greater than 100 is
an set as the positive integers which greater than 100 are infinite.
(v)
The set of
prime numbers less than 99 is a finite set because the positive integers less
than 99 are finite.
3. State whether each of
the following set is finite or infinite :
(i)
The set of
lines which are parallel to the x – axis
(ii)
The set of
letters in the English alphabet
(iii)
The set of
numbers which are multiple of 5.
(iv)
The set of
animals living on the earth
(v)
The set of
circles passing through the origin (0,0)
Solution:-
(i)
The set of
lines which are parallel to the x – axis is an infinite set because we can draw
infinite number of lines parallel to x – axis.
(ii)
The set of
letters in the English alphabet is a finite set because there are 26 letters in
the English alphabet.
(iii)
The set of
numbers which are multiple of 5 is an
infinite set because there are infinite multiples of 5.
(iv)
The set of
animals living on the earth is a finite set because the number of animals
living on the earth is very large but finite.
(v)
The set of
circles passing through the origin (0,0) is an infinite set because we can draw
infinite number of circles passing through origin of different radius.
4.
In the following, state whether A = B or not :
(i)
A = { a,
b, c, d } B = { d, c, b, a
}
(ii)
A = { 4,
8, 12, 16 } B = { 8, 4, 16, 18 }
(iii)
A = { 2,
4, 6, 8, 10 } B = { x : x is possible
even integer and x ≤ 10 }
(iv)
A = { x :
x is a multiple of 10 }, B = { 10, 15, 20, 25, 30, . . . . }
Solution:-
(i)
We have,
A = { a, b, c, d }
B = { d, c, b, a }
Then, A and B are equal sets as
repetition of elements in a set do not change a set. Thus
A = B
(ii)
We have,
A = { 4,
8, 12, 16 }
B = { 8, 4, 16, 18 }
It seen that A and B are not equal.
Therefore A is not equal to B.
(iii)
We have,
A = {
2, 4, 6, 8, 10 }
B = { x : x is possible even integer and x ≤
10 }
B = { 2, 4, 6, 8, 10 }
Therefore A = B
(iv)
We have
A = { x : x is a
multiple of 10 }
B = { 10, 15, 20, 25,
30, . . . . }
It can be seen that B ϵ 15 but A ∉ 15.
Therefore A is not equal
to B
5. Are the following pair
of sets equal ? Give reasons .
(i)
A = { 2, 3
}, B = { x : x is solution of x2
+ 5x + 6 = 0}
(ii)
A = { x :
x is a letter in the word FOLLOW }
B = { y : y is a letter in the word WOLF }
Solution:-
(i)
We Have,
A = { 2, 3
}
B = { x : x is a
solution of x2 + 5x + 6 = 0}
The equation x2 + 5x +
6 = 0
We can solved as ,
(x+2)(x+3) = 0
x = -2, -3
Therefore A = {2,3} ; B = {-2,-3}
Therefore A is not equal to B
(ii)
We have
A = { x : x is a
letter in the word FOLLOW }
B = { y : y is a letter in the word WOLF }
A = { F, O, L, W } , B = { W, O, L, F }
Therefore A and B are equal sets
as repetition of element in a set do not change .
Therefore, A = B
6. From the sets given
below, select equal sets :
A = { 2, 4, 8, 12 }, B = { 1, 2, 3, 4 } C = { 4, 8, 12, 14 } D = { 3, 1, 4, 2 }, E = { -1, 1 }, F = { 0, a }, G = { 1, -1 } , H = { 0, 1 }
Solution:-
A = { 2, 4, 8, 12 }
B = { 1, 2, 3, 4 }
C = { 4, 8, 12, 14 }
D = { 3, 1, 4, 2 }
E = { -1, 1 }
F = { 0, a }
G = { 1, -1 }
H = { 0, 1 }
We know that,
8 ϵ A , 8 ∉ B
, 8 ∉ D , 8 ∉ E ,
8 ∉ F , 8 ∉ G, 8 ∉ H
⇒ A ≠ B , A ≠ D,
A ≠ E , A ≠ F , A ≠ G , A ≠ H
It can be written as
2 ϵ A , 2 ∉ C
∴ A ≠ C
3 ϵ B , 3 ∉ C ,
3 ∉ E, 3 ∉ F , 3 ∉ G
, 3 ∉ H,
∴ B ≠ C , B ≠ E , B ≠ F ,
B ≠ G, B ≠ H
It can be written as
12 ϵ C , 12 ∉ D, 12 ∉ E , 12 ∉ F,
12 ∉ G, 12 ∉ H
∴ C ≠ D, C ≠ E , C ≠ F ,
C ≠ G, C ≠ H
4 ϵ D, 4 ∉ E , 4 ∉ F , 4 ∉ G ,
4 ∉ H
Here, E ≠ F , E ≠ G ,
E ≠ H , F ≠ G , F ≠ H ,
G ≠ H
Order in which the
elements of a set are listed is not significant .
∴ B = D
and E = G
Therefore, among the given
sets, B = D and E = G
Published by Lokesh Das
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CLASS 11 ENGLISH MEDIUM ALL BOOK SOLUTIONS. CLASS 11 MATHEMATICS SOLUTIONS IN ENGLISH MEDIUM. CLASS 11 MATHEMATICS CHAPTER ONE SETS SOLUTIONS IN ENGLISH MEDIUM. CLASS 11 MATHS SETS EXERCISE 1.2 SOLUTIONS IN ENGLISH MEDIUM.
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