Class 11 Maths Exercise 1.3 Solution

Class 11 maths exercise 1.3 solution | class 11 ch 1 exercise solutions | class 11 chapter 1 exercise solution | sets class 11 ncert solutions | ncert solution for class 11 maths chapter 1 | ncert exemplar class 11 maths

Looking for NCERT Class 11 Maths Exercise 1.3 solutions? You’ve landed at the right place! Here, you’ll find detailed and easy-to-follow solutions for all the questions in Exercise 1.3 of Chapter 1 – Sets. These Class 11 Maths solutions are prepared according to the latest NCERT syllabus and are ideal for clarifying concepts like union, intersection, and difference of sets. Whether you’re studying for exams or revising the chapter, these NCERT solutions for Class 11 Maths Chapter 1 will help you build a strong foundation in set theory. View or download the complete solutions now and study smarter!

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Exercise 1.3

1. Make correct statements by filling in the symbols â or $\not \subset$ in the blank spaces:

(i)

$\{2,3,4\} \ldots\{1,2,3,4,5\}$

Answer

$\{2,3,4\} \subset\{1,2,3,4,5\}$
Since,

$2,3,4$ comes in the second set.

(ii)

$\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \ldots\{\mathrm{b}, \mathrm{c}, \mathrm{d}\}$

Answer

$\{\mathrm{a}, \mathrm{b}, \mathrm{c}\} \not \subset\{\mathrm{b}, \mathrm{c}, \mathrm{d}\}$
Since, 'a' is not in the second set.

(iii)

$\{x$ : x is a student of Class XI of your school

$\}$ .

$\{x: x$ student of your school

$\}$

Answer

$\{x$ : x is a student of Class XI of your school

$\} \subset\{x: x$ student of your school

$\}$
Since,

$x$ is a part of the school too.

(iv)

$\{x: x$ is a circle in the plane

$\} .\{x: x$ is a circle in the same plane with radius 1 unit

$\}$

Answer

$\{x: x$ is a circle in the plane

$\} \not \subset \{x: x$ is a circle in the same plane with radius 1 unit}
Since, first set has no fixed radius circles whereas the second set has only circles with radius 1 unit

(v)

$\{x: x$ is a triangle in a plane

$\} \ldots\{x: x$ is a rectangle in the plane

$\}$

Answer

$\{x: x$ is a triangle in a plane

$\} \not \subset\{x: x$ is a rectangle in the plane

$\}$
Since, set 1 has triangle whereas set 2 has rectangle.

(vi)

$\{x: x$ is an equilateral triangle in a plane

$\} \ldots\{x: x$ is a triangle in the same plane

$\}$

Answer

$\{x: x$ is an equilateral triangle in a plane

$\} \subset\{x: x$ is a triangle in the same plane

$\}$
Since, equilateral triangle is a type of triangle itself.

(vii)

$\{x: x$ is an even natural number

$\} \ldots\{x: x$ is an integer

$\}$

Answer

$\{x: x$ is an even natural number

$\} \subset\{x: x$ is an integer

$\}$
Since, all the even natural numbers are a type of integers.

2. Examine whether the following statements are true or false:

(i)

$\{\mathrm{a}, \mathrm{b}\} \not \subset\{\mathrm{b}, \mathrm{c}, \mathrm{a}\}$

Answer

Let us assume that

$\mathrm{A}=\{\mathrm{a}, \mathrm{b}\}$ and

$\mathrm{B}=\{\mathrm{b}, \mathrm{c}, \mathrm{a}\}$
Now, we can observe that every element of

$A$ is an element of

$B$ .
Thus, A

$\subset$ B

$\therefore$ The statement is false.

(ii)

$\{\mathrm{a}, \mathrm{e}\} \subset\{x: x$ is a vowel in the English alphabet

$\}$

Answer

Let us assume that

$\mathrm{A}=\{\mathrm{a}, \mathrm{e}\}$ and

$\mathrm{B}=\{x: x$ is a vowel in the English alphabets

$\}$

$=\{\mathrm{a}, \mathrm{e}, \mathrm{i}, \mathrm{o}, \mathrm{u}\}$
Now, we can observe that every element of

$A$ is an element of

$B$ .
Thus,

$\mathrm{A} \subset \mathrm{B}$

$\therefore$ The statement is true.

(iii)

$\{1,2,3\} \subset\{1,3,5\}$

Answer

Let us assume that

$\mathrm{A}=\{1,2,3\}$ and

$\mathrm{B}=\{1,3,5\}$ ,
Now, we can observe that 2 belongs to A but 2 does not belongs to B .
Thus, A B

$\therefore$ The statement is false.

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(iv)

$\{\mathrm{a}\} \subset\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$

Answer

Let us assume that

$\mathrm{A}=\{\mathrm{a}\}$ and

$\mathrm{B}=\{\mathrm{b}, \mathrm{c}, \mathrm{a}\}$
Now, we can observe that every element of

$A$ is an element of

$B$ .
Thus,

$\mathrm{A} \subset \mathrm{B}$

$\therefore$ The statement is true.

(v)

$\{\mathrm{a}\} \in\{\mathrm{a}, \mathrm{b}, \mathrm{c}\}$

Answer

Let us assume that

$\mathrm{A}=\{\mathrm{a}\}$ and

$\mathrm{B}=\{\mathrm{b}, \mathrm{c}, \mathrm{a}\}$
Now, we can observe that every element of

$A$ is an element of

$B$
Thus,

$\mathrm{A} \subset \mathrm{B}$

$\therefore$ The statement is false.

(vi)

$\{x: x$ is an even natural number less than 6

$\} \subset\{x: x$ is a natural number which divides 36

$\}$

Answer

Let us assume that

$A=\{x: x$ is an even natural number less than 6

$\}$

$=\{2,4\}$
and

$\mathrm{B}=\{x: x$ is a natural number which divide 36

$\}$

$=\{1,2,3,4,6,9,12,18,36\}$
Now, we can observe that every element of

$A$ is an element of

$B$ .
Thus,

$\mathrm{A} \subset \mathrm{B}$

$\therefore$ The statement is true.

3. Let $\mathrm{A}=\{1,2,\{3,4\}, 5\}$ . Which of the following statements are incorrect and why?

(i)

$\{3,4\} \subset \mathrm{A}$

Answer

Here, we know that

$\{3,4\}$ is a member of set A .

$=\{3,4\} \in \mathrm{A}$
Thus, the given statement is incorrect.

(ii)

$\{3,4\} \in \mathrm{A}$

Answer

Here, we know that

$\{3,4\}$ is a member of set A .
Thus, the given statement is incorrect.

(iii)

$\{\{3,4\}\} \subset \mathrm{A}$

Answer

Here, we know that

$\{3,4\}$ is a member of set

$A$ .

$=\{3,4\}$ is a set.
Thus, the given statement is correct.

(iv)

$1 \in \mathrm{A}$

Answer

Here, we can observe that 1 is a member of set

$A$ .
Thus, the given statement is correct.

(v)

$1 \subset \mathrm{A}$

Answer

Here, we can see that 1 is a member of set A but is not any set itself. Thus, the given statement is incorrect.

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(vi)

$\{1,2,5\} \subset \mathrm{A}$

Answer

Here, we can see that

$1,2,5$ is a member of set A
Thus, the given statement is correct.

(vii)

$\{1,2,5\} \in A$

Answer

Here, we can see that

$1,2,5$ is a member of set A

$=\{1,2,5\}$ is a subset of A
Thus, the given statement is incorrect.

(viii)

$\{1,2,3\} \subset \mathrm{A}$

Answer

Here, we can see that 3 is not a member of set

$A$

$=\{1,2,3\}$ is not a subset of A
Thus, the given statement is incorrect.

(ix)

$\phi \in \mathrm{A}$

Answer

Here, we can see that

$\phi$ is not a member of set A
Thus, the given statement is correct.

(x)

$\phi \subset \mathrm{A}$

Answer

Here, as we know the null set is a subset of every set.
Thus, the given statement is correct.

(xi)

$\{\phi\} \subset \mathrm{A}$

Answer

$\{\phi\}$ is the set containing the null set.

$\{\phi\} \subset \mathrm{A}$ is only possible if

$\phi$ is in set A . But it is not there. So, the statement is incorrect.

4. Write down all the subsets of the following sets:

(i)

$\{a\}$

Answer

The subsets of

$\{a\}$ are

$\phi$ and

$\{a\}$ .

(ii)

$\{\mathrm{a}, \mathrm{b}\}$

Answer

The subsets of

$\{a, b\}$ are

$\phi,\{a\},\{b\}$ , and

$\{a, b\}$ .

(iii)

$\{1,2,3\}$

Answer

$\text{The subsets of} \ \{1,2,3\}$

$\text{are} \ \phi,\{1\},\{2\},\{3\},\{1,2\},\{2,3\},\{1,3\}$ and

$\{1,2,3\}$

(iv)

$\phi$

Answer

The only subset of

$\phi$ is

$\phi$ .

5. How many elements has

$\mathrm{P}(\mathrm{A})$ , if

$\mathrm{A}=\phi$ ?

Answer

$\mathrm{P}(\mathrm{A})$ is the power set of set A .
Number of element of

$P(A)=2^{n}$
Where n is the number of elements of the set

$A$ .
Given

$A=\phi$
Then the number of elements of set

$A=0$

$\therefore$ Number of elements of

$\mathrm{P}(\mathrm{A})=2^{\mathrm{n}}$

$=2^{0}$

$=1$

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6. Write the following as intervals:

(i)

$\{x: x \in R,-4 < x \leq 6\}$

Answer

Let us assume that

$\mathrm{A}=\{x: x \in \mathrm{R},-4 < x \leq 6\}$

$\therefore$ Set A can be written in the form of intervals as follows:

$=(-4,6)$

(ii)

$\{x: x \in R,-12 < x < -10\}$

Answer

Let us assume that

$\mathrm{A}=\{x: x \in \mathrm{R},-12 < x < -10\}$

$\therefore$ Set A can be written in the form of intervals as follows:

$=(-12,-10)$

(iii)

$\{x: x \in R, 0 \leq x < 7\}$

Answer

Let us assume that

$\mathrm{A}=\{x: x \in R, 0 \leq x < 7\}$

$\therefore$ Set A can be written in the form of intervals as follows:

$=(0,7)$

(iv)

$\{x: x \in R, 3 \leq x \leq 4\}$

Answer

Let us assume that

$\mathrm{A}=\{x: x \in R, 3 \leq x \leq 4\}$

$\therefore$ Set A can be written in the form of intervals as follows:

$=[3,4]$

7. Write the following intervals in set-builder form:

(i)

$(-3,0)$

Answer

Here, the given interval

$=(-3,0)$
for finding the set builder form for the given set the values range from -3 to 0
This means value of any variable

$x$ is greater than -3 and less than 0
Now, set builder form of the given interval can be written as follows:

$A=\{x: x \in R,-3 < x < 0\}$

(ii)

$[6,12]$

Answer

Here, the given interval

$=[6,12]$
Now the set of values of any variable for set given will be such that

$x$ is greater than 6 (including 6 as the bracket is closed) and less than 12 (including 12)
set builder form of the given interval can be written as follows:

$A=\{x: x \in R, 6 \leq x \leq 12\}$

(iii)

$(6,12]$

Answer

Here, the given interval

$=(6,12]$
Now the set of values of any variable for set given is such that

$x$ is greater than 6 (excluding 6 as the bracket is open) and less than 12 (including 12, as the bracket is closed)
set builder form of the given interval can be written as follows:

$A=\{x: x \in R, 6 < x \leq 12\}$

(iv)

$[-23,5)$

Answer

Here, the given interval

$=[-23,5)$
Now the set of values of any variable for given set is such that

$x$ is greater than -23(including 23 as the bracket is closed) and less than 5 (as the bracket is open)
set builder form of the given interval can be written as follows.

8. What universal set(s) would you propose for each of the following:

(i) The set of right triangles.

Answer

We know that right triangles are a type of triangle with an angle

$90^{\circ}$
Thus, a set of triangles will contain all the right triangles.

$\therefore$ universal set,

$\mathrm{U}=\{x: x$ is a triangle in plane

$\}$

(ii) The set of isosceles triangles.

$A=\{x: x \in R,-23 \leq x < 5\}$

Answer

We know that isosceles triangles are a type of triangle with any two of the angles equal in measure.
Thus, a set of triangles will contain all the isosceles triangles.

$\therefore$ universal set,

$\mathrm{U}=\{x: x$ is a triangle in plane

$\}$

9. Given the sets $\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and $\mathrm{C}=\{0,2,4,6,8\}$ , which of the following may be considered as universal set (s) for all the three sets $\mathrm{A}, \mathrm{B}$ and C ?

(i)

$\{0,1,2,3,4,5,6\}$

Answer

A universal set is a set which contains all the elements of its subsets. So if there are two sets with some members then the universal set containing those two sets, will have elements will all the members of both sets without repetition.
Given:

$\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and

$\mathrm{C}=\{0,2,4,6,8\}$
Now, let

$\mathrm{D}=\{0,1,2,3,4,5,6\}$
Since, 8 belongs to C then its universal set must contain 8 , but D does not contain 8 .

$\therefore \mathrm{D}$ is not a universal set for

$\mathrm{A}, \mathrm{B}, \mathrm{C}$ .

(ii)

$\phi$

Answer

$\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and

$\mathrm{C}=\{0,2,4,6,8\}$
Now, let

$\mathrm{D}=\phi$
Since,

$D$ is an empty set it does not contain any element

$\therefore \mathrm{D}$ is not a universal set for

$\mathrm{A}, \mathrm{B}, \mathrm{C}$ .

(iii)

$\{0,1,2,3,4,5,6,7,8,9,10\}$

Answer

$\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and

$\mathrm{C}=\{0,2,4,6,8\}$
Now, let

$\mathrm{D}=\{0,1,2,3,4,5,6,7,8,9,10\}$
Since, all the members of

$\mathrm{A}, \mathrm{B}, \mathrm{C}$ belongs to D .

$\therefore \mathrm{D}$ is a universal set for

$\mathrm{A}, \mathrm{B}, \mathrm{C}$ .

(iv)

$\{1,2,3,4,5,6,7,8\}$

Answer

$\mathrm{A}=\{1,3,5\}, \mathrm{B}=\{2,4,6\}$ and

$\mathrm{C}=\{0,2,4,6,8\}$
Now, let

$\mathrm{D}=\{1,2,3,4,5,6,7,8\}$
Since, 0 belongs to C but it is not a member of D .

$\therefore \mathrm{D}$ is not a universal set.

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