# NCERT Solutions For Class 11 Maths Chapter 1 Exercise 1.2

Sets

## Chapter 1 Ex.1.2 Question 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 number, \(x < 5\) 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) A set of even prime numbers is not a null set because \(2\) is an even prime number.

(iii) {\(x:x\) is a natural number, \(x < 5\) and \(x > 7\)} is a null set because a number cannot be simultaneously 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 intersect. Hence, they have no common point.

## Chapter 1 Ex.1.2 Question 2

Which of the following sets are finite or infinite?

(i) The set of months of a year

(ii) \(\left\{{1, 2, 3 \dots} \right\}\)

(iii) \(\left\{{1, 2, 3 \dots 99, 100} \right\}\)

(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) \(\left\{{1, 2, 3 \dots} \right\}\) is an infinite set as it has infinite number of natural numbers.

(iii) \(\left\{{1, 2, 3 \dots 99, 100} \right\}\) is a finite set because the numbers from \(1\) to \(100\) are finite.

(iv) The set of positive integers greater than \(100\) is an infinite set because positive integers greater than \(100\) are infinite in number.

(v) The set of prime numbers less than \(99\) is a finite set because prime numbers less than \(99\) are finite in number.

## Chapter 1 Ex.1.2 Question 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 lines parallel to the *\(x\)*-axis are infinite in number.

(ii) The set of letters in the English alphabet is a finite set because it has \(26\) elements.

(iii) The set of numbers which are multiple of \(5\) is an infinite set because multiples of \(5\) are infinite in number.

(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (but it is quite a large number).

(v) The set of circles passing through the origin \((0,0)\) is an infinite set because infinite number of circles can pass through the origin.

## Chapter 1 Ex.1.2 Question 4

In the following, state whether \(A = B\) or not:

(i) \(A = \left\{ {a,b,c,d} \right\} \qquad B = \left\{ {d,c,b,a} \right\}\)

(ii) \(A = \left\{ {4,8,12,16} \right\} \qquad B = \left\{ {8,4,16,18} \right\}\)

(iii) \(A = \left\{ {2,4,6,8,10} \right\} \quad B =\) {\(x:x\) is positive even integer and \(x \le 10\) }

(iv) \(A = \) {\(x:x\) is a multiple of \(10\)} B = {\(10,15,20,25,30,\dots\)}

**Solution**

(i) \(A = \left\{ {a,b,c,d} \right\}\qquad B = \left\{ {d,c,b,a} \right\}\)

The elements of sets \(A\) and \(B\) are same.

Therefore, \(A = B\)

(ii) \(A =\) {\(4, 8, 12, 16\)}; \(\qquad B = {8, 4, 16, 18}\)

We see that \(12 \in A\) but \(12 \notin B\).

Therefore, \(A \ne B\)

(iii) \(A = \left\{ {2,4,6,8,10} \right\} \qquad B = \left\{ {x:x{\text { is positive even integer and }}x \le 10} \right\}\)

We see that \(B = \left\{ {2,4,6,8,10} \right\}\)

Therefore, \(A = B\)

(iv) \(A = \left\{ {x:x\;\;{\text{ is a multiple of }}10} \right\} \qquad B = \left\{ {10,15,20,25,30...} \right\}\)

We see that \(15 \in B\) but \(15 \notin A\).

Therefore, \(A \ne B\)

## Chapter 1 Ex.1.2 Question 5

Are the following pair of sets equal? Give reasons.

(i) \(A = \left\{ {2,3} \right\};{\rm{ }}B = \left\{ {x:x{\text{is solution of}}{x^2} + 5x + 6 = 0} \right\}\)

(ii) \(A = \left\{ {x:x{\text{ is a letter in the word }}FOLLOW} \right\}\)

\(B = \left\{ {y:y{\text{ is a letter in the word }}WOLF} \right\}\)

**Solution**

(i) \(A = \left\{ {2,3} \right\};{\rm{ }}B = \left\{ {x:x{\text{ is solution of }}{x^2} + 5x + 6 = 0} \right\}\)

The equation \({x^2} + 5x + 6 = 0\) can be solved as:

\[\begin{align}x\left( {x + 3} \right) + 2\left( {x + 3} \right) &= 0\\\left( {x + 2} \right)\left( {x + 3} \right) &= 0\\x = -2\;\;\;or\;\;\;x &= -3\end{align}\]

Hence,\(A = \left\{ {2,3} \right\};{\rm{ }}B = \left\{ { - 2, - 3} \right\}\)

Therefore, \(A \ne B\)

(ii) \(A = \left\{ {x:x{\text{ is a letter in the word }}FOLLOW} \right\}\)

\(B = \left\{ {y:y{\text{ is a letter in the word }}WOLF} \right\}\)

Hence,\(A = \left\{ {F,L,O,W} \right\};{\rm{ }}B = \left\{ {W,O,L,F} \right\}\)

The elements of sets \(A\) and \(B\) are same.

Therefore, \(A = B\)

## Chapter 1 Ex.1.2 Question 6

From the sets given below, select equal sets:

\[\begin{align}A &= \left\{ {2,4,8,12} \right\}{\rm{,}}\;\;\;B = \left\{ {1,2,3,4} \right\}{\rm{,}}\;\;\;C = \left\{ {4,8,12,14} \right\}\;\;\;D = \left\{ {3,1,4,2} \right\}\\E &= \left\{ {-1,1} \right\},\;\;\;\;\;\;\;\;\;F = \left\{ {0,a} \right\},\;\;\;\;\;\;\;\;G = \left\{ {1,-1} \right\},\;\;\;\;\;\;\;\;\;\;H = \left\{ {0,1} \right\}\end{align}\]

**Solution**

\[\begin{align}A &= \left\{ {2,4,8,12} \right\}{\rm{,}}\;\;\;B = \left\{ {1,2,3,4} \right\}{\rm{,}}\;\;\;C = \left\{ {4,8,12,14} \right\}\;\;\;D = \left\{ {3,1,4,2} \right\}\\E &= \left\{ {-1,1} \right\},\;\;\;\;\;\;\;\;\;F = \left\{ {0,a} \right\},\;\;\;\;\;\;\;\;G = \left\{ {1,-1} \right\},\;\;\;\;\;\;\;\;\;\;H = \left\{ {0,1} \right\}\end{align}\]

We see that

\[8 \in A,\;\;8 \notin B,\;\;8 \in C,\;\;8 \notin D,\;\;8 \notin E,\;\;8 \notin F,\;\;8 \notin G,\;\;8 \notin H\]

Therefore,

\[A \ne B,\;\;A \ne D,\;\;A \ne E,\;\;A \ne F,\;\;A \ne G,\;\;A \ne H\]

Also, \(2 \in A,\;\;2 \notin C\)

Therefore, \(A \ne C\)

\[3 \in B,\;\;3 \notin C,\;\;3 \notin E,\;\;3 \notin F,\;\;3 \notin G,\;\;3 \notin H\]

Therefore, \(B \ne C,\;\;B \ne E,\;\;B \ne F,\;\;B \ne G,\;\;B \ne H\)

Again, \(12 \in C,\;\;12 \notin D,\;\;12 \notin E,\;\;12 \notin F,\;\;12 \notin G,\;\;12 \notin H\)

Therefore, \(C \ne D,\;\;C \ne E,\;\;C \ne F,\;\;C \ne G,\;\;C \ne H\)

Now,

\(4 \in D,\;\;4 \notin E,\;\;4 \notin F,\;\;4 \notin G,\;\;4 \notin H\)

Therefore, \(D \ne E,\;\;D \ne F,\;\;{\rm{ }}D \ne G,\;\;D \ne H\)

Similarly, \(E \ne F,\;\;E \ne G,\;\;E \ne H,\;\;F \ne G,\;\;F \ne H,\;\;G \ne H\)

The elements of sets \(B\) and \(D\) are same also, the elements of sets \(E\) and \(G\) are same.

Therefore, \(B = D\) and \(E = G\)

Hence, sets, \(B = D\) and \(E = G\)