Class 11 NCERT Solutions – Chapter 1 Sets – Exercise 1.2

Last Updated : 04 Dec, 2020

i) Set of odd natural numbers divisible by 2

Solution:

Yes, this is a null set because three is no odd natural number divisible by 2.

Note: A set which does not contain any element is called the null set.

ii) Set of even prime numbers

Solution:

No, this is not a null set because 2 is an even number which is prime.

iii) {x : x is a natural numbers, x < 5 and x > 7}

Solution:

Yes, this is a null set because three is no natural number which is less than 5 and greater than 7.

iv) {y : y is a point common to any two parallel lines}

Solution:

Yes, this is a null set because two parallel lines have no points in common because they do not intersect.

i) The set of months of a year

Solution:

This is a finite set because there is only 12 months in a year.

Note: A set which is empty or consists of a definite number of elements is called finite otherwise, the set is called infinite.

ii) {1, 2, 3, …}

Solution:

This is an infinite set because there will always be a new number if we add 1 to the previous number, or we can say that it is a set of natural numbers and there are infinite natural numbers.

iii) {1, 2, 3, …, 99, 100}

Solution:

This is a finite set because there is only 100 numbers in the set.

iv) The set of positive integers greater than 100

Solution:

This is an infinite set because there are infinite positive integers greater than 100 which can be generated by adding 1 to the previous number.

v) The set of prime numbers less than 99

Solution:

This is a finite set because prime numbers which are less than 99 are finite.

i) The set of lines which are parallel to the x-axis

Solution:

Infinite. We can draw infinite parallel lines with respect to the x-axis.

ii) The set of letters in the English alphabet

Solution:

Finite. There are only 26 letters in the English alphabet.

iii) The set of numbers which are multiples of 5

Solution:

Infinite. There are infinite numbers which are multiple of 5 namely {5, 10, 15 ——-}

iv) The set of animals living on the earth

Solution:

Finite. The animals Living on the earth can be counted they are not infinite.

v) The set of circles passing through the origin (0, 0)

Solution:

Infinite. We can draw infinite number of circles passing through the origin with different radius.

i) A = {a, b, c, d} B = {d, c, b, a}

Solution:

Yes. Every element of A is also an element of B and every element of B is also an element of A namely {a, b, c, d}.

Note: Two sets A and B are said to be equal if they have exactly the same elements, and we write A = B. Otherwise, the sets are said to be unequal, and we write A â‰  B. Order in which elements appear does not matter.

ii) A = {4, 8, 12, 16} B = {8, 4, 16, 18}

Solution:

No. 12 is an element that is present in A but not in B and similarly 18 is an element present in B not in A.

iii) A = {2, 4, 6, 8, 10} B = {x : x is positive even integer and x â‰¤ 10}

Solution:

Yes. If we define set B it can be written like this {2, 4, 6, 8, 10} and therefore every element of A is also an element of B and every element of B is also an element of A.

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

Solution:

No. If we define B we can clearly see that {-40, -30, -20, -10, 0}. All these numbers are also multiples of 10, and they are not in set B. Hence A â‰  B.

i) A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0}

Solution:

No.

By solving the equation x2 + 5x + 6,

x2 + 5x + 6 = 0

x2 + 2x + 3x + 6 = 0

(x + 2)(x + 3) = 0

x = -2, -3

Now it is clear that B can be defined as {-2, -3}.

Now -2, -3 are not in A and also 2, 3 is not in set B. Hence A â‰  B.

ii) A = {x : x is a letter in the word FOLLOW} B = {y : y is a letter in the word WOLF}

Solution:

Yes.

It is clear that A can be defined as {F, O, L, W} an if we define B, {F, O, L, W}. Now it is clear that every element of A is also an element of B and every element of B is also an element of A namely {W, O, L, F}.

A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D = {3, 1, 4, 2} E = {â€“1, 1}, F = {0, a}, G = {1, â€“1}, H = {0, 1}

Solution:

B = D & E = G.

Note: Two sets A and B are said to be equal if they have exactly the same elements.

For A, 8âˆˆ A, but 8âˆ‰ B,8âˆ‰ B,8âˆ‰ D,8âˆ‰ E, 8âˆ‰ F,8âˆ‰ G,8âˆ‰ H,

Hence, A is not equal to B, D, E, F, G, H.

Also, 2âˆˆ A, but 2âˆ‰ C Hence A, â‰  C.

For B, 2âˆˆ B, but 2âˆ‰ C,2âˆ‰ E,2âˆ‰ F,2âˆ‰ G, 2âˆ‰ H.

Hence, B is not equal to C, E, F, G, H.

Also, Every element of B can be found in D namely {1, 2, 3, 4} and vice-versa is also true. Hence B = D.

For C, 14âˆˆ C, but 14âˆ‰ D,14âˆ‰ E,14âˆ‰ F,14âˆ‰ G, 14âˆ‰ H.

Hence C is not equal to D, E, F, G, H.

For D, 2âˆˆ D, but 12âˆ‰ E,2âˆ‰ F,2âˆ‰ G, 2âˆ‰ H.

Hence D is not equal to E, F, G, H.

For E, -1âˆˆ E, but -1âˆ‰ F, -1âˆ‰ H.

Hence E is not equal to F, H.

Also, Every element of E can be found in G namely {-1, 1} and vice-versa is also true. Hence E = G.

For F, 0âˆˆ F, but 0âˆ‰ G, 0âˆ‰ H.

Hence F is not equal to G, H.

For G, -1âˆˆ G, but -1âˆ‰ H.

Hence G is not equal to H.

So, we can observe that only B = D & E = G.

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