## MATHS PRATICE QUESTION PAPERS RELEASED BY APOLLO IAS ACADEMY

For this cluster a pair of and cluster, 2A test the question book banks area unit given by Greek deity ACADEMY. create use of those inquiries to clear the exams.

These queries that you just see they were free from value. The queries that we tend to area unit given you to follow well, for this test every day you need to follow then solely you'll be able to bring home the bacon your goal. The queries that area unit given by Greek deity ACADEMY helps you to clear the exams.

For this test, these queries area unit over enough to attain your goal. think about these queries. These queries area unit straightforward to find out if you follow well. These queries area unit over enough for you to clear the exams. For the exams, they're going to raise twenty five queries.

each test the queries that arouse the exams won't be with the exception of this question book bank. If the notes wherever useful for you share along with your friend's circle.

**POINTS to recollect**

**Diamond State Morgan’s Laws**

Augustus Diamond State Morgan (1806 – 1871) was a British man of science.

He was born on the twenty-seventh of June 1806 in Madurai, Tamilnadu, India. His father was denoted in an Asian country by the archipelago Company. once he was seven months recent, his family captive back to England. Diamond State Morgan was educated at Trinity School, Cambridge, London. He developed laws forset distinction and complementation. This area unit is known as Diamond State Morgan’s laws.

## De Morgan’s Laws for Set distinction

These laws relate to the set activities association, crossing purpose, and set distinction.

Let us place confidence in three sets A, B, and C as A = − 5 2,, − 1 3,, B = − 3,−203,,,5 and

C = − 2 one , , − 045 , , .

Now, B CÈ =

A B − ∪ ( ) C = −5 one, ... (1)

Then, A B - = and A C− = −5 one, , 3

( ) A B − ∪( ) A C− = −5 one, , 3 ... (2)

( ) A B − ∩( ) A C− = −5 one, ... (3)

From (1) and (2), we tend to check that

A B − ∪ ( ) C ≠ − ( ) A B ∪ − ( ) A C

But note that from (1) and (3), we tend to check that

A B − ∪ C = A B − ∩ A C

Now, B CÇ = −2 zero, ,5

A B − ∩ ( ) C = −5 one, , 3 ... (4)

From (3) and (4) we tend to check that

A B − ∩ ( ) C ≠ − ( ) A B ∩ − ( ) A C

But note that from (2) and (4), we tend to induce A B − ∩ ( ) C A =−∪ ( ) B A( ) −C

De Morgan's laws for set contrast: For any three sets A, B, and C.

(i) A B − ∪ ( ) C A =−∩ ( ) B A( ) −C (ii) A B − ∩ ( ) C A =−∪ ( ) B A( ) −C

( ) A B − ∪( ) A C− ∪( ) A B ∩ =____

American state Morgan’s Laws for Complementation

These laws relate the set operations on union, intersection, and complementation.

Let us place confidence in the universal set U=, A=, and B=.

Check whether or not or not

A B − = A B ∩ ′

Now, A B È = Thinking Corner

Then, ( ) A B ∪ ′

= .....(1)

Next, A¢

= and B¢

= 6

Then, A B ′ ∩ ′

= .....(2)

From (1) and (2), we tend to induce ( ) A B ∪ ′ = A B ′ ∩ ′

A B − ∪B A − ′

=____

Also, A B ∩ = 3 5, , Thinking Corner

( ) A B ∩ , , , , ′ = zero one 246 .....(3)

A′ = 0, , 246, andB′ = one 2, ,6

A B ′ ∪ ′ = zero one, ,246 , , .....(4)

From (3) and (4), we tend to induce ( ) A B ∩ ′ = A B ′ ∪ ′

De Morgan's laws for complementation: Let 'U' be the set containing

finite sets A and B. Then (i) ( ) A B ∪ ′ = A B ′ ∩ ′ (ii) ( ) A B ∩ ′ = A B ′ ∪ ′

Example 1.26 If U = ∈ x x: , − ≤2 one x ≤ zero ,

A x = = : , x p2 one + ∈p p , − ≤1 4 ≤ , B x = = : , x letter3 one + ∈q alphabetic character , − ≤1 4 < ,

verify American state Morgan’s laws for complementation.

A A ( ) B =____

Thinking Corner resolution

Given U = − ten ,

A = −9 and B = −

Law (i) ( ) A B ∪ ′ =A B ′ ∩ ′

Now, A B È = ten

( ) A B ∪ ′ = ..... (1)

( ) A B ∪ ′ ∪( ) A B ′ ∩ =___

Thinking Corner

Then, A¢ = 10 and B¢ =

A B ′ ∩ ′ = ..... (2)

From (1) and (2), it's verified that ( ) A B ∪ ′ = A B ′ ∩ ′

Law (ii) ( ) A B ∩ ′ = A B ′ ∪ ′

Now, A B Ç =

( ) A B ∩ ′ = ..... (3)

Then, A B ′ ∪ ′ = ..... (4)

From (3) and (4), it's verified that ( ) A B ∩ ′ = A B ′ ∪ ′

## No comments:

## Post a Comment