MATHS PRATICE QUESTION PAPERS RELEASED BY APOLLO IAS ACADEMY - DAILY JOB ALERT FOR FREEE

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MATHS PRATICE QUESTION PAPERS RELEASED BY APOLLO IAS ACADEMY

MATHS PRATICE QUESTION PAPERS RELEASED BY APOLLO IAS ACADEMY



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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 ′ ∪ ′


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