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

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Monday, June 1, 2020

MATHS PRATICE QUESTION PAPERS RELEASED BY APOLLO IAS ACADEMY


For this group 2 and group, 2A exam the question book banks are given by APOLLO ACADEMY. Make use of these questions to clear the exams. 

These questions that you see they were free from cost. The questions which we are given you to practice well, for this exam day_by_day you must practice then only you can achieve your goal. The questions that are given by APOLLO ACADEMY is helping you to clear the exams. 

For this exam, these questions are more than enough to achieve your goal. Concentrate on these questions. These questions are easy to learn if you practice well. These questions are more than enough for you to clear the exams. For the exams, they will ask 25 questions.

 Every exam the questions that ask for the exams will not be apart from this question book bank. If the notes where helpful for you share with your friend's circle. 

POINTS TO REMEMBER

 De Morgan’s Laws


Augustus De Morgan (1806 – 1871) was a British mathematician. 


He was born on 27th June 1806 in Madurai, Tamilnadu, India. His
father was posted in India by the East India Company. When he was
seven months old, his family moved back to England. De Morgan was
educated at Trinity College, Cambridge, London. He formulated laws for
set difference and complementation. These are called De Morgan’s laws.

De Morgan’s Laws for Set Difference

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


Let us consider three sets A, B and C as A = −{ } 5 2 , , − 1 3, , B = −{ } 3,−203 ,,,5 and
C = −{ } 2 1 , , − 045 , , .
Now, B CÈ = { , --- 321 , , , 0345 ,,}
 A B − ∪ ( ) C = { } −5 1, ... (1)
Then, A B - = { , -5 1} and A C− = { } −5 1, , 3
 ( ) A B − ∪( ) A C− = { } −5 1, , 3 ... (2)
 ( ) A B − ∩( ) A C− = { } −5 1, ... (3)
From (1) and (2), we see that
 A B − ∪ ( ) C ≠ − ( ) A B ∪ − ( ) A C
But note that from (1) and (3), we see that

 A B − ∪ C = A B − ∩ A C

Now, B CÇ = { } −2 0, ,5
 A B − ∩ ( ) C ={ } −5 1, , 3 ... (4)
From (3) and (4) we see that
 A B − ∩ ( ) C ≠ − ( ) A B ∩ − ( ) A C
But note that from (2) and (4), we get 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 ∩ =____

 De Morgan’s Laws for Complementation


These laws relate the set operations on union, intersection, and complementation.
Let us consider universal set U={0,1,2,3,4,5,6}, A={1,3,5} and B={0,3,4,5}.
Check whether
A B − = A B ∩ ′
Now, A B È = { , 0 1, , 345, } Thinking Corner
Then, ( ) A B ∪ ′
= { , 2 6} .....(1)
Next, A¢
= { , 0 246 ,,} and B¢
 = { ,1 2, }6
Then, A B ′ ∩ ′
= { , 2 6} .....(2)
From (1) and (2), we get ( ) A B ∪ ′ = A B ′ ∩ ′

A B − ∪B A − ′
=____
Also, A B ∩ = { } 3 5, , Thinking Corner
 ( ) A B ∩ , , , , ′ = { } 0 1 246 .....(3)
 A′ = { } 0, , 246, andB′ = { } 1 2, ,6
 A B ′ ∪ ′ = { } 0 1, ,246 , , .....(4)
From (3) and (4), we get ( ) A B ∩ ′ = A B ′ ∪ ′

De Morgan's laws for complementation: Let 'U' be the general 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 1 x ≤ 0 ,
A x = = { } : , x p2 1 + ∈p p , − ≤1 4 ≤ , B x = = { } : , x q3 1 + ∈q q  , − ≤1 4 < ,
verify De Morgan’s laws for complementation.
A A ( ) B =____
Thinking Corner Solution
Given U = −{ , 2 1 − , , 0 1, , 234, , , 567,,, 8 9, } 10 ,

A = −{ ,1 1, , 357,,}9 and B = −{ , 2 1,,, 4 7 10}

Law (i) ( ) A B ∪ ′ =A B ′ ∩ ′
Now, A B È = { , - -2 1, ,1 345 , , , , 7 9, } 10

 ( ) A B ∪ ′ = { , 0 2, , 6 8} ..... (1)
( ) A B ∪ ′ ∪( ) A B ′ ∩ =___
Thinking Corner
Then, A¢ = { , -2 0, , 2468 , , , } 10 and B¢ = { , -1 0, , 2 3 5, ,689 ,,}
 A B ′ ∩ ′ = { , 0 2, , 6 8} ..... (2)
 From (1) and (2), it is verified that ( ) A B ∪ ′ = A B ′ ∩ ′
Law (ii) ( ) A B ∩ ′ = A B ′ ∪ ′
Now, A B Ç = { ,1 7}

 ( ) A B ∩ ′ = { , - -2 1, , 0 2345 , , ,,, 689, ,10} ..... (3)

Then, A B ′ ∪ ′ = { , - -2 1, , 0 2345 , , ,,, 689, ,10} ..... (4)

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

MODEL 1 QUESTIONS PAPER WITH 
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MODEL 2 QUESTIONS PAPER WITH 
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MODEL 3 QUESTIONS PAPER WITH 
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MODEL 4 QUESTIONS PAPER WITH 
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