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S seema garhwal
No, it is false. To prove : P ( A ) ∪ P ( B )  P ( A ∪ B ) Let,  A = {1,3}    and    B = {3,4}         A  B = {1,3,4} P(A) = { {},{1},{3},{1,3}} P(B) = { {},{3},{4},{3,4}} L.H.S =       P(A)  P(B) = { {},{1},{3},{1,3},{3,4},{4}} R.H.S.=      P(A  B) = {  {},{1},{3},{1,3},{3,4},{4},{1,4},{1,3,4}} Hence, L.H.S. R.H.S          

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D Divya Prakash Singh
Universal set = U = {1,2,3,4,5,6,7....................} (i) {x : x is an even natural number} = {2,4,6,8,..........}    {x : x is an even natural number}'= U - {x : x is an even natural number}  = {1,3,5,7,9,..........} = {x : x is an odd natural number} (ii) { x : x is an odd natural number }' = U - { x : x is an odd natural number } = {x : x is an even natural number} (iii)  {x : x is a...

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D Divya Prakash Singh
Given,  A  X B  X     and  A  X  B  X To prove:   A = B A = A (AX)              (A  X B  X)     = A (BX)      = (AB)  (AX)     =  (AB)              (A  X  )     =  (AB)  B = B (BX)              (A  X B  X)     = B (AX)      = (BA)  (BX)     =  (BA)              (B  X  )     =  (BA)  We know that    (AB) =  (BA) = A = B Hence, A = B        

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S seema garhwal
Let,       A = {0,1,2}              B = {1,2,3}              C = {1,2,3,4,5}  Given,   A  B = A  C L.H.S :     A  B = {1,2} R.H.S :    A  C = {1,2} and here   {1,2,3}  {1,2,3,4,5}  =   B  C. Hence,  A  B = A  C need not imply B = C.  

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S seema garhwal
This can be solved as follows (ii) A  ( A  B ) = A    A  ( A  B ) =  (A  A)  ( A  B )     A  ( A  B ) =  A  ( A  B )                          {  A  ( A  B ) = A  proved in 9(i)}    A  ( A  B ) =   A

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D Divya Prakash Singh
Given,    A  B, B  C and A  C are non-empty sets To prove : A   B  C   Let A = {1,2}       B = {1,3}       C = {3,2} Here,   A  B = {1}              B  C = {3}              A  C = {2} and   A   B  C                     

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S seema garhwal
(i) A  ( A  B ) = A We know that  A  A             and       A  B  A           A  ( A  B )    A and also , A   A  ( A  B ) Hence,  A  ( A  B ) = A           

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S seema garhwal
A = ( A  B )  ( A – B ) L.H.S = A = Red coloured area R.H.S =  ( A  B )  ( A – B ) ( A  B ) = green coloured ( A – B ) =  yellow coloured ( A  B )  ( A – B ) = coloured part Hence, L.H.S = R.H.S = Coloured part   A  ( B – A ) = ( A  B ) A = sky blue coloured ( B – A )=pink coloured  L.H.S = A  ( B – A ) = sky blue coloured + pink coloured  R.H.S = ( A  B ) = brown coloured part L.H.S =...

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D Divya Prakash Singh
n ( taking tea) = 150 n (taking coffee) = 225 n ( taking both ) = 100 n(people taking tea or coffee) = n ( taking tea) +  n (taking coffee) -  n ( taking both )                                                 = 150 + 225 - 100                                                 =375 - 100                                                  = 275 Total students = 600 n(students  taking neither tea nor...

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D Divya Prakash Singh
n (hindi ) = 100 n (english ) = 50 n(both) = 25 n(students  in the group ) = n (hindi ) +  n (english ) - n(both)                                           = 100 + 50 - 25                                            = 125 Hence,there are 125 students in the group.

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S seema garhwal
Given,    P ( A ) = P ( B ) To prove :   A =  B  Let,    x A          A   P ( A ) = P ( B ) For some C  P ( B )  ,  x  C  Here, C  B  Therefore, x   B      and    A  B Similarly we can say B  A. Hence,   A =  B 

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D Divya Prakash Singh
n(H) = 25 n(T) = 26 n(I) = 26 n(H  I) = 9 n( T  I ) = 8 n( H  T ) = 11 n(H  T  I ) = 3   the number of people who read at least one of the newspapers = n(HTI) = n(H) + n(T) + n(I) - n(H  I) -  n( T  I ) - n( H  T ) + n(H  T  I )                                                                                                      = 25 + 26 + 26 - 9 - 8 - 11 + 3                                    ...

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S seema garhwal
Given ,      A  B To prove :  C – B   C – A Let, x  C - B means    x C  but  xB A  B  so x C but   xA      i.e.   x  C - A Hence,    C – B   C – A

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S seema garhwal
First, we need  to show  A B    A – B = Let  A  B  To prove : A – B = Suppose  A – B   this means, x  A and x  B , which is not possible as  A  B . SO,   A – B = . Hence, A  B  A – B = . Now, let A – B = To prove :  A  B  Suppose, x  A  A – B =  so x  B  Since,  x  A    and   x  B  and A – B =   so  A  B   Hence, A B    A – B = .   Let    A B To prove :  A  B = B We can say B   A  B Suppose,...

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D Divya Prakash Singh
n(A) = 21                                                                                                       n(B) = 26                                                                             n(C) = 29  n( A  B) = 14 n( A  C) = 12 n (B  C ) = 14 n( A  B  C) = 8       n(liked product C only) = 29 - 4 -8 - 6 = 11 11 people like only product C.

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S seema garhwal
Let A, B, and C be the sets such that A  B = A  C and A  B = A  C To prove : B = C.  A  B = A + B - A  B =  A  C = A + C - A  C                A + B - A  B =  A + C - A  C                    B - A  B =   C - A  C              ( since  A  B = A  C )                          B = C Hence proved that  B= C.  

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S seema garhwal
The given statement is true, Let,  A  B and x  B  Suppose, x  A Then, x  B , which is contradiction to x  B Hence, x  A.    

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D Divya Prakash Singh
The following are the answers for the questions (i) A  A′  U (ii) ′  A   A (iii) A  A′   (iv) U′  A  

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S seema garhwal
The given statement is false, Let        x be 2               A = { 1,2,3}              B = { 4,5,6,7} Here, 2   { 1,2,3} = x  A   and   { 1,2,3}   { 4,5,6,7} = A B But,  2 { 4,5,6,7}  implies  x  B

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S seema garhwal
The given statement is false Let , A = {1,2}         B = {3,4,5 }         C = { 1,2,6,7,8} Here, {1,2}  {3,4,5 } = A  B  and  {3,4,5 }  { 1,2,6,7,8} = B  C  But ,   {1,2}    { 1,2,6,7,8}  =  A  C
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