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Binary operations on the set are is a function from   i.e. * is a function from    Hence, the total number of binary operations on set   is  Hence, option B is correct.  
 is defined as    is defined as  Let     Then we have ,  if x=1   and                                                            Hence,for  ,     and . Hence , gof and fog do not coincide with .                                    
The number of equivalence relations containing  is given by  We are left with four  pairs  ,, .   , so relation R is reflexive.     and     , so relation R is not symmetric.    , so realation R is not transitive. Hence , equivalence relation is bigger than R  is universal relation. Thus the total number of equivalence relations cotaining   is two. Thus, option B is correct.
The smallest  relations containing and which are reflexive and symmetric but not transitive is given by    , so relation R is reflexive.     and     , so relation R is symmetric.   but   , so realation R is not transitive. Now, if we add any two pairs    and  to relation R, then relation R  will become transitive. Hence, the total number of the desired relation is one. Thus, option A is...
Given : ,       are defined  by  and . It can be observed that                                 Hence, f and g are equal functions.                      
X = as    An element     is identity element for operation *,  if  For  ,                                             Hence, 0 is identity element of operation *. An element    is invertible if there exists ,  such that         i.e.     means       or     But since we have  X =  and . Then .    is inverse of a  for . Hence, inverse of element  ,   is  6-a   i.e. ,                 
Let   be defined as          Let  . Then                                                                 for all    Thus,  is identity element for operation *. An element     will be invertible if there exists   , such that .     (here  is identity element)       Hence, all elements A  of P(X) are invertible with               
It is given that   be defined as Now, let  . Then,  And Therefore, Therefore, we can say that    is the identity element for the given operation *. Now, an element A  P(X) will be invertible if there exists B  P(X) such that Now, We can see that   such that  Therefore, by this we can say that all the element A of P(X) are invertible with  
Given     and    is defined as                and    For  , we have                                                                                                                          the operation is commutative.                              where                     the operation is not associative Let   . Then we have :                                                               ...
 is defined as      , F is not one-one. So inverse of F does not exists. Hence, F is not invertible i.e.  does not exists.  
The number of all onto functions from the set to itself is  permutations on n symbols  1,2,3,4,5...............n.  Hence, permutations  on n symbols  1,2,3,4,5...............n = n Thus, total number of all onto maps from the set to itself is same as  permutations on n symbols  1,2,3,4,5...............n which is  n.  
Given      is defined as  . As we know that                  Hence, X is the identity element of binary operation *. Now, an element  is invertible if there exists a  , such that              (X is identity element)   i.e.              This is possible only if  . Hence, X is only invertible element  in  with respect to operation *                                                                 ...
Given a non empty set X, consider P(X) which is the set of all subsets of X. Since, every set is subset of itself , ARA  for all    R is reflexive. Let  This is not same as  If        and     then we cannot say that B is related to A.  R is not symmetric. If  this implies     R is transitive. Thus, R is not an equivalence relation because it is not symmetric.                          
        and               and         Onto :            Consider element in codomain N . It is clear that this element is not an image of any of element  in domain N .     f is not onto.      Now, it is clear that   , there exists       such that  . Hence,  is onto.      
One - one: Since     As we can see    but   so    is not one-one. Thus , g(x) is not injective. Let                             Since,     so x and y are both positive.          Hence, gof is injective.                   
One-one:    Let                   We need to prove .So, Let     then there cubes will not be equal i.e. .  It will contradict given condition of cubes being equal.  Hence,   and it is one -one which means it is injective.            
The function    defined by    ,  One- one:          Let                  ,                      It is observed that if x is positive and y is negative.           Since x is positive and y is negative.     but 2xy is negative. Thus, the case of x is positive and y is negative is removed. Same happens in the case  of y is positive and x is negative so this case is also removed. When x and y...
This can be solved as following f : R → R                                   
    if n is odd     if n is even. For one-one:    Taking x as odd number and y as even number.                                                            Now,   Taking y as odd number and x  as even number.                                                     This is also impossible. If both x and y are odd :                                                                                 If...