Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q

Question 6. Let ∗ be the binary operation on N given by a ∗ b = L.C.M. of a and b. Find

(v) Which elements of N are invertible for the operation ∗?

An element a is invertible in N  if  Here a is inverse of b. a*b=1=b*a a*b=L.C.M. od a and b a=b=1 So 1 is the only invertible element of N

Q. 19 Number of binary operations on the set \{a, b\} are
(A) 10
(B) 16
(C) 20
(D ) 8

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.  

Q. 18 Let f : R \rightarrow R be the Signum Function defined as

        f(x) = \left\{\begin{matrix} 1 & x> 0 \\ 0 &x = 0 \\ -1& x < 0 \end{matrix}\right.

and g : R \rightarrow R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

 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 .                                    

Q. 17 Let A = \{1, 2, 3\}. Then number of equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C) 3
(D) 4

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.

Q.16 Let A = \{1, 2, 3\}. Then number of relations containing (1, 2) and (1, 3) which are
reflexive and symmetric but not transitive is

(A) 1
(B) 2
(C) 3
(D) 4

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...

Q. 15 Let A = \{- 1, 0, 1, 2\}, B = \{- 4, - 2, 0, 2\} and f, g : A \rightarrow B be functions defined  by f(x) = x^2 -x, x\in A and g(x) = 2\left |x - \frac{1}{2} \right | - 1, x\in A. Are f and g equal?

Justify your answer. (Hint: One may note that two functions f : A \rightarrow B and
g : A \rightarrow B such that f (a) = g (a) \;\forall a \in A, are called equal functions).

Given : ,       are defined  by  and . It can be observed that                                 Hence, f and g are equal functions.                      

Q. 14  Define a binary operation ∗ on the set \{0, 1, 2, 3, 4, 5\} as

a * b = \left\{\begin{matrix} a + b &if\;a+b < 6 \\ a+ b -6 & if\;a+b\geq6 \end{matrix}\right.

Show that zero is the identity for this operation and each element a\neq 0of the set
is invertible with 6 - a being the inverse of a.

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. ,                 

Q. 13   Given a non-empty set X, let* : P(X) \times P(X) \rightarrow P(X) be defined as
A * B = (A - B) \cup (B -A), \;\forall A, B \in P(X). Show that the empty set \phi is the
identity for the operation ∗ and all the elements A of P(X) are invertible with
A^{-1} =A. (Hint : (A - \phi) \cup (\phi - A) = Aand (A -A) \cup (A - A) = A * A = \phi).

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               

 

Q 13 ) Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A–1 = A. (Hint : (A – φ) ∪ (φ – A) = A and (A – A) ∪ (A – A) = A ∗ A = φ).

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  

Q. 12  Consider the binary operations * : R \times R \rightarrow R and \circ : R \times R \rightarrow R defined as
a *b = |a - b| and a \circ b = a \;\forall a \in R. Show that ∗ is commutative but not
associative, \circ is associative but not commutative. Further, show that \forall a,b,c \in R,
a*(b\circ c) = (a*b)\circ (a*c). [If it is so, we say that the operation ∗ distributes
over the operation \circ]. Does \circ distribute over ∗? Justify your answer.

Given     and    is defined as                and    For  , we have                                                                                                                          the operation is commutative.                              where                     the operation is not associative Let   . Then we have :                                                               ...

Q. 11 Let S = \{a, b, c\} and T = \{1, 2, 3\}. Find F^{-1} of the following functions F from S
to T, if it exists.

(ii)   F = \{(a, 2), (b, 1), (c, 1)\}

 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.  

Q. 11 Let S = \{a, b, c\} and T = \{1, 2, 3\}. Find F^{-1} of the following functions F from S
to T, if it exists. 

(i) F = \{(a, 3), (b, 2), (c, 1)\}

 is defined as      is given by   

Q. 10 Find the number of all onto functions from the set \{1, 2, 3, ... , n\} to itself.

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.  

Q. 9 Given a non-empty set X, consider the binary operation * : P(X) \times P(X) \rightarrow P(X)
given by A * B = A \cap B\;\; \forall A ,B\in P(X), where P(X) is the power set of X.
Show that X is the identity element for this operation and X is the only invertible
element in P(X) with respect to the operation ∗.

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 *                                                                 ...

Q.8 Given a non empty set X, consider P(X) which is the set of all subsets of X. Define the relation R in P(X) as follows:
For subsets A, B in P(X), ARB if and only if A \subset B. Is R an equivalence relation
on P(X)? Justify your answer.

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.                          

Q. 7 Give examples of two functions f : N\rightarrow N and g : N\rightarrow N such that gof is onto
but f is not onto.

(Hint : Consider f(x) = x + 1 and g(x) = \left\{\begin{matrix} x -1 & if x > 1\\ 1 & if x = 1 \end{matrix}\right.

        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.      

Q. 6 Give examples of two functions f : N \rightarrow Zand g : Z \rightarrow Z such that gofis
injective but g is not injective.

(Hint : Consider f (x) = x and g (x) = | x |).

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.                   

Q.5 Show that the function f : R \rightarrow R given by f (x) = x ^3 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.            

Q. 4 Show that the function f : R \rightarrow \{x \in R : - 1 < x < 1\} defined byf(x) = \frac{x}{1 + |x|}   x \in R   is one one and onto function.

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...

Q. 3 If f : R → R is defined by f(x) = x 2– 3x + 2, find f (f (x)).

This can be solved as following f : R → R                                   
Exams
Articles
Questions