**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 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 be the Signum Function defined as

and be the Greatest Integer Function given by , where is greatest integer less than or equal to . Then, does and coincide in ?

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 .

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

Q. 17 Let . Then number of equivalence relations containing 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 . Then number of relations containing and 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 , and be functions defined by and . Are and equal?

Justify your answer. (Hint: One may note that two functions and

such that , 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 as

Show that zero is the identity for this operation and each element of the set

is invertible with being the inverse of .

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 be defined as

Show that the empty set is the

identity for the operation ∗ and all the elements A of P(X) are invertible with

. (Hint : and ).

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 and defined as

and . Show that ∗ is commutative but not

associative, is associative but not commutative. Further, show that ,

. [If it is so, we say that the operation ∗ distributes

over the operation ]. Does 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 and . Find of the following functions F from S

to T, if it exists.

(ii)

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 and . Find of the following functions F from S

to T, if it exists.

(i)

Q. 10 Find the number of all onto functions from the set 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

given by , 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 . 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 and such that is onto

but is not onto.

(Hint : Consider and

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 and such that is

injective but g is not injective.

(Hint : Consider and ).

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 given by 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 defined by 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...

If f defined from R to R is defined by f(x) equals x to the power 2 minus 3x plus 2, find f (f (x)).

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

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