NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

NCERT Solutions for Class 12 relations and functions NCERT solutions: Exercise 1.1

Question1(i) . Determine whether each of the following relations are reflexive, symmetric and transitive:

(i) Relation in the set defined as

Answer:

Since, so is not reflexive.

Since, but so is not symmetric.

Since, but so is not transitive.

Hence, is neither reflexive nor symmetric and nor transitive.

Question 1(ii) . Determine whether each of the following relations are reflexive, symmetric and transitive:

(ii) Relation R in the set N of natural numbers defined as

Answer:

Since,

so is not reflexive.

Since, but

so is not symmetric.

Since there is no pair in such that so this is not transitive.

Hence, is neither reflexive nor symmetric and
nor transitive.

Question1(iii) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iii) Relation R in the set as

Answer:

Any number is divisible by itself and .So it is reflexive.

but .Hence,it is not symmetric.

and 4 is divisible by 2 and 4 is divisible by 4.

Hence, it is transitive.

Hence, it is reflexive and transitive but not symmetric.

Question.1(iv) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(iv). Relation R in the set Z of all integers defined as

Answer:

For , as which is an integer.

So,it is reflexive.

For , and because are both integers.

So, it is symmetric.

For , as are both integers.

Now, is also an integer.

So, and hence it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(a)

Answer:

,so it is reflexive

means .

i.e. so it is symmetric.

means also .It states that i.e. .So, it is transitive.

Hence, it is reflexive, symmetric and transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(b)

Answer:

as and is same human being.So, it is reflexive.

means .

It is same as i.e. .

So,it is symmetric.

means and .

It implies that i.e. .

Hence, it is reflexive, symmetric and
transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(c)

Answer:

means but i.e. .So, it is not reflexive.

means but i.e .So, it is not symmetric.

means and .

i.e. .

Hence, it is not reflexive,not symmetric and
not transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v). Relation R in the set A of human beings in a town at a particular time given by

(d)

Answer:

means but i.e. .

So, it is not reflexive.

means but i.e. .

So, it is not symmetric.

Let, means and .

This case is not possible so it is not transitive.

Hence, it is not reflexive, symmetric and
transitive.

Question:1(v) Determine whether each of the following relations are reflexive, symmetric and
transitive:

(v) Relation R in the set A of human beings in a town at a particular time given by

(e)

Answer:

means than i.e. .So, it is not reflexive..

means than i.e. .So, it is not symmetric.

Let, means and than i.e. .

So, it is not transitive.

Hence, it is neither reflexive nor symmetric and nor transitive.

Question:2 Show that the relation R in the set R of real numbers defined as
is neither reflexive nor symmetric nor transitive.

Answer:

Taking

and

So, R is not reflexive.

Now,

because .

But, i.e. 4 is not less than 1

So,

Hence, it is not symmetric.

as

Since because

Hence, it is not transitive.

Thus, we can conclude that it is neither reflexive, nor symmetric, nor transitive.

Question:3 Check whether the relation R defined in the set as
is reflexive, symmetric or transitive.

Answer:

R defined in the set

Since, so it is not reflexive.

but

So, it is not symmetric

but

So, it is not transitive.

Hence, it is neither reflexive, nor symmetric, nor transitive.

Question:4 Show that the relation R in R defined as , is reflexive and

transitive but not symmetric.

Answer:

As so it is reflexive.

Now we take an example

as

But because .

So,it is not symmetric.

Now if we take,

Than, because

So, it is transitive.

Hence, we can say that it is reflexive and transitive but not symmetric.

Question:5 Check whether the relation R in R defined by is reflexive,
symmetric or transitive.

Answer:

because

So, it is not symmetric

Now, because

but because

It is not symmetric

as .

But, because

So it is not transitive

Thus, it is neither reflexive, nor symmetric, nor transitive.

Question:6 Show that the relation R in the set given by is
symmetric but neither reflexive nor transitive.

Answer:

Let A=

We can see so it is not reflexive.

As so it is symmetric.

But so it is not transitive.

Hence, R is symmetric but neither reflexive nor transitive.

Question:7 Show that the relation R in the set A of all the books in a library of a college,
given by is an equivalence
relation.?

Answer:

A = all the books in a library of a college

because x and x have the same number of pages so it is reflexive.

Let means x and y have same number of pages.

Since y and x have the same number of pages so .

Hence, it is symmetric.

Let means x and y have the same number of pages.

and means y and z have the same number of pages.

This states,x and z also have the same number of pages i.e.

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence
relation.?

Question:8 Show that the relation R in the set given by , is an equivalence relation. Show that all the elements of are related to each other and all the elements of are related to each other. But no element of is related to any element of .

Answer:

Let there be then as which is even number. Hence, it is reflexive

Let where then as

Hence, it is symmetric

Now, let

are even number i.e. are even

then, is even (sum of even integer is even)

So, . Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The elements of are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of is not related to because a difference of odd and even number is not even.

Question:9(i) Show that each of the relation R in the set , given by

(i) is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

For , as which is multiple of 4.

Henec, it is reflexive.

Let, i.e. is multiple of 4.

then is also multiple of 4 because = i.e.

Hence, it is symmetric.

Let, i.e. is multiple of 4 and i.e. is multiple of 4 .

is multiple of 4 and is multiple of 4

is multiple of 4

is multiple of 4 i.e.

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is

is multiple of 4.

is multiple of 4.

is multiple of 4.

Question:9(ii) Show that each of the relation R in the set , given by

(ii) is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer:

For , as

Henec, it is reflexive.

Let, i.e.

i.e.

Hence, it is symmetric.

Let, i.e. and i.e.

i.e.

Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The set of all elements related to 1 is {1}

Question:10(i) Give an example of a relation.

(i) Which is Symmetric but neither reflexive nor transitive.

Answer:

Let

so it is not reflexive.

and so it is symmetric.

but so it is not transitive.

Hence, symmetric but neither reflexive nor transitive.

Question:10(ii) Give an example of a relation.

(ii) Which is transitive but neither reflexive nor symmetric.

Answer:

Let

Now for , so it is not reflexive.

Let i.e.

Then is not possible i.e. . So it is not symmetric.

Let i.e. and i.e.

we can write this as

Hence, i.e. . So it is transitive.

Hence, it is transitive but neither reflexive nor symmetric.

Question:10(iii) Give an example of a relation.

(iii) Which is Reflexive and symmetric but not transitive.

Answer:

Let

Define a relation R on A as

If , i.e. . So it is reflexive.

If , and i.e. . So it is symmetric.

and i.e. . and

But So it is not transitive.

Hence, it is Reflexive and symmetric but not transitive.

Question:10(iv) Give an example of a relation.

(iv) Which is Reflexive and transitive but not symmetric.

Answer:

Let there be a relation R in R

because

Let i.e.

But i.e.

So it is not symmetric.

Let i.e. and i.e.

This can be written as i.e. implies

Hence, it is transitive.

Thus, it is Reflexive and transitive but not symmetric.

Question:10(v) Give an example of a relation.

(v) Which is Symmetric and transitive but not reflexive.

Answer:

Let there be a relation A in R

So R is not reflexive.

We can see and

So it is symmetric.

Let and

Also

Hence, it is transitive.

Thus, it Symmetric and transitive but not reflexive.

Question:11 Show that the relation R in the set A of points in a plane given by
, is an equivalence relation. Further, show that the set of
all points related to a point is the circle passing through P with origin as
centre.

Answer:

The distance of point P from the origin is always the same as the distance of same point P from origin i.e.

R is reflexive.

Let i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e.

R is symmetric.

Let and

i.e. the distance of point P from the origin is the same as the distance of point Q from the origin, and also the distance of point Q from the origin is the same as the distance of the point S from the origin.

We can say that the distance of point P, Q, S from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e.

R is transitive.

Hence, R is an equivalence relation.

The set of all points related to a point are points whose distance from the origin is the same as the distance of point P from the origin.

In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these sets of points form a circle with the centre as the origin and this circle passes through the point.

Question:12 Show that the relation R defined in the set A of all triangles as , is equivalence relation. Consider three right angle triangles T ₁ with sides 3, 4, 5, T ₂ with sides 5, 12, 13 and T ₃ with sides 6, 8, 10. Which triangles among T ₁ , T ₂ and T ₃ are related?

Answer:

All triangles are similar to itself, so it is reflexive.

Let,

i.e.T ₁ is similar to T2

T ₁ is similar to T2 is the same asT2 is similar to T ₁ i.e.

Hence, it is symmetric.

Let,

and i.e. T ₁ is similar to T2 and T2 is similar toT ₃ .

T ₁ is similar toT ₃ i.e.

Hence, it is transitive,

Thus, , is equivalence relation.

Now, we see the ratio of sides of triangle T ₁ andT ₃ are as shown

i.e. ratios of sides of T ₁ and T ₃ are equal.Hence, T ₁ and T ₃ are related.

Question:13 Show that the relation R defined in the set A of all polygons as , is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?

Answer:

The same polygon has the same number of sides with itself,i.e. , so it is reflexive.

Let,

i.e.P ₁ have same number of sides as P ₂

P ₁ have the same number of sides as P ₂ is the same as P ₂ have same number of sides as P ₁ i.e.

Hence,it is symmetric.

Let,

and i.e. P ₁ have the same number of sides as P ₂ and P ₂ have same number of sides as P ₃

P ₁ have same number of sides as P ₃ i.e.

Hence, it is transitive,

Thus, , is an equivalence relation.

The elements in A related to the right angle triangle T with sides 3, 4 and 5 are those polygons which have 3 sides.

Hence, the set of all elements in A related to the right angle triangle T is set of all triangles.

Question:14 Let L be the set of all lines in XY plane and R be the relation in L defined as . Show that R is an equivalence relation. Find the set of all lines related to the line

Answer:

All lines are parallel to itself, so it is reflexive.

Let,

i.e.L ₁ is parallel to L ₂ .

_L1 is parallel to L ₂ is same as L ₂ is parallel to L ₁ i.e.

Hence, it is symmetric.

Let,

and i.e. _L1 is parallel to L ₂ and L ₂ is parallel to L ₃ .

L ₁ is parallel to L ₃ i.e.

Hence, it is transitive,

Thus, , is equivalence relation.

The set of all lines related to the line are lines parallel to

Here, Slope = m = 2 and constant = c = 4

It is known that the slope of parallel lines are equal.

Lines parallel to this ( ) line are ,

Hence, set of all parallel lines to are .

Question:15 Let R be the relation in the set A= {1,2,3,4}

given by . Choose the correct answer.

(A) R is reflexive and symmetric but not transitive.

(B) R is reflexive and transitive but not symmetric.

(C) R is symmetric and transitive but not reflexive.

(D) R is an equivalence relation.

Answer:

A = {1,2,3,4}

For every there is .

R is reflexive.

Given, but

R is not symmetric.

For there are

R is transitive.

Hence, R is reflexive and transitive but not symmetric.

The correct answer is option B.

Question:16 Let R be the relation in the set N given by . Choose the correct answer.

(A)
(B)
(C)
(D)

Answer:

(A) Since, so

(B) Since, so

(C) Since, and so

(d) Since, so

The correct answer is option C.

Relations and Functions Class 12 NCERT Solutions: Exercise 1.2

Question:1 Show that the function defined by is one-one and onto,where R _∗ is the set of all non-zero real numbers. Is the result true, if the domain R _∗ is replaced by N with co-domain being same as R _∗ ?

Answer:

Given, is defined by .

One - One :

f is one-one.

Onto:

We have , then there exists ( Here ) such that

.

Hence, the function is one-one and onto.

If the domain R _∗ is replaced by N with co-domain being same as R _{∗ i.e.} defined by

g is one-one.

For ,

but there does not exists any x in N.

Hence, function g is one-one but not onto.

Question:2(i) Check the injectivity and surjectivity of the following functions:

(i) given by

Answer:

One- one:

then

f is one- one i.e. injective.

For there is no x in N such that

f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:2(ii) Check the injectivity and surjectivity of the following functions:

(ii) given by

Answer:

One- one:

For then

but

f is not one- one i.e. not injective.

For there is no x in Z such that

f is not onto i.e. not surjective.

Hence, f is neither injective nor surjective.

Question:2(iii) Check the injectivity and surjectivity of the following functions:

(iii) given by

Answer:

One- one:

For then

but

f is not one- one i.e. not injective.

For there is no x in R such that

f is not onto i.e. not surjective.

Hence, f is not injective and not surjective.

Question:2(iv) Check the injectivity and surjectivity of the following functions:

(iv) given by

Answer:

One- one:

then

f is one- one i.e. injective.

For there is no x in N such that

f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:2(v) Check the injectivity and surjectivity of the following functions:

(v) given by

Answer:

One- one:

For then

f is one- one i.e. injective.

For there is no x in Z such that

f is not onto i.e. not surjective.

Hence, f is injective but not surjective.

Question:3 Prove that the Greatest Integer Function , given by , is neither one-one nor onto, where denotes the greatest integer less than or equal to .

Answer:

One- one:

For then and

but

f is not one- one i.e. not injective.

For there is no x in R such that

f is not onto i.e. not surjective.

Hence, f is not injective but not surjective.

Question:4 Show that the Modulus Function f : R → R, given by , is neither one-one nor onto, where is if is positive or 0 and is , if is negative.

Answer:

One- one:

For then

f is not one- one i.e. not injective.

For ,

We know is always positive there is no x in R such that

f is not onto i.e. not surjective.

Hence, , is neither one-one nor onto.

Question:5 Show that the Signum Function , given by

is neither one-one nor onto.

Answer:

is given by

As we can see , but

So it is not one-one.

Now, f(x) takes only 3 values (1,0,-1) for the element -3 in codomain ,there does not exists x in domain such that .

So it is not onto.

Hence, signum function is neither one-one nor onto.

Question:6 Let , and let be a function from A to B. Show that f is one-one.

Answer:

Every element of A has a distant value in f.

Hence, it is one-one.

Question:7(i) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(i) defined by

Answer:

Let there be such that

f is one-one.

Let there be ,

Puting value of x,

f is onto.

f is both one-one and onto hence, f is bijective.

Question:7(ii) In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.

(ii) defined by

Answer:

Let there be such that

For and

f is not one-one.

Let there be (-2 in codomain of R)

There does not exists any x in domain R such that

f is not onto.

Hence, f is neither one-one nor onto.

Question:8 Let A and B be sets. Show that such that is
bijective function.

Answer:

Let

such that

and

f is one- one

Let,

then there exists such that

f is onto.

Hence, it is bijective.

Question:9 Let be defined by for all . State whether the function f is bijective. Justify your answer.

Answer:

,

Here we can observe,

and

As we can see but

f is not one-one.

Let, (N=co-domain)

case1 n be even

For ,

then there is such that

case2 n be odd

For ,

then there is such that

f is onto.

f is not one-one but onto

hence, the function f is not bijective.

Question:10 Let and . Consider the function defined by . Is f one-one and onto? Justify your answer.

Answer:

Let such that

f is one-one.

Let, then

such that

For any there exists such that

f is onto

Hence, the function is one-one and onto.

Question:11 Let be defined as . Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

One- one:

For then

does not imply that

example: and

f is not one- one

For there is no x in R such that

f is not onto.

Hence, f is neither one-one nor onto.

Option D is correct.

Question:12 Let be defined as . Choose the correct answer.

(A) f is one-one onto

(B) f is many-one onto

(C) f is one-one but not onto

(D) f is neither one-one nor onto.

Answer:

One - One :

Let

f is one-one.

Onto:

We have , then there exists such that

.

Hence, the function is one-one and onto.

The correct answer is A .

Relation and Function Class 12 maths chapter 1 question answer: Exercise 1.3

Question:1 Let and be given by and . Write down .

Answer:

Given : and

and

Hence, =

Question:2 Let , and be functions from to . Show that

Answer:

To prove :

Hence,

To prove:

Hence,

Question:3(i) Find and , if

(i) and

Answer:

and

Question:3(ii) Find gof and fog, if

(ii) and

Answer:

The solution is as follows

(ii) and

Question:4 If show that , for all . What is the inverse of ?

Answer:

, for all

Hence,the given function is invertible and the inverse of is itself.

Question:5(i) State with reason whether following functions have inverse

(i)

with

Answer:

(i) with

From the given definition,we have:

f is not one-one.

Hence, f do not have an inverse function.

Question:5(ii) State with reason whether following functions have inverse

(ii) with

Answer:

(ii) with

From the definition, we can conclude :

g is not one-one.

Hence, function g does not have inverse function.

Question:5(iii) State with reason whether following functions have inverse

(iii) with

Answer:

(iii) with

From the definition, we can see the set have distant values under h.

h is one-one.

For every element y of set ,there exists an element x in such that

h is onto

Thus, h is one-one and onto so h has an inverse function.

Question:6 Show that , given by is one-one. Find the inverse of the function

Answer:

One -one:

f is one-one.

It is clear that is onto.

Thus,f is one-one and onto so inverse of f exists.

Let g be inverse function of f in

let y be an arbitrary element of range f

Since, is onto, so

for

,

Question:7 Consider given by . Show that f is invertible. Find the inverse of .

Answer:

is given by

One-one :

Let

f is one-one function.

Onto:

So, for there is ,such that

f is onto.

Thus, f is one-one and onto so exists.

Let, by

Now,

and

Hence, function f is invertible and inverse of f is .

Question:8 Consider f : R+ → [4, ∞) given by . Show that f is invertible with the inverse of f given by , where R+ is the set of all non-negative real numbers.

Answer:

It is given that
, and

Now, Let f(x) = f(y)

⇒ x ² + 4 = y ² + 4

⇒ x ² = y ²

⇒ x = y

⇒ f is one-one function.

Now, for y [4, ∞), let y = x ² + 4.

⇒ x ² = y -4 ≥ 0

⇒ for any y R, there exists x = R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x ² + 4) =

And, fog(y) = f(g(y)) = =

Therefore, gof = gof = I _R .

Therefore, f is invertible and the inverse of f is given by

f-1(y) = g(y) =

Question:9 Consider given by . Show that is invertible with

Answer:

One- one:

Let

Since, x and y are positive.

f is one-one.

Onto:

Let for ,

f is onto and range is .

Since f is one-one and onto so it is invertible.

Let by

Hence, is invertible with the inverse of given by

Question:10 Let be an invertible function. Show that f has a unique inverse. (Hint: suppose and are two inverses of . Then for all ,
. Use one-one ness of f).

Answer:

Let be an invertible function

Also, suppose f has two inverse

For , we have

[f is invertible implies f is one - one]

[g is one-one]

Thus,f has a unique inverse.

Question:11 Consider given by , and . Find and show that .

Answer:

It is given that

Now,, lets define a function g :
such that

Now,

Similarly,

And

Hence, and , where and

Therefore, the inverse of f exists and

Now,
is given by

Now, we need to find the inverse of ,

Therefore, lets define such that

Now,

Similarly,

Hence, and , where and

Therefore, inverse of exists and

Therefore,

Hence proved

Question:12 Let be an invertible function. Show that the inverse of is , i.e.,

Answer:

To prove:

Let be a invertible function.

Then there is such that and

Also,

and

and

Hence, is invertible function and f is inverse of .

i.e.

Question:13 If be given by , then is

(A)

(B)

(C)

(D)

Answer:

Thus, is x.

Hence, option c is correct answer.

Question:14 Let be a function defined as . The inverse of is the map given by

(A)

(B)

(C)

(D)

Answer:

Let f inverse

Let y be the element of range f.

Then there is such that

Now , define as

Hence, g is inverse of f and

The inverse of f is given by .

The correct option is B.

Class 12 maths chapter 1 NCERT solutions: Exercise 1.4

Question:1(i) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(i) On , define ∗ by

Answer:

(i) On , define ∗ by

It is not a binary operation as the image of under * is .

Question:1(ii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(ii) On , define ∗ by

Answer:

(ii) On , define ∗ by

We can observe that for ,there is a unique element ab in .

This means * carries each pair to a unique element in .

Therefore,* is a binary operation.

Question:1(iii) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(iii) On , define ∗ by

Answer:

(iii) On , define ∗ by

We can observe that for ,there is a unique element in .

This means * carries each pair to a unique element in .

Therefore,* is a binary operation.

Question:1(iv) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this.

(iv) On , define ∗ by

Answer:

(iv) On , define ∗ by

We can observe that for ,there is a unique element in .

This means * carries each pair to a unique element in .

Therefore,* is a binary operation.

Question:1(v) Determine whether or not each of the definition of ∗ given below gives a binary operation. In the event that ∗ is not a binary operation, give justification for this

(v) On , define ∗ by

Answer:

(v) On , define ∗ by

* carries each pair to a unique element in .

Therefore,* is a binary operation.

Question:2(i) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(i)On , define

Answer:

a*b=a-b

b*a=b-a

so * is not commutative

(a*b)*c=(a-b)-c

a*(b*c)=a-(b-c)=a-b+c

(a*b)*c not equal to a*(b*c), so * is not associative

Question:2(ii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(ii) On , define

Answer:

(ii) On , define

ab = ba for all

ab+1 = ba + 1 for all

for

where

operation * is not associative.

Question:2(iii) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(iii) On , define

Answer:

(iii) On , define

ab = ba for all

for all

for

operation * is commutative.

operation * is associative.

Question:2(iv) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(iv) On , define

Answer:

(iv) On , define

ab = ba for all

2ab = 2ba for all

for

the operation is commutative.

where

operation * is not associative.

Question:2(v) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(v) On , define

Answer:

(v) On , define

and

for

the operation is not commutative.

where

operation * is not associative.

Question:2(vi) For each operation ∗ defined below, determine whether ∗ is binary, commutative or associative.

(vi) On , define

Answer:

(iv) On , define

and

for

the operation is not commutative.

where

operation * is not associative.

Question:3

Consider the binary operation Λ on the set {1, 2, 3, 4, 5} defined by

a Λ b = min {a, b}. Write the operation table of the operation Λ .

Answer:

for

The operation table of the operation is given by :

Question:4(i) Consider a binary operation ∗ on the set given by the following multiplication table (Table 1.2).

(i) Compute and

(Hint: use the following table)

Answer:

(i)

Question:4(ii) Consider a binary operation ∗ on the set given by the following multiplication table (Table 1.2).

(ii) Is ∗ commutative?

(Hint: use the following table)

Answer:

(ii)

For every , we have . Hence it is commutative.

Question:4(iii) Consider a binary operation ∗ on the set { given by the following multiplication table (Table 1.2).

(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).

(Hint: use the following table)

Answer:

(iii) (2 ∗ 3) ∗ (4 ∗ 5).

from the above table

Question:5 Let ∗′ be the binary operation on the set defined by . Is the operation ∗′ same as the operation ∗ defined
in Exercise 4 above? Justify your answer.

Answer:

for

The operation table is as shown below:

The operation ∗′ same as the operation ∗ defined in Exercise 4 above.

Question:6(i) let ∗ be the binary operation on N given by . Find

(i) 5 ∗ 7, 20 ∗ 16

Answer:

a*b=LCM of a and b

(i) 5 ∗ 7, 20 ∗ 16

Question:6(ii) Let ∗ be the binary operation on N given by . Find

(ii) Is ∗ commutative?

Answer:

(ii) for all

Hence, it is commutative.

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

(iii) Is ∗ associative?

Answer:

a b = L.C.M. of a and b

(iii)

Hence, the operation is associative.

Question:6(iv) Let ∗ be the binary operation on N given by . Find

(iv) the identity of ∗ in N

Answer:

(iv) the identity of ∗ in N

We know that

for

Hence, 1 is the identity of ∗ in N.

Question 6(v) 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 ∗?

Answer:

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

Question:7 Is ∗ defined on the set by a binary operation? Justify your answer.

Answer:

A =

Operation table is as shown below:

From the table, we can observe that

Hence, the operation is not a binary operation.

Question:8 Let ∗ be the binary operation on N defined by a ∗ b = H.C.F. of a and b. Is ∗ commutative? Is ∗ associative? Does there exist identity for this binary
operation on N?

Answer:

a ∗ b = H.C.F. of a and b for all

H.C.F. of a and b = H.C.F of b and a for all

Hence, operation ∗ is commutative.

For ,

Hence, ∗ is associative.

An element will be identity for operation * if for .

Hence, the operation * does not have any identity in N.

Question:9(i) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(i) Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defined as .It is observed that:

here

Hence, the * operation is not commutative.

It can be observed that

for all

The operation * is not associative.

Question:9(ii) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(ii) Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as .It is observed that:

For

Hence, the * operation is commutative.

It can be observed that

for all

The operation * is not associative.

Question:9(iii) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(iii) Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as .It is observed that:

For

for

Hence, the * operation is not commutative.

It can be observed that

for all

The operation * is not associative.

Question:9(iv) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(iv) Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defined as .It is observed that:

For

for

Hence, the * operation is commutative.

It can be observed that

for all

The operation * is not associative.

Question:9(v) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(v) Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as .It is observed that:

For

for

Hence, the * operation is commutative.

It can be observed that

for all

The operation * is associative.

Question:9(vi) Let ∗ be a binary operation on the set Q of rational numbers as follows:

(vi) Find which of the binary operations are commutative and which are associative.

Answer:

On the set Q ,the operation * is defines as .It is observed that:

For

for

Hence, the * operation is not commutative.

It can be observed that

for all

The operation * is not associative.

Question:10 Find which of the operations given above has identity.

Answer:

An element will be identity element for operation *

if for all

when .

Hence, has identity as 4.

However, there is no such element which satisfies above condition for all rest five operations.

Hence, only (v) operations have identity.

Question:11 Let and ∗ be the binary operation on A defined by Show that ∗ is commutative and associative. Find the identity element for ∗ on A, if any.

Answer:

and ∗ be the binary operation on A defined by

Let

Then,

We have

Thus it is commutative.

Let

Then,

Thus, it is associative.

Let will be a element for operation * if for all .

i.e.

This is not possible for any element in A .

Hence, it does not have any identity.

Question:12(i) State whether the following statements are true or false. Justify.

(i) For an arbitrary binary operation ∗ on a set N,

Answer:

(i) For an arbitrary binary operation ∗ on a set N,

An operation * on a set N as

Then , for b=a=2

Hence, statement (i) is false.

Question:12(ii) State whether the following statements are true or false. Justify.

(ii) If ∗ is a commutative binary operation on N, then

Answer:

(ii) If ∗ is a commutative binary operation on N, then

R.H.S

(* is commutative)

( as * is commutative)

= L.H.S

Hence, statement (ii) is true.

Question:13 Consider a binary operation ∗ on N defined as . Choose the correct answer.

(A) Is ∗ both associative and commutative?
(B) Is ∗ commutative but not associative?
(C) Is ∗ associative but not commutative?
(D) Is ∗ neither commutative nor associative?

Answer:

A binary operation ∗ on N defined as .

For

Thus, it is commutative.

where

Hence, it is not associative.

Hence, B is the correct option.

NCERT solutions for class 12 maths chapter 1 Relations and Functions: Miscellaneous Exercise

Question:1 Let be defined as . Find the function such that .

Answer:

and

For one-one:

Thus, f is one-one.

For onto:

For ,

Thus, for , there exists such that

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Let as

and

Hence, defined as

Question:2 Let be defined as , if n is odd and , if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer:

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 both x and y are even :

f is one-one.

Onto:

Any odd number 2r+1 in codomain of N is an image of 2r in domain N and any even number 2r in codomain N is the image of 2r+1 in domain N.

Thus, f is onto.

Hence, f is one-one and onto i.e. it is invertible.

Sice, f is invertible.

Let as if m is even and if m is odd.

When x is odd.

When x is even

Similarly, m is odd

m is even ,

and

Hence, f is invertible and the inverse of f is g i.e. , which is the same as f.

Hence, inverse of f is f itself.

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

Answer:

This can be solved as following

f : R → R

Question:4 Show that the function defined by is one one and onto function.

Answer:

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 both are positive:

When x and y both are negative :

f is one-one.

Onto:

Let such that

If y is negative, then

If y is positive, then

Thus, f is onto.

Hence, f is one-one and onto.

Question:5 Show that the function given by is injective.

Answer:

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

Question:6 Give examples of two functions and such that is injective but g is not injective. (Hint : Consider and ).

Answer:

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.

Question:7 Give examples of two functions and such that is onto but is not onto.

(Hint : Consider and

Answer:

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.

Question: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.

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.

Question: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 ∗.

Answer:

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 *

Question:10 Find the number of all onto functions from the set to itself.

Answer:

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.

Question:11(i) Let and . Find of the following functions F from S to T, if it exists.

(i)

Answer:

is defined as

is given by

Question:11(ii) Let and . Find of the following functions F from S to T, if it exists.

(ii)

Answer:

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.

Question: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.

Answer:

Given and is defined as
and

For , we have

the operation is commutative.

where

the operation is not associative

Let . Then we have :

Hence,

Now,

for

Hence, operation o does not distribute over operation *.

Question: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).

Answer:

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

Question: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 .

Answer:

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. ,

Question: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).

Answer:

Given :

,

are defined by and .

It can be observed that

Hence, f and g are equal functions.

Question: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

Answer:

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 correct.

Question:17 Let . Then number of equivalence relations containing is
(A) 1
(B) 2
(C) 3
(D) 4

Answer:

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.

Question: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 ?

Answer:

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 .

Question:19 Number of binary operations on the set are(A) 10(B) 16(C) 20(D ) 8

Answer:

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