cbse ncert solutions for class 11 mathematics chapter 02 · 2019-11-26 · class – xi – cbse...

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Class – XI – CBSE - Mathematics Relations and Functions

CBSE NCERT Solutions for Class 11 Mathematics Chapter 02

Back of Chapter Questions

Exercise 2.1

1. If find the values of x and y.

Solution: Step

1:

Given,

By comparing the above step we get,

Overall hint: When two coordinates are equal their abscissa and ordinate are also equal

2. If the set A has elements and the set , then find the number of elements in

and

Step 2:

Therefore,

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Class – XI – CBSE - Chemistry Relations and Functions

Solution: Step

1:

Given that, Set

Number of elements in set

Step 2:

Overall hint: The total elements in product of two sets is The product of elements in individual

sets.

3. If

Solution: Step1:

Given that,

As we know that the Cartesian product of two non-empty sets and is defined as

Step 2:

4. State whether each of the following statement is true or false. If the statement is false, rewrite the

given statement correctly

(i) If

Number of elements in

Therefore, the number of elements in is 9

and , find

Therefore,

Overall hint:

and then

has elements and

3



Solution:

Step 1:

False

If


given statement correctly Solution:

(ii) If and are non-empty sets, then is a non-empty set of ordered pairs such

that and

Step1:

True

As we know that, If and are non-empty sets, then is a non-empty set of ordered

pairs such that and

Overall hint: If and are non-empty sets, then is a non-empty set of ordered pairs such that and


given statement correctly

(iii) If .

Overall Hint: If

then

4



Solution:

Step1: True

As,

7. If

Solution:

8. If

Solution:

Step 1:

Given that,

As we know, the Cartesian product of two non-empty sets and is defined as

is the set of all second elements.

find

Find and

Step 2:

Therefore, is the set of all first elements and

Hence, and

Overall hint:

Overall hint:

Step 1:

We know that fo r any non - empty set is defined as

Given,

Overall hint:

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9. Let . Verify that

(i)

Solution:

Step 1;

…(i)

10. Let . Verify that

(ii)

Solution:

Step 1:

Given that, and

Step 2:

…..(ii)

Overall Hint: is Associative property of 3 Sets

To prove is a subset of

and

and

is a subset of

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Overall Hint: For a set to be subset all its elements must be present in another set which may

consist of other elements also.

11. Let . Write . How many subsets will have? List them.

Solution: Step 1:

12. Let and be two sets such that

are in

find and , where

Solution:

Step2:

By this, it is clear that all the elements of are the elements of . Thus, is a

subset of

and

Given that,

Step 2:

It is known that if C is a set with n then n

Thus, the set has subsets. These are

Over all hint: If C is a set with n then n

and . If

a nd are distinct elements.

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Step 1:

Given, are in

It is known that,

Set of first elements of the ordered pair elements of

Set of second elements of the ordered pair elements of .

Step 2:

and z are the elements of

the elements of that are

. Since,

Overall hint: The first elements are in first set and the second elements are in the second set

13. The Cartesian product has elements among which are found . Find the

set and the remaining elements of .

Solution:

Step 2:

and ; and

and

Step 1:

We know that, if and , then .

.

is

that is and and

and .

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Exercises 2.2

1. Let . Define a relation from to by

. Write down its domain, co-domain and range.

Solution:

Step 2:

The domain of will be the set of all first elements of the ordered pairs in the relation

Therefore, Domain of

Step 3:

The whole set

Therefore, Co-domain of

The range of R is the set of all second elements of the ordered pairs in the relation.

Therefore, Range of

The ordered pair and are two of the nine elements of .

It is known that

So, the set A is

The other elements of are and

O verall Hint:

Step 1:

Given relation from to is given as

i.e.,

is the co - domain of relation

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Overall Hint: The domain of will be the set of all first elements of the ordered pairs in the

relation and the range of R is the set of all second elements of the ordered pairs in the relation.

. Depict this relationship using roster form. Write down the domain and the range.

Solution:

The domain of is the set of all first elements of the ordered pairs in the relation


Step 3:

The Range of is the set of all second elements of the ordered pairs in the relation

Therefore, Range of

Overall hint: The domain of is the set of all first elements of the ordered pairs in the relation

and the Range of is the set of all second elements of the ordered pairs in the relation

3.

. Write in roster

form.

2. Define a relation on the set of natural n umbers by

Step 1

Given,

The natural numbers less than are and .

Therefore,

Step 2:

and Define a relation from to by

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Solution:

Step 1:

Therefore, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

Overall Hint: x-y=n/2 +1

4. The figure shows a relationship between the sets and Write this relation

(i) In set-builder form (ii) Roster form.

What is it’s domain and range?

Solution:

Domain of

Range of

Overall hint: In Set builder form we write the generalized equation but in roster form we write all

the elements

Given, and

Step 1:

From the given figure,

( i ) From the figure we can write the relation in set builder form as

Step 2:

( ii )

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5. Let defined by

.

(ii) Find the

domain of

(iii) Find the range of

Solution:

(iii) The Range of is the set of all second elements of the ordered pairs in the relation

Therefore, Range of



6. Determine the domain and range of the relation defined by

Solution:

Step 1:

Given relation,

Therefore,

Step 2:

Step 1:

Given,

( i )

Step 2:

( ii ) The domain of is the set of all first elements of the or dered pairs in the relation


Step 3:

Let be the relation on

( i) Wri te in roster form

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and

and

and

The domain of is the set of all first elements of the ordered pairs in the relation


Step 3:

The Range of is the set of all second elements of the ordered pairs in the relation

Therefore Range of

Overall Hint: The domain of is the set of all first elements of the ordered pairs in the relation


7. Write the relation in roster form.

Solution:

Step 1:

Given relation is,

We know that

Therefore,

Overall hint: The prime number below 10 makes up domain and the co domain is made up of the

cube of prime numbers below 10

8. Let Find the number of relations from

Solution:

Step 1:

Given that, .

We know that, If are non-empty sets, then is a non-empty set of ordered pairs

such that Step

2:

Therefore,

From the above step, . So, the number of subsets of .

and to

is

is a prime number less than

and are the prime numbers less than

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Thus, the number of relations from to is

Overall hint: , So, the number of subsets of .

9. Let be the relation on defined by . Find the

domain and range of Solution:

Step 1:

Given relation is

We know that the difference between any two integers is always an integer.


And, Range of



Exercise 2.3

1. Is the following relation are function? Give reasons. If it is a function, determine its domain and

range.

(i)

Solution:

Step 1:

Are the elements of the domain of the given relation.

Step 2:

The elements of the domain of the relation have unique images. So, this relation is function

From the given relation, Domain and Range

is

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and

Overall Hint For a relation to be function it should have a unique image

Is the following relation are function? Give reasons. If it is a function, determine its domain and

range.

(ii)

Solution:

Step 1:

are the elements of the domain of the given relation.

Step 2:

The elements of the domain of the relation have unique images. So, this relation is a function

From the given relation we get, Domain and Range

Overall Hint: For a relation to be function it should have a unique image

Is the following relation are function? Give reasons. If it is a function, determine its domain and

range.

(iii)

Solution:

Step 1:

has two images and . So, this relation is not function.

Whereas for a relation to be a function, the domain elements should have unique images Overall

hint: For a relation to be function it should have a unique image

2. Find the domain and range of the following real function:

(i)

The element

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Solution: Step 1:

(i) Given,

We can also write

is all real numbers except positive real numbers.

Overall hint: Domain is R and range is the where the function exists

Find the domain and range of the following real function:

(ii)

Step2:

Therefore,

As is defined for , the domain of is

It is observed that the range of

Therefore, the range of is

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Solution:

will always lie between and

Overall hint: Domain is R and range is the where the function exists

3. A function is defined by Write down the values of

(i)

Solution:

Step 1:

Overall hint: Substitute x values in given f(x) equation

is defined by Write down the values of

(ii)

Step 1:

Given,

Step 2:

For any value of such that ,

Therefore, the range of

Given function is

Putting in

A function

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Solution:

Step 1:


Write down the values of


4. The function which maps temperature in degree Celsius into temperature in degree Fahrenheit

is defined by Find

Given function is

Putting in

A fun ction is defined by

( iii)

Given function is

Putting in

( i)

( ii)

( iii)

( iv) The value of when

Solution:

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Overall hint: Substitute c values in t(c) equation

Step 1:

Given,


Solution:

Step 1:

Given,

Step 1 :

Given,

= 32

The function which maps temperature in degree Celsius into temperature in degree Fahrenheit

is defined by Find

( ii)

Solution:


is defined by Find

( ii )

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Find

when



is defined by

(iv) The value of

Solution:

Step 1:

Given,

Given,

Therefore, the value of C when

Overall hint: Substitute t(c) values in c equation

5. Find the range of each of the following functions:

(i)

, is

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Solution:

Overall Hint: Range is where the function exists

Find the range of each of the following functions:

(ii) is a real number

Solution:

Therefore, Range of


Find the range of each of the following functions: (iii) is a real number

Solution:

Step 1:

Step 1:

If

[ by adding 2 both the sides ]

Therefore, Range of

Step 1:

is a real number

Let be any real number

⇒ [ by adding 2 both the sides ]

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is a real number

From the given function, the range of is the set of all real numbers.

Therefore, Range of


Miscellaneous Exercise

Solution:

1. The relation is defined by, .

Show that is a function and is not a function.

Step 1:

The relation is defined by

For

For

And at or

We can observe that for , the images of are unique. Therfore, the given relation

is a function.

The relation is defined by

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Step 3:

The element as it’s images

Therefore, the given relation is not a function.

Overall Hint: A function must have unique images

2. If .

Solution:

Overall hint: Take x=1.1 and x=1

3. Find the domain of the function

Solution:

Step 1:

For

For

And at or

has

find

Step 1:

Given,

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is defined for all real numbers except for

.

Overall hint: Domain is where f(x) can be defined

4. Find the domain and the range of the real function defined by Solution:

Step 1:

The given function is

Thus, the domain of is the set of all real numbers greater than or equal to .


is the set of all real numbers greater than or equal to

.

Overall hint: Domain is where f(x) can be defined and range is where we get f(x) values

Given function is

It i s seen that function


and

Function

Step 2:

As

Thus, the range of

Therefore, Range of

is defined only when

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5. Find the domain and the range of the real function defined by Solution:

Step 1:

is . The given function

given, | is defined for all real values of . We know that from

Therefore, Domain

of Step 2:

Also, for | assumes all the real values,

Therefore, the range of is the set of all non-negative real numbers.

Overall Hint: Domain is where f(x) can be defined and range is where we get f(x) values

6. Let be a function from into Determine the range of

Solution:

7. Let be defined, respectively by

Step 1:

Giv en function is

For

The range of is the set of all second elements. It can be observed that all these elements are

greater than or equal to but less than .

Therefore, range of

Overall hint: Range of f is where the function exists

Find

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and Solution:

Step 1:

8. Let be a function from defined by

Given, is defined as

Therefore,

Step 2:

Therefore,

Therefore,

Overall hint: (f - g)(x)=f(x) - g(x) and (f+g)(x)=f(x)+g(x)

for some integers Determine t o

26



Solution:

in equation (1)

Therefore, the values of and are and respectively.

Overall Hint: Take a,b values from the relation using f(x)=ax+b

from defined by . Are the 9. Let be a relations

following true?

(i) for all

Solution:

Step 1:

is not true.

Overall hint: If one (a,a) isn’t present in the relation then the given statement is false

9. Let be a relations from defined by

. Are the following true?

(ii) implies

Step 1:

Given function is

……….. (1)

…………..(2)

Step 2:

Put the value of

to

As, ; however,

Thus, the statement for all

to

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Solution:

Step 1:

Mark1:1

DL1: E

Overall Hint: when (a,b) is there (b,a) need not be present

10. Let

. Are

the

(iii)

Solution:

We see that because

Now,

implies is not true

Overall hint: If (a,b),(b,c) exists (a,c) may or may not exist

11. Let are the

(i)

We see that because and

We know that, Therefore,

Thus, the statement , implies is not true.

and and

be a relations from to defined by

following true?

implies

Thus, .

Thus, the statement

and

following true?

is a relations from to

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Justify your answer Solution:

Step 1:

Given,

Overall Hint: A relation from a non-empty set to a non-empty set is a subset of the

Cartesian product of

12. Let are the

(ii)

Justify your answer Solution:

Step 1:

As the same first element i.e., has two different images i.e., and

The relation is not a function.

Overall Hint: A pre image must have only one image

and

A relation from a non - empty set to a non - empt y set

is a subset of the Cartesian product of

Therefore,

It is seen that is a subset of

Thus, f is a relation from A to B.

and

following true?

is function from to

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13. Let be the subset of defined by a function

from to

? Justify your answer.

Solution:

Overall hint: function must have single pre image for a image

14. Let and let be defined by the highest prime factor of

Solution:

is defined as the highest prime factor of

Therefore, The highest prime factor of

Is

Step 1:

The relation f is defined as

We know that a relation from se t to set is said to be a function if every element of set

has unique image in set

As i.e., and .

The element has two different images i.e., and .

Therefore, the given relation is not a function.

Find the range of

Step 1:

Given, and

.

Prime factor of

Prime factor of

Prime fact or of

Prime factor of

Prime factor of

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Step 2:

The highest prime factor of




The range of is the set of all where

Therefore, Range of

Overall hint: The range of is the set of all where

cbse ncert solutions for class 11 mathematics chapter 02 · 2019-11-26 · class – xi – cbse...

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