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# RD Sharma Solution for Class 12 Science Mathematics Chapter 2 - Functions

Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 12 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 2 - Functions.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 12 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 12 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

Exercise/Page

## RD Sharma Solution for Class 12 Science Mathematics Chapter 2 - Functions Page/Excercise 2.1

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 2

Solution 3

Solution 4

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 5(iv)

Solution 5(v)

Solution 5(vi)

Solution 5(vii)

Solution 5(viii)

Solution 5(ix)

Solution 5(x)

Solution 5(xi)

Solution 5(xii)

Solution 5(xiii)

Solution 5(xiv)

Solution 5(xv)

Solution 5(xvi)

Solution 6

Solution 7

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f.

Hence, f has to be onto.

Solution 15

Suppose f is not one-one.

Then, there exists two elements, say 1 and 2 in the domain whose image in the co-domain is same.

Also, the image of 3 under f can be only one element.

Therefore, the range set can have at most two elements of the co-domain {1, 2, 3}

i.e f is not an onto function,  a contradiction.

Hence, f must be one-one.

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 22

Solution 23

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 1(iv)

Solution 1(v)

Solution 1(vi)

Solution 1(vii)

Solution 1(viii)

Solution 1(ix)

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11(i)

Solution 11(ii)

Solution 11(iii)

Solution 11(iv)

Solution 12

## RD Sharma Solution for Class 12 Science Mathematics Chapter 2 - Functions Page/Excercise 2.5

Solution 1

Thus, h is a bijection and is invertible.

Solution 2(i)

Solution 2(ii)

Solution 3

Solution 4

f-1 = {(3, 1), (5, 2), (7, 3), (9, 4)}

f-1 0g-1 = {(7, 1), (23, 2), (47, 3), (79, 4)} ……. (A)

Now (A) & (B) we have (g0f)-1 = f-10g-1

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42