Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15 Important

Example 16

Example 17 Important

Example 18

Example 19

Example 20 Important You are here

Example 21

Example 22

Example 23 Important

Example 24

Example 25

Example 26 Important

Example 27

Example 28

Example 29 (i)

Example 29 (ii) Important

Example 29 (iii) Important

Example 29 (iv)

Example 30 Important

Example 31

Example 32 Important

Example 33 Important

Example 34

Example 35

Example 36 Important

Example 37 Important

Example 38

Example 39 Important

Example 40

Example 41 Important

Example 42 Important Deleted for CBSE Board 2022 Exams

Example 43 Deleted for CBSE Board 2022 Exams

Example 44 (i)

Example 44 (ii) Important

Example 44 (iii) Important

Example 45 (i)

Example 45 (ii) Important

Example 45 (iii) Important

Example 46

Example 47 Important

Example 48

Last updated at April 13, 2021 by Teachoo

Example 20 Show that the function f defined by f (x) = |1β π₯ + | π₯ ||, where x is any real number is a continuousGiven π(π₯) = |(1βπ₯+|π₯|)| Let π(π) = 1βπ₯+|π₯| & π(π) = |π₯| Then , πππ(π) = β(π(π₯)) = β(1βπ₯+|π₯|) = |(1βπ₯+|π₯|)| = π(π) We know that, Modulus function is continuous β΄ π(π) = |π₯| is continuous Also, π(π) = (πβπ)+|π| Since (1βπ₯) is a polynomial & every polynomial function is continuous β΄ (πβπ) is continuous Also, |π| is also continuous Since Sum of two continuous function is also continuous Thus, π(π₯) = 1βπ₯+|π₯| is continuous . Hence, π(π₯) & β(π₯) are both continuous . We know that If two function of π(π₯) & β(π₯) both continuous, then their composition πππ(π) is also continuous Hence, π(π) is continuous .