Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10 You are here

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

Example 21

Example 22

Example 23

Example 24 Important

Example 25

Example 26 (i)

Example 26 (ii) Important

Example 26 (iii) Important

Example 26 (iv)

Example 27 Important

Example 28

Example 29 Important

Example 30 Important

Example 31

Example 32

Example 33 Important

Example 34 Important

Example 35

Example 36 Important

Example 37

Example 38 Important

Example 39 (i)

Example 39 (ii) Important

Example 40 (i)

Example 40 (ii) Important

Example 40 (iii) Important

Example 41

Example 42 Important

Example 43

Question 1 Deleted for CBSE Board 2024 Exams

Question 2 Important Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Last updated at April 16, 2024 by Teachoo

Example 10 Discuss the continuity of the function f defined by π(π₯)={β(&π₯+2, ππ π₯β€1@&π₯β2, ππ π₯>1)β€ π(π₯)={β(&π₯+2, ππ π₯β€1@&π₯β2, ππ π₯>1)β€ Since we need to find continuity at of the function We check continuity for different values of x When x = 1 When x < 1 When x > 1Case 1 : When x = 1 f(x) is continuous at π₯ =1 if L.H.L = R.H.L = π(1) if limβ¬(xβ1^β ) π(π₯)=limβ¬(xβ1^+ ) " " π(π₯)= π(1) Since there are two different functions on the left & right of 1, we take LHL & RHL . LHL at x β 1 limβ¬(xβ1^β ) f(x) = limβ¬(hβ0) f(1 β h) = limβ¬(hβ0) (1ββ)+2 = limβ¬(hβ0) (3ββ) = 3 β 0 = 3 RHL at x β 1 limβ¬(xβ1^+ ) f(x) = limβ¬(hβ0) f(1 + h) = limβ¬(hβ0) (1+β)β2 = limβ¬(hβ0) (β1+β) = β1 + 0 = β1 Since L.H.L β R.H.L f(x) is not continuous at x = 1 Case 2 : When x < 1 For x < 1, f(x) = x + 2 Since this a polynomial It is continuous β΄ f(x) is continuous for x < 1 Case 3 : When x > 1 For x > 1, f(x) = x β 2 Since this a polynomial It is continuous β΄ f(x) is continuous for x > 1 Hence, only π₯=1 is point of discontinuity. β΄ f is continuous at all real numbers except 1 Thus, f is continuous for πβ R β {1}