# Ex 5.1, 34

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 5.1, 34 Find all the points of discontinuity of f defined by ๐ ๐ฅ๏ทฏ= ๐ฅ๏ทฏ โ ๐ฅ+1๏ทฏ. When ๐ฅโฅ0 ๐ ๐ฅ๏ทฏ = ๐ฅโ ๐ฅ+1๏ทฏ โx โ x + 1 = x โ x โ 1 = ๐ฅโ๐ฅ+1 = x โ x โ 1 = โ1 When ๐ฅ<โ1 ๐ ๐ฅ๏ทฏ = โ๐ฅโ โ ๐ฅ+1๏ทฏ๏ทฏ = โ๐ฅ+๐ฅ+1 = 1 When โ1โค๐ฅ<0 ๐ ๐ฅ๏ทฏ = โ๐ฅโ ๐ฅ+1๏ทฏ = โ๐ฅโ๐ฅ+1 = โ2๐ฅ+1 Now ๐ ๐ฅ๏ทฏ= 1 ๐๐ ๐ฅโคโ1๏ทฎโ2๐ฅโ1 ๐๐ โ1โค๐ฅ<0๏ทฎโ1 ๐๐ ๐ฅโฅ0๏ทฏ๏ทฏ Checking continuity Case 1 At x = 0 A function is continuous at x = 0 if L.H.L = R.H.L = ๐ 0๏ทฏ i.e. lim๏ทฎxโ 0๏ทฎโ๏ทฏ๏ทฏ ๐ ๐ฅ๏ทฏ= lim๏ทฎxโ 0๏ทฎ+๏ทฏ๏ทฏ ๐ ๐ฅ๏ทฏ= ๐ 0๏ทฏ & ๐ ๐ฅ๏ทฏ = โ1 ๐ 0๏ทฏ = โ1 Thus L.H.L = R.H.L = ๐ 0๏ทฏ Hence ๐ ๐๏ทฏ is continuous at x = 0 Case 2 At x > 0 ๐ ๐ฅ๏ทฏ = โ1 it is a constant function. & Every constant function is continuous for all real number. โ ๐ ๐๏ทฏ is continuous for x > 0 . Case 3 At x = โ 1 A function is continuous at x = โ1 if L.H.L = R.H.L = ๐ โ1๏ทฏ & ๐ ๐ฅ๏ทฏ = โ2๐ฅโ1 ๐ โ1๏ทฏ = โ2 โ1๏ทฏโ1 = 2 โ 1 = 1 Thus L.H.L = R.H.L = ๐ โ1๏ทฏ Hence ๐ ๐๏ทฏ is continuous at x = โ1 Case 4 At x < โ1 ๐ ๐ฅ๏ทฏ = 1 , it is a constant function & Every constant function is continuous for all real number โ ๐ ๐๏ทฏ is continuous for x < โ1 Case 5 At โ1โค๐ฅ<0 ๐ ๐ฅ๏ทฏ = โ2๐ฅโ1, is a polynomial & we know that every polynomial function is continuous for all real values . Hence ๐ ๐๏ทฏ = โ๐๐โ๐ is continuous at โ๐โค๐<๐ โ There is no point of discontinuity Hence ๐ ๐๏ทฏ is continuous for all ๐โ๐น

Ex 5.1, 9
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Ex 5.1, 13 Important

Ex 5.1, 16 Important

Ex 5.1, 18 Important

Ex 5.1, 28 Important

Ex 5.1, 30 Important

Ex 5.1, 34 Important You are here

Ex 5.2, 5 Important

Ex 5.2, 9 Important

Ex 5.2, 10 Important

Ex 5.3, 10 Important

Ex 5.3, 14 Important

Example 32 Important

Example 33 Important

Ex 5.5,6 Important

Ex 5.5, 7 Important

Ex 5.5, 11 Important

Ex 5.5, 16 Important

Ex 5.6, 7 Important

Ex 5.6, 11 Important

Example 41 Important

Ex 5.7, 14 Important

Example 42 Important

Ex 5.8, 5 Important

Example 44 Important

Example 45 Important

Example 47 Important

Misc 6 Important

Misc 15 Important

Misc 16 Important

Misc 23 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.