Ex 5.1, 7 - Find all points of discontinuity of f(x) = {|x| + 3 - Ex 5.1

Slide30.JPG
Slide31.JPG Slide32.JPG Slide33.JPG Slide34.JPG Slide35.JPG

  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
Ask Download

Transcript

Ex 5.1, 7 Find all points of discontinuity of f, where f is defined by = +3, 3 2 , 3< <3 6 +2, 3 We have, = +3, 3 2 , 3< <3 6 +2, 3 Case 1 At = 3 f is continuous at x = 3 if L.H.L = R.H.L = 3 & = +3 3 = 3 + 3 = 3 + 3 = 6 Thus, L.H.L = R.H.L = 3 f is continuous at x = 3 Case 2 At =3 f is continuous at x = 3 if L.H.L = R.H.L = 3 Since L.H.L R.H.L i.e. 6 20 f is not continuous at x = 3 Case 3 Let = , where 3 < c < 3 = 2 f is continuous at x = c if lim x = ( ) f is continuous at = & 3 < x < 3 Thus, f is continuous at each x ( 3, 3) Case 4 Let = where c < 3) = +3 f is continuous at x = c (where c < 3) if lim x = ( ) Hence lim x = ( ) f is continuous at = ( < 3) f is continuous at all real numbers less than 3. Case 5 If = & >3 = 6x+2 f is continuous at x = c where c > 3 if lim x = ( ) Thus lim x = ( ) f is continuous at = ( >3) f is continuous at all real numbers greater than 3. Hence, f is discontinuous at only =3 f is continuous at all real numbers except 3. f is continuous at x R 3

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.