# Ex 5.1, 12 - Chapter 5 Class 12 Continuity and Differentiability

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.1, 12 Find all points of discontinuity of f, where f is defined by ๐ ๐ฅ๏ทฏ= ๐ฅ10โ1, ๐๐ ๐ฅโค1๏ทฎ&๐ฅ2 , ๐๐ ๐ฅ>1๏ทฏ๏ทฏ We have ๐ ๐ฅ๏ทฏ= ๐ฅ10โ1, ๐๐ ๐ฅโค1๏ทฎ&๐ฅ2 , ๐๐ ๐ฅ>1๏ทฏ๏ทฏ Case 1 At x = 1 f is continuous at x = 1 if L.H.L = R.H.L = ๐ 1๏ทฏ i.e. if lim๏ทฎxโ 1๏ทฎโ๏ทฏ๏ทฏ ๐ ๐ฅ๏ทฏ = lim๏ทฎxโ 1๏ทฎ+๏ทฏ๏ทฏ ๐ ๐ฅ๏ทฏ = ๐ 1๏ทฏ Thus, L.H.L โ R.H.L โ f is discontinuous at ๐ =๐ Case 2 Let x = c , where c < 1 โด ๐ ๐ฅ๏ทฏ=๐ฅ10โ1 f is continuous at x = c if lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) โ f is continuous for ๐ฅ =๐ less than 1. โ f is at continuous for all real numbers less than 1. Case 3 Let x = c (where c > 1) ๐ ๐ฅ๏ทฏ=๐ฅ2 f is continuous at x = c if lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) Thus lim๏ทฎxโ๐๏ทฏ ๐ ๐ฅ๏ทฏ=๐(๐) โ f is continuous at ๐ฅ =๐ (c is greater than 1) โ f is continuous at all real numbers greater than 1. Hence, only x = 1 point of discontinuity โ f is continuous for all real point except 1. Thus, f is continuous for all xโR โ{1}

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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.