Ex 5.1, 12 - Find all points of discontinuity f(x) = {x10 - 1 - Ex 5.1

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  1. Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.1, 12 Find all points of discontinuity of f, where f is defined by ๐‘“ ๐‘ฅ๏ทฏ= ๐‘ฅ10โˆ’1, ๐‘–๐‘“ ๐‘ฅโ‰ค1๏ทฎ&๐‘ฅ2 , ๐‘–๐‘“ ๐‘ฅ>1๏ทฏ๏ทฏ We have ๐‘“ ๐‘ฅ๏ทฏ= ๐‘ฅ10โˆ’1, ๐‘–๐‘“ ๐‘ฅโ‰ค1๏ทฎ&๐‘ฅ2 , ๐‘–๐‘“ ๐‘ฅ>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๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = lim๏ทฎxโ†’ 1๏ทฎ+๏ทฏ๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ = ๐‘“ 1๏ทฏ Thus, L.H.L โ‰  R.H.L โ‡’ f is discontinuous at ๐’™ =๐Ÿ Case 2 Let x = c , where c < 1 โˆด ๐‘“ ๐‘ฅ๏ทฏ=๐‘ฅ10โˆ’1 f is continuous at x = c if lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) โ‡’ f is continuous for ๐‘ฅ =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‘“ ๐‘ฅ๏ทฏ=๐‘ฅ2 f is continuous at x = c if lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) Thus lim๏ทฎxโ†’๐‘๏ทฏ ๐‘“ ๐‘ฅ๏ทฏ=๐‘“(๐‘) โ‡’ f is continuous at ๐‘ฅ =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 xโˆˆR โˆ’{1}

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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.