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Example 11 - Find all points of discontinuity f(x) = {x+2, 0, x-2 - Examples

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Example 11 Find all the points of discontinuity of the function f defined by ๐‘“๏ท๐‘ฅ๏ทฏ=๏ท๏ท&๐‘ฅ+2 ,๐‘–๐‘“ ๐‘ฅ<1๏ทฎ0 , ๐‘–๐‘“ ๐‘ฅ=1๏ทฎ&๐‘ฅโˆ’2 ,๐‘–๐‘“ ๐‘ฅ>1๏ทฏ๏ทฏ Given ๐‘“๏ท๐‘ฅ๏ทฏ=๏ท๏ท&๐‘ฅ+2 ,๐‘–๐‘“ ๐‘ฅ<1๏ทฎ0 , ๐‘–๐‘“ ๐‘ฅ=1๏ทฎ&๐‘ฅโˆ’2 ,๐‘–๐‘“ ๐‘ฅ>1๏ทฏ๏ทฏ Case 1 Checking continuity at x = 1 f is continuous at x = 1 if, L.H.L = R.H.L = ๐‘“(1) i.e. ๏ทlim๏ทฎxโ†’๏ท1๏ทฎโˆ’๏ทฏ๏ทฏ ๐‘“๏ท๐‘ฅ๏ทฏ=๏ทlim๏ทฎxโ†’๏ท1๏ทฎ+๏ทฏ๏ทฏ ๐‘“๏ท๐‘ฅ๏ทฏ=๐‘“(1) Since, L.H.L โ‰  R.H.L โ‡’ f is not continuous at ๐‘ฅ=1. Case 2 Let c be any real number greater than 1. So, ๐‘ฅ =๐‘ where c >1 โˆด ๐‘“๏ท๐‘ฅ๏ทฏ=๐‘ฅโˆ’2 f is continuous at x = c if ๏ทlim๏ทฎxโ†’๐‘๏ทฏ ๐‘“๏ท๐‘ฅ๏ทฏ= ๐‘“๏ท๐‘๏ทฏ Hence, f is continuous at ๐‘ฅ =๐‘ (c is greater than 1) โ‡’ f is continuous at all point ๐‘ฅ>1 Case 3 Let c be any real number less than 1. So, ๐‘ฅ =๐‘ where c < 1 ๐‘“๏ท๐‘ฅ๏ทฏ=๐‘ฅ+2 f is continuous at x = c if ๏ทlim๏ทฎxโ†’๐‘๏ทฏ ๐‘“๏ท๐‘ฅ๏ทฏ= ๐‘“๏ท๐‘๏ทฏ Hence ๏ทlim๏ทฎxโ†’๐‘๏ทฏ ๐‘“๏ท๐‘ฅ๏ทฏ= ๐‘“๏ท๐‘๏ทฏ โˆด f is continuous at for all real number less than 1. Hence only ๐‘ฅ=1 is point of discontinuty. โ‡’ f is continuous at all real point except 1. Thus, f is continuous for ๐’™ โˆˆ๐‘โˆ’{๐Ÿ}

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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