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Ex 5.2, 9 Prove that the function f given by ๐‘“ (๐‘ฅ) = | ๐‘ฅ โ€“ 1|, ๐‘ฅ โˆˆ ๐‘… is not differentiable at x = 1. f(x) = |๐‘ฅโˆ’1| = {โ–ˆ((๐‘ฅโˆ’1), ๐‘ฅโˆ’1โ‰ฅ0@โˆ’(๐‘ฅโˆ’1), ๐‘ฅโˆ’1<0)โ”ค = {โ–ˆ((๐‘ฅโˆ’1), ๐‘ฅโ‰ฅ1@โˆ’(๐‘ฅโˆ’1), ๐‘ฅ<1)โ”ค Now, f(x) is a differentiable at x = 1 if LHD = RHD (๐’๐’Š๐’Ž)โ”ฌ(๐กโ†’๐ŸŽ) (๐’‡(๐’™) โˆ’ ๐’‡(๐’™ โˆ’ ๐’‰))/๐’‰ = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (๐‘“(1) โˆ’ ๐‘“(1 โˆ’ โ„Ž))/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (|1 โˆ’ 1|โˆ’|(1 โˆ’ โ„Ž)โˆ’1|)/โ„Ž = (๐‘™ ๐‘–๐‘š)โ”ฌ(hโ†’0) (0 โˆ’|โˆ’โ„Ž|)/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (0 โˆ’ โ„Ž)/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (โˆ’โ„Ž)/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (โˆ’1) = โˆ’1 (๐’๐’Š๐’Ž)โ”ฌ(๐กโ†’๐ŸŽ) (๐’‡(๐’™ + ๐’‰) โˆ’ ๐’‡(๐’™))/๐’‰ = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (๐‘“(1 + โ„Ž) โˆ’ ๐‘“(1))/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (|(1 + โ„Ž) โˆ’ 1|โˆ’|1 โˆ’ 1|)/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (|โ„Ž| โˆ’ 0)/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (โ„Ž โˆ’ 0)/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) โ„Ž/โ„Ž = (๐‘™๐‘–๐‘š)โ”ฌ(hโ†’0) (1) = 1 Since LHD โ‰  RHD โˆด f(x) is not differentiable at x = 1 Hence proved

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

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo