Ex 5.2

Chapter 5 Class 12 Continuity and Differentiability
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Ex 5.2, 9 Prove that the function f given by ๐ (๐ฅ) = | ๐ฅ โ 1|, ๐ฅ โ ๐ is not differentiable at x = 1. f(x) = |๐ฅโ1| = {โ((๐ฅโ1), ๐ฅโ1โฅ[email protected]โ(๐ฅโ1), ๐ฅโ1<0)โค = {โ((๐ฅโ1), ๐ฅโฅ[email protected]โ(๐ฅโ1), ๐ฅ<1)โค Now, f(x) is a differentiable at x = 1 if LHD = RHD (๐๐๐)โฌ(๐กโ๐) (๐(๐) โ ๐(๐ โ ๐))/๐ = (๐๐๐)โฌ(hโ0) (๐(1) โ ๐(1 โ โ))/โ = (๐๐๐)โฌ(hโ0) (|1 โ 1|โ|(1 โ โ)โ1|)/โ = (๐ ๐๐)โฌ(hโ0) (0 โ|โโ|)/โ = (๐๐๐)โฌ(hโ0) (0 โ โ)/โ = (๐๐๐)โฌ(hโ0) (โโ)/โ = (๐๐๐)โฌ(hโ0) (โ1) = โ1 (๐๐๐)โฌ(๐กโ๐) (๐(๐ + ๐) โ ๐(๐))/๐ = (๐๐๐)โฌ(hโ0) (๐(1 + โ) โ ๐(1))/โ = (๐๐๐)โฌ(hโ0) (|(1 + โ) โ 1|โ|1 โ 1|)/โ = (๐๐๐)โฌ(hโ0) (|โ| โ 0)/โ = (๐๐๐)โฌ(hโ0) (โ โ 0)/โ = (๐๐๐)โฌ(hโ0) โ/โ = (๐๐๐)โฌ(hโ0) (1) = 1 Since LHD โ  RHD โด f(x) is not differentiable at x = 1 Hence proved