Ex 5.2, 10 - Prove that greatest integer function f(x) = [x]

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Ex 5.2, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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Ex 5.2, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 3 Ex 5.2, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 4 Ex 5.2, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 5 Ex 5.2, 10 - Chapter 5 Class 12 Continuity and Differentiability - Part 6

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

Transcript

Ex 5.2, 10 (Introduction) Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at π‘₯=1 and π‘₯= 2. Ex 5.2, 10 Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at π‘₯=1 and π‘₯= 2. f (x) = [x] Let’s check for both x = 1 and x = 2 At x = 1 f (x) is differentiable at x = 1 if LHD = RHD (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙) βˆ’ 𝒇(𝒙 βˆ’ 𝒉))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ([1] βˆ’ [(1 βˆ’ β„Ž)])/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (1 βˆ’ 0)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) 1/β„Ž = 1/0 = Not defined (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙 + 𝒉) βˆ’ 𝒇(𝒙))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ([(1 + β„Ž)] βˆ’ [1])/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (1 βˆ’ 1)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) 0/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) 0 = 0 For greatest integer function [1] = 1 [1 βˆ’ h] = 0 [1 + h] = 1 Since LHD β‰  RHD ∴ f(x) is not differentiable at x = 1 Hence proved At x = 2 f (x) is differentiable at x = 2 if LHD = RHD L H D (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙) βˆ’ 𝒇(𝒙 βˆ’ 𝒉))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(2) βˆ’ 𝑓(2 βˆ’ β„Ž))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ([2] βˆ’ [(2 βˆ’ β„Ž)])/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (2 βˆ’ 1)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) 1/β„Ž = 1/0 = Not defined R H D (π’π’Šπ’Ž)┬(π‘β†’πŸŽ) (𝒇(𝒙 + 𝒉) βˆ’ 𝒇(𝒙))/𝒉 = (π‘™π‘–π‘š)┬(hβ†’0) (𝑓(2 + β„Ž) βˆ’ 𝑓(2))/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) ([(2 + β„Ž)] βˆ’ [2])/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) (2 βˆ’ 2)/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) 0/β„Ž = (π‘™π‘–π‘š)┬(hβ†’0) 0 = 0 For greatest integer function [2] = 2 [2 βˆ’ h] = 1 [2 + h] = 2 Since LHD β‰  RHD ∴ f(x) is not differentiable at x = 2 Hence proved

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.