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

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

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

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

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 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, Social Science, Physics, Chemistry, Computer Science at Teachoo.