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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

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

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Ex 5.2, 10 (Introduction) Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at 𝑥=1 and 𝑥= 2. Ex 5.2, 10 Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at 𝑥=1 and 𝑥= 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 (𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙) − 𝒇(𝒙 − 𝒉))/𝒉 = (𝑙𝑖𝑚)┬(h→0) (𝑓(1) − 𝑓(1 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) ([1] − [(1 − ℎ)])/ℎ = (𝑙𝑖𝑚)┬(h→0) (1 − 0)/ℎ = (𝑙𝑖𝑚)┬(h→0) 1/ℎ = 1/0 = Not defined (𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙 + 𝒉) − 𝒇(𝒙))/𝒉 = (𝑙𝑖𝑚)┬(h→0) (𝑓(1 + ℎ) − 𝑓(1))/ℎ = (𝑙𝑖𝑚)┬(h→0) ([(1 + ℎ)] − [1])/ℎ = (𝑙𝑖𝑚)┬(h→0) (1 − 1)/ℎ = (𝑙𝑖𝑚)┬(h→0) 0/ℎ = (𝑙𝑖𝑚)┬(h→0) 0 = 0 For greatest integer function [1] = 1 [1 − h] = 0 [1 + h] = 1 Since LHD ≠ 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 (𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙) − 𝒇(𝒙 − 𝒉))/𝒉 = (𝑙𝑖𝑚)┬(h→0) (𝑓(2) − 𝑓(2 − ℎ))/ℎ = (𝑙𝑖𝑚)┬(h→0) ([2] − [(2 − ℎ)])/ℎ = (𝑙𝑖𝑚)┬(h→0) (2 − 1)/ℎ = (𝑙𝑖𝑚)┬(h→0) 1/ℎ = 1/0 = Not defined R H D (𝒍𝒊𝒎)┬(𝐡→𝟎) (𝒇(𝒙 + 𝒉) − 𝒇(𝒙))/𝒉 = (𝑙𝑖𝑚)┬(h→0) (𝑓(2 + ℎ) − 𝑓(2))/ℎ = (𝑙𝑖𝑚)┬(h→0) ([(2 + ℎ)] − [2])/ℎ = (𝑙𝑖𝑚)┬(h→0) (2 − 2)/ℎ = (𝑙𝑖𝑚)┬(h→0) 0/ℎ = (𝑙𝑖𝑚)┬(h→0) 0 = 0 For greatest integer function [2] = 2 [2 − h] = 1 [2 + h] = 2 Since LHD ≠ RHD ∴ f(x) is not differentiable at x = 2 Hence proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.