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

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Ex 5.2, 10 (Introduction) Greatest Integer Function f(x) = [x] than or equal to x. Greatest Integer Function [x] Going by same Concept 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. (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 + h] = 1 [1 − h] = 0 [1] = 1 (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 (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 + h] = 2 [2 − h] = 1 [2] = 2 Since LHD ≠ RHD ∴ f(x) is not differentiable at x = 1 Hence proved (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 (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 + h] = 2 [2 − h] = 1 [2] = 2 For greatest integer function [2 + h] = 2 [2 − h] = 1 [2] = 2 For greatest integer function [2 + h] = 2 [2 − h] = 1 [2] = 2 Since LHD ≠ RHD ∴ f(x) is not differentiable at x = 2 Hence proved