# Ex 5.2, 10

Last updated at June 19, 2017 by Teachoo

Last updated at June 19, 2017 by Teachoo

Transcript

Ex 5.2, 10 (Introduction) Greatest Integer Function F(x) = [x] Example: [1] = 1 [1.2] = 1 [1.9999] = 1 [2] = 2 [2.01] = 2 [20001] = 2 ∴ , [1 + h] = 1 [1 − h] = 0 [1] = 1 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] At x = 1 f (x) is differentiable at x = 1 y if L H D = R H D (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 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 Since LHD ≠ RHD ∴ f(x) is not differentiable at x = 2 Hence proved

Chapter 5 Class 12 Continuity and Differentiability

Ex 5.1, 9
Important

Ex 5.1, 13 Important

Ex 5.1, 16 Important

Ex 5.1, 18 Important

Ex 5.1, 28 Important

Ex 5.1, 30 Important

Ex 5.1, 34 Important

Ex 5.2, 5 Important

Ex 5.2, 9 Important

Ex 5.2, 10 Important You are here

Ex 5.3, 10 Important

Ex 5.3, 14 Important

Example 32 Important

Example 33 Important

Ex 5.5,6 Important

Ex 5.5, 7 Important

Ex 5.5, 11 Important

Ex 5.5, 16 Important

Ex 5.6, 7 Important

Ex 5.6, 11 Important

Example 41 Important

Ex 5.7, 14 Important

Example 42 Important

Ex 5.8, 5 Important

Example 44 Important

Example 45 Important

Example 47 Important

Misc 6 Important

Misc 15 Important

Misc 16 Important

Misc 23 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.