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

Last updated at April 16, 2024 by Teachoo

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]
Letβs check for both x = 1 and x = 2
At x = 1
f (x) is differentiable at x = 1 if
LHD = RHD
(πππ)β¬(π‘βπ) (π(π) β π(π β π))/π
= (πππ)β¬(hβ0) (π(1) β π(1 β β))/β
= (πππ)β¬(hβ0) ([1] β [(1 β β)])/β
= (πππ)β¬(hβ0) (1 β 0)/β
= (πππ)β¬(hβ0) 1/β
= 1/0
= Not defined
(πππ)β¬(π‘βπ) (π(π + π) β π(π))/π
= (πππ)β¬(hβ0) (π(1 + β) β π(1))/β
= (πππ)β¬(hβ0) ([(1 + β)] β [1])/β
= (πππ)β¬(hβ0) (1 β 1)/β
= (πππ)β¬(hβ0) 0/β
= (πππ)β¬(hβ0) 0
= 0
For greatest integer function
[1] = 1
[1 β h] = 0
[1 + h] = 1
Since LHD β RHD
β΄ 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
(πππ)β¬(π‘βπ) (π(π) β π(π β π))/π
= (πππ)β¬(hβ0) (π(2) β π(2 β β))/β
= (πππ)β¬(hβ0) ([2] β [(2 β β)])/β
= (πππ)β¬(hβ0) (2 β 1)/β
= (πππ)β¬(hβ0) 1/β
= 1/0
= Not defined
R H D
(πππ)β¬(π‘βπ) (π(π + π) β π(π))/π
= (πππ)β¬(hβ0) (π(2 + β) β π(2))/β
= (πππ)β¬(hβ0) ([(2 + β)] β [2])/β
= (πππ)β¬(hβ0) (2 β 2)/β
= (πππ)β¬(hβ0) 0/β
= (πππ)β¬(hβ0) 0
= 0
For greatest integer function
[2] = 2
[2 β h] = 1
[2 + h] = 2
Since LHD β RHD
β΄ f(x) is not differentiable at x = 2
Hence proved

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.

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