Misc 21 - Does there exist a function which is continuous but not

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Misc  21 - Chapter 5 Class 12 Continuity and Differentiability - Part 2

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  1. Chapter 5 Class 12 Continuity and Differentiability (Term 1)
  2. Serial order wise

Transcript

Misc 21 Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.Consider the function 𝑓(π‘₯)=|π‘₯|+|π‘₯βˆ’1| 𝑓 is continuous everywhere , but it is not differentiable at π‘₯ = 0 & π‘₯ = 1 𝑓(π‘₯)={β–ˆ( βˆ’π‘₯βˆ’(π‘₯βˆ’1) π‘₯≀0@π‘₯βˆ’(π‘₯βˆ’1) 0<π‘₯<1@π‘₯+(π‘₯βˆ’1) π‘₯β‰₯1)─ = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ Checking both continuity and differentiability at For x < 0 For x > 1 For 0 < x < 1 For x = 0 For x = 1 Case 1: For π‘₯<0 𝑓(π‘₯)=βˆ’2π‘₯+1 𝑓(π‘₯) is polynomial ∴ 𝑓(π‘₯) is continuous & differentiable Case 2: For π‘₯>1 𝑓(π‘₯)=2π‘₯βˆ’1 𝑓(π‘₯) is polynomial ∴ 𝑓(π‘₯) is continuous & differentiable Case 3: For 0<π‘₯<1 𝑓(π‘₯)=1 𝑓(π‘₯) is a constant function ∴ 𝑓(π‘₯) is continuous & differentiable Case 4: At π‘₯=0 𝑓(π‘₯) = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ Checking continuity A function is continuous at π‘₯=0 if LHL = RHL = 𝑓(0) i.e. (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^βˆ’ ) 𝒇(𝒙) = (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^+ ) 𝒇(𝒙) = 𝑓(0) (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^βˆ’ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(0βˆ’β„Ž) =lim┬(β„Ž β†’0) 𝑓 (βˆ’β„Ž) = lim┬(β„Ž β†’0) βˆ’2(βˆ’β„Ž)+1 = 1 (π₯𝐒𝐦)┬(𝒙 β†’πŸŽ^+ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(0+β„Ž) = lim┬(β„Ž β†’0) 𝑓(β„Ž) = lim┬(β„Ž β†’0) 1 = 1 And, 𝑓(0)= βˆ’2(0)+1= 1 Hence, LHL = RHL = f (0) ∴ 𝑓 is continuous Checking Differentiability at x = 0 𝑓 is differentiable at π‘₯ = 0 if L.H.D = R.H.D i.e., lim┬(β„Ž β†’0 ) (𝑓(0) βˆ’ 𝑓(0 βˆ’ β„Ž))/β„Ž = lim┬(β„Ž β†’0 ) (𝑓(0 + β„Ž) βˆ’ 𝑓(0))/β„Ž (π₯𝐒𝐦)┬(𝒉 β†’πŸŽ ) (𝒇(𝟎) βˆ’ 𝒇(𝟎 βˆ’ 𝒉))/𝒉 =lim┬(β„Ž β†’0 ) (𝑓(0) βˆ’ 𝑓(βˆ’β„Ž))/(β„Ž ) =lim┬(β„Ž β†’0) ((βˆ’2(0) + 1) βˆ’ (2(βˆ’β„Ž)+1))/β„Ž =lim┬(β„Ž β†’0 ) (1 + 2β„Ž βˆ’1)/β„Ž =lim┬(β„Ž β†’0 ) 2β„Ž/β„Ž =lim┬(β„Ž β†’0) 2 = 2 (π₯𝐒𝐦)┬(𝒉 β†’πŸŽ ) (𝒇(𝟎 + 𝒉) βˆ’ 𝒇(𝟎))/𝒉 =lim┬(β„Žβ†’0) (𝑓 (β„Ž) βˆ’ 𝑓(0))/β„Ž =lim┬(β„Žβ†’0 ) (1 βˆ’ (βˆ’2(0) + 1))/β„Ž =lim┬(β„Žβ†’0 ) (1 βˆ’ 1)/β„Ž =lim┬(β„Žβ†’0) 0/β„Ž = 0 Since L.H.D β‰  R.H.D ∴ 𝑓(π‘₯) is not differentiable at π‘₯=0 Case 5: At π‘₯=1 𝑓(π‘₯) = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ Checking continuity A function is continuous at π‘₯=1 if LHL = RHL = 𝑓(1) i.e. (π₯𝐒𝐦)┬(𝒙 β†’πŸ^βˆ’ ) 𝒇(𝒙) = (π₯𝐒𝐦)┬(𝒙 β†’πŸ^+ ) 𝒇(𝒙) = 𝑓(1) (π₯𝐒𝐦)┬(𝒙 β†’πŸ^βˆ’ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(1βˆ’β„Ž) =lim┬(β„Ž β†’0) 1 = 1 (π₯𝐒𝐦)┬(𝒙 β†’πŸ^+ ) 𝒇(𝒙)=lim┬(β„Ž β†’0 ) 𝑓(1+β„Ž) = lim┬(β„Ž β†’0) 2(1+β„Ž)βˆ’1 = 2(1 + 0) βˆ’ 1 = 1 And, 𝑓(0)= 2(1)βˆ’1= 1 Hence, LHL = RHL = f (1) ∴ 𝑓 is continuous Checking Differentiability at x = 1 𝑓 is differentiable at π‘₯ =1 if L.H.D = R.H.D i.e., lim┬(β„Ž β†’0 ) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž))/β„Ž = lim┬(β„Ž β†’0 ) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž lim┬(β„Ž β†’0 ) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž) )/β„Ž =lim┬(β„Ž β†’0 ) ((2(1) βˆ’ 1) βˆ’ 1)/(β„Ž ) =lim┬(β„Ž β†’0 ) (1 βˆ’ 1)/(β„Ž ) =lim┬(β„Ž β†’0 ) 0/β„Ž = 0 lim┬(β„Ž β†’0 ) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž =lim┬(β„Ž β†’0 ) ((2(1 + β„Ž) βˆ’ 1) βˆ’ (2(1) βˆ’1))/β„Ž = lim┬(β„Ž β†’0 ) ((2 + 2β„Ž βˆ’1) βˆ’ (2 βˆ’1))/β„Ž = lim┬(β„Ž β†’0 ) ((1 + 2β„Ž) βˆ’ 1)/β„Ž = lim┬(β„Ž β†’0 ) 2β„Ž/β„Ž =lim┬(β„Ž β†’0 ) 2 = 2Since L.H.D β‰  R.H.D ∴ 𝑓 is not differentiable at π‘₯=1 Thus , 𝑓 is not differentiable at π‘₯=0 & π‘₯=1 , but continuous at all points Note :- Here we can take function |𝒙|=|π’™βˆ’π’‚|+|π’™βˆ’π’ƒ| where a & b can have any constant value . 𝑓 will be continuous at all points , but 𝑓 is not differentiable at 𝒙=𝒂 & 𝒙=𝒃

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.