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  1. Chapter 5 Class 12 Continuity and Differentiability
  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 every where , 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 continuity Case 1 At π‘₯<0 𝑓(π‘₯)=βˆ’2π‘₯+1 𝑓(π‘₯) is polynomial β‡’ 𝑓(π‘₯) is continuous Case 2 At π‘₯>1 𝑓(π‘₯)=2π‘₯βˆ’1 𝑓(π‘₯) is polynomial β‡’ 𝑓(π‘₯) is continuous [Every polynomial function is continuous] Case 3 At 0<π‘₯<1 𝑓(π‘₯)=1 𝑓(π‘₯) is constant," " β‡’ 𝑓(π‘₯) is continuous Case 4 At π‘₯=0 𝑓(π‘₯) = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ A function is continuous at π‘₯=0 if LHL = RHL = 𝑓(0) i.e. lim┬(π‘₯ β†’0^βˆ’ ) 𝑓 (π‘₯) = lim┬(π‘₯ β†’0^+ ) 𝑓 (π‘₯) = 𝑓(0) & 𝑓(π‘₯)= βˆ’2π‘₯+1 𝑓(0)= βˆ’2(0)+1= 1 Hence LHL = RHL = f (0) β‡’ 𝑓 is continuous . lim┬(π‘₯ β†’0^βˆ’ ) 𝑓 (π‘₯) =lim┬(β„Ž β†’0^βˆ’ ) 𝑓 (βˆ’2π‘₯+1) Putting π‘₯ = 0 = βˆ’2(0)+1 = 1 lim┬(π‘₯ β†’0^+ ) 𝑓 (π‘₯) =lim┬(π‘₯ β†’0^+ ) 1 = 1 Case 5 At π‘₯=1 A function is continuous at π‘₯=1 if LHL = RHL = 𝑓(1) i.e. lim┬(π‘₯ β†’1^βˆ’ ) 𝑓 (π‘₯) = lim┬(π‘₯ β†’1^+ ) 𝑓 (π‘₯) = 𝑓(1) lim┬(π‘₯ β†’1^βˆ’ ) 𝑓 (π‘₯) =lim┬(π‘₯ β†’1^βˆ’ )1 = 1 (𝐴𝑠 π‘₯<1) lim┬(π‘₯ β†’1^+ ) 𝑓 (π‘₯) =lim┬(π‘₯ β†’1^+ ) 2π‘₯βˆ’1 Putting π‘₯ =1 = 2(1)βˆ’1 = 2βˆ’1 = 1 (𝐴𝑠 π‘₯>1) & 𝑓 (π‘₯)=2π‘₯βˆ’1 𝑓 (1)=2(1)βˆ’1=2βˆ’1 =1 Hence LHL = RHL = 𝑓(1) β‡’ 𝑓 is continuous at π‘₯=1 Thus 𝑓(π‘₯)=|π‘₯|+|π‘₯βˆ’1| is continuous for all value of π‘₯ . To check differentiability Case 1 At π‘₯<1 𝑓(π‘₯)=βˆ’2π‘₯+1 𝑓(π‘₯) is polynomial β‡’ 𝑓(π‘₯) is differentiable Case 2 At π‘₯>1 𝑓(π‘₯)=2π‘₯βˆ’1 𝑓(π‘₯) is polynomial β‡’ 𝑓(π‘₯) is differentiable Case 3 At 0<π‘₯<1 𝑓(π‘₯)=1 𝑓(π‘₯) is constant β‡’ 𝑓(π‘₯) is 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)─ We know that 𝑓 is differentiate at π‘₯ = 0 If L.H.D = R.H.D = 𝑓^β€² (0) i.e., lim┬(β„Ž β†’0^βˆ’ ) (𝑓(0) βˆ’ 𝑓(0 βˆ’ β„Ž))/β„Ž = lim┬(β„Ž β†’0^+ ) (𝑓(0 + β„Ž) βˆ’ 𝑓(0))/β„Ž = 𝑓^β€² (𝑐) lim┬(β„Ž β†’0 ) (𝑓(0) βˆ’ 𝑓(0 βˆ’ β„Ž))/β„Ž =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 lim┬(β„Ž β†’0 ) (𝑓(0 + β„Ž) βˆ’ 𝑓(0))/β„Ž =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 Hence L.H.D β‰  R.H.D β‡’ 𝑓(π‘₯) is not differentiable at π‘₯=0 Case 5 At π‘₯=1 𝑓(π‘₯) is differentiable at π‘₯=1 if L.H.D = R.H.D = 𝑓′(1) lim┬(β„Ž β†’1^βˆ’ ) (𝑓(1) βˆ’ 𝑓(1 βˆ’ β„Ž))/β„Ž = lim┬(β„Ž β†’1^+ ) (𝑓(1 + β„Ž) βˆ’ 𝑓(1))/β„Ž = 𝑓′(1) 𝑓(π‘₯) = {β–ˆ( βˆ’2π‘₯+1 π‘₯≀0@ 1 0<π‘₯<1@2π‘₯βˆ’1 π‘₯β‰₯1)─ 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†’1^+ ) (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 = 2 Thus , 𝑓 is not differentiable at π‘₯=0 & π‘₯=1 , but continuous at all points Note :- Here we can take function |π‘₯|=|π‘₯βˆ’π‘Ž|+|π‘₯βˆ’π‘| where a & b can have any constant value . 𝑓 will be continuous at all points , but 𝑓 is not differentiable at π‘₯=π‘Ž & π‘₯=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 9 years. He provides courses for Maths and Science at Teachoo.