Miscellaneous

Chapter 5 Class 12 Continuity and Differentiability
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Misc 20 Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.Consider the function π(π₯)=|π₯|+|π₯β1| π is continuous everywhere , but it is not differentiable at π₯ = 0 & π₯ = 1 π(π₯)={β( βπ₯β(π₯β1) π₯β€0@π₯β(π₯β1) 0<π₯<1@π₯+(π₯β1) π₯β₯1)β€ = {β( β2π₯+1 π₯β€0@ 1 0<π₯<1@2π₯β1 π₯β₯1)β€ 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 π(π₯)=β2π₯+1 π(π₯) is polynomial β΄ π(π₯) is continuous & differentiable Case 2: For π₯>1 π(π₯)=2π₯β1 π(π₯) is polynomial β΄ π(π₯) is continuous & differentiable Case 3: For 0<π₯<1 π(π₯)=1 π(π₯) is a constant function β΄ π(π₯) is continuous & differentiable Case 4: At π₯=0 π(π₯) = {β( β2π₯+1 π₯β€0@ 1 0<π₯<1@2π₯β1 π₯β₯1)β€ Checking continuity A function is continuous at π₯=0 if LHL = RHL = π(0) i.e. (π₯π’π¦)β¬(π βπ^β ) π(π) = (π₯π’π¦)β¬(π βπ^+ ) π(π) = π(0) (π₯π’π¦)β¬(π βπ^β ) π(π)=limβ¬(β β0 ) π(0ββ) =limβ¬(β β0) π (ββ) = limβ¬(β β0) β2(ββ)+1 = 1 (π₯π’π¦)β¬(π βπ^+ ) π(π)=limβ¬(β β0 ) π(0+β) = limβ¬(β β0) π(β) = limβ¬(β β0) 1 = 1 And, π(0)= β2(0)+1= 1 Hence, LHL = RHL = f (0) β΄ π is continuous Checking Differentiability at x = 0 π is differentiable at π₯ = 0 if L.H.D = R.H.D i.e., limβ¬(β β0 ) (π(0) β π(0 β β))/β = limβ¬(β β0 ) (π(0 + β) β π(0))/β (π₯π’π¦)β¬(π βπ ) (π(π) β π(π β π))/π =limβ¬(β β0 ) (π(0) β π(ββ))/(β ) =limβ¬(β β0) ((β2(0) + 1) β (2(ββ)+1))/β =limβ¬(β β0 ) (1 + 2β β1)/β =limβ¬(β β0 ) 2β/β =limβ¬(β β0) 2 = 2 (π₯π’π¦)β¬(π βπ ) (π(π + π) β π(π))/π =limβ¬(ββ0) (π (β) β π(0))/β =limβ¬(ββ0 ) (1 β (β2(0) + 1))/β =limβ¬(ββ0 ) (1 β 1)/β =limβ¬(ββ0) 0/β = 0 Since L.H.D β  R.H.D β΄ π(π₯) is not differentiable at π₯=0 Case 5: At π₯=1 π(π₯) = {β( β2π₯+1 π₯β€0@ 1 0<π₯<1@2π₯β1 π₯β₯1)β€ Checking continuity A function is continuous at π₯=1 if LHL = RHL = π(1) i.e. (π₯π’π¦)β¬(π βπ^β ) π(π) = (π₯π’π¦)β¬(π βπ^+ ) π(π) = π(1) (π₯π’π¦)β¬(π βπ^β ) π(π)=limβ¬(β β0 ) π(1ββ) =limβ¬(β β0) 1 = 1 (π₯π’π¦)β¬(π βπ^+ ) π(π)=limβ¬(β β0 ) π(1+β) = limβ¬(β β0) 2(1+β)β1 = 2(1 + 0) β 1 = 1 And, π(0)= 2(1)β1= 1 Hence, LHL = RHL = f (1) β΄ π is continuous Checking Differentiability at x = 1 π is differentiable at π₯ =1 if L.H.D = R.H.D i.e., limβ¬(β β0 ) (π(1) β π(1 β β))/β = limβ¬(β β0 ) (π(1 + β) β π(1))/β limβ¬(β β0 ) (π(1) β π(1 β β) )/β =limβ¬(β β0 ) ((2(1) β 1) β 1)/(β ) =limβ¬(β β0 ) (1 β 1)/(β ) =limβ¬(β β0 ) 0/β = 0 limβ¬(β β0 ) (π(1 + β) β π(1))/β =limβ¬(β β0 ) ((2(1 + β) β 1) β (2(1) β1))/β = limβ¬(β β0 ) ((2 + 2β β1) β (2 β1))/β = limβ¬(β β0 ) ((1 + 2β) β 1)/β = limβ¬(β β0 ) 2β/β =limβ¬(β β0 ) 2 = 2Since L.H.D β  R.H.D β΄ π is not differentiable at π₯=1 Thus , π is not differentiable at π₯=0 & π₯=1 , but continuous at all points Note :- Here we can take function |π|=|πβπ|+|πβπ| where a & b can have any constant value . π will be continuous at all points , but π is not differentiable at π=π & π=π