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Question 1 Important

Last updated at April 16, 2024 by Teachoo

Misc 18 If π (π₯)=|π₯|^3, show that π β³(π₯) exists for all real π₯ and find it. We know that |π₯|={β( π₯ π₯β₯0@βπ₯ π₯<0)β€ Therefore, π (π₯)=|π₯|^3 = {β( (π₯)^3 , π₯β₯0@(βπ₯)^3 , π₯<0)β€ = {β( π₯^3 , π₯β₯0@γβπ₯γ^3 , π₯<0)β€ Case 1: When πβ₯π π (π₯)=π₯^3 Differentiating π€.π.π‘.π₯. πβ²(π₯)=γ3π₯γ^2 Again Differentiating π€.π.π‘.π₯. πβ²β²(π₯)= (3π₯^2 )^β² πβ²β²(π)=" " 6π₯ Hence, πβ²β²(π) exists for all value of π₯ greater than 0. Case 2: When π<π π (π₯)=γβπ₯γ^3 Differentiating π€.π.π‘.π₯. πβ²(π₯)=γβ3π₯γ^2 Again Differentiating π€.π.π‘.π₯. πβ²β²(π₯)= (γβ3π₯γ^2 )^β² π^β²β² (π)=" "β6π₯ Hence, πβ²β²(π) exists for all value of π₯ less than 0. Case 3: At x = 0 To check if πβ²β²(π) exists for x = 0, We need to check differentiability of πβ²(π) at π = π Here, π(π₯)= {β( π₯^3 , π₯β₯0@γβπ₯γ^3 , π₯<0)β€ πβ²(π)= {β( γ3π₯γ^2 , π₯β₯0@γβ3π₯γ^2 , π₯<0)β€ We know that πβ²(π₯) is differentiate at π₯ = 0 if L.H.D = R.H.D(π₯π’π¦)β¬(π βπ ) (π^β² (π) β π^β² (π β π))/π = limβ¬(β β0 ) (π^β² (0) β π^β² (ββ))/β = limβ¬(β β0 ) (γ3(0)γ^2 β(βγ3(ββ)γ^2))/β = limβ¬(β β0 ) γ3βγ^2/β = limβ¬(h β0 ) (3β) Putting β =0 = 3(0) = 0 (π₯π’π¦)β¬(π βπ ) (π^β² (π + π) βπ(π))/π = limβ¬(β β0 ) (γππγ^β² (β) β π(0))/(β ) = limβ¬(β β0 ) (γ3(β)γ^2 β γ3(0)γ^2)/β = limβ¬(β β0 ) γ3βγ^2/β = limβ¬(β β0 ) 3β Putting β =0 = 3(0) = 0 Thus, LHD = RHD Therefore, π^β² (π) is differentiable at π₯ = 0 So, we can say that π^β²β² (π) exists for x = 0 a Thus, π^β²β²(π) exists for all real values of π₯ Hence proved