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Question

Let f:RR be defined as f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪x5sin(1x)+5x2,x<00x=0x5cos(1x)+λx2,x>0 The value of λ for which f′′(0) exists, is

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Solution

f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪x5sin(1x)+5x2,x<00x=0x5cos(1x)+λx2,x>0

f(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪5x4sin(1x)x3cos(1x)+10x,x<00x=05x4cos(1x)+x3sin(1x)+2λx,x>0

f′′(x)=⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪20x3sin(1x)5x2cos(1x)3x2cos(1x)xsin(1x)+10,x<00x=020x3cos(1x)+5x2sin(1x)+3x2sin(1x)xcos(1x)+2λ,x>0
if f′′(0) exists then
f′′(0+)=f′′(0)
2λ=10λ=5

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