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Question Number 117066 by bemath last updated on 09/Oct/20

Given f(x)= ((x−1)/(x+1)) . If f^2 (x)=f(f(x)), f^3 (x)=f(f(f(x))) , f^(1998) (x) = g(x) then ∫_(1/e) ^1 g(x) dx = _?

Givenf(x)=x−1x+1.Iff2(x)=f(f(x)),f3(x)=f(f(f(x))),f1998(x)=g(x)then∫1e1g(x)dx=_?

Answered by 1549442205PVT last updated on 09/Oct/20

f^2 (x)=((((x−1)/(x+1))−1)/(((x−1)/(x+1))+1))=((−2)/(2x))=−(1/x) f^3 (x)=−(1/((x−1)/(x+1)))=−((x+1)/(x−1)) f^4 (x)=−((((x−1)/(x+1))+1)/(((x−1)/(x+1))−1))=−((2x)/(−2))=x f^5 (x)=((x−1)/(x+1))=f(x) From that we get the sequence f (x)=f^5 (x)=f^9 (x)=...f^(4k+1) (x) Since 1998=4.499+2,we infer g(x)=f^(1998(x)) =f^2 (x)=−(1/x).Therefore, ∫_(1/e) ^( 1) g(x)dx=∫_(1/e) ^( 1) (−(1/x))dx=−ln∣x∣_(1/e) ^1 =ln((1/e))=ln1−lne=0−1=−1

f2(x)=x−1x+1−1x−1x+1+1=−22x=−1xf3(x)=−1x−1x+1=−x+1x−1f4(x)=−x−1x+1+1x−1x+1−1=−2x−2=xf5(x)=x−1x+1=f(x)Fromthatwegetthesequencef(x)=f5(x)=f9(x)=...f4k+1(x)Since1998=4.499+2,weinferg(x)=f1998(x)=f2(x)=−1x.Therefore,∫1e1g(x)dx=∫1e1(−1x)dx=−ln∣x∣1/e1=ln(1e)=ln1−lne=0−1=−1

Commented by bemath last updated on 09/Oct/20

yes...santuyy

yes...santuyy

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