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Question

If v(x) is larger of ex1 and (1+x)log(1+x) for xϵ(0,) then log(v(8)+1) is equal to

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Solution

atx=0bothfuctionsareequalandderivateofex1=d(ex1)dx=ex>0derivateof(1+x)log(1+x)=d((1+x)log(1+x))dx
=(1+x)d(log(1+x))dx+log(1+x)d(1+x)dx=(1+x).1(1+x)+log(1+x)=1+log(1+x)>0andforx>0exisgreaterthan1+log(1+x).thatmeansthatex1>(1+x)log(1+x)soV(x)=ex1V(8)=e81V(8)+1=e8log(V(8)+1)=8.

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