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Question

Let F(x) = f(x) + f(1x), where f(x) = x1 log t1+tdt . Then F(e) =

A
12
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B
0
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C
1
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D
2
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Solution

The correct option is A 12
Given f(x)=x1 log t1+tdt. Put t=1zdt=1zzdz
f(1x)=1x1 log t1+tdt=x1 log(1z)1+(1z)(1z2)dz=x1 log zz(z+1) dz=x1 log tt(t+1)dt
F(x)=f(x)+f(1x)=x1[log t1+t+log tt(t+1)]dt
=x1 log ttdt=[(log t)2t]x1=(log x)22F(e)=12

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