Let F(x) = f(x) + f(1x), where f(x) = ∫x1logt1+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 A12 Givenf(x)=∫x1logt1+tdt.Putt=1z⇒dt=−1zzdz f(1x)=∫1x1logt1+tdt=∫x1log(1z)1+(1z)(−1z2)dz=∫x1logzz(z+1)dz=∫x1logtt(t+1)dt F(x)=f(x)+f(1x)=∫x1[logt1+t+logtt(t+1)]dt =∫x1logttdt=[(logt)2t]x1=(logx)22⇒F(e)=12