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Question

For x>0, let f(x)=x1logt1+tdt. Then f(x)+f(1x) is equal to

A
14logx2
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B
14(logx)2
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C
logx
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D
12(logx)2
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Solution

The correct option is D 12(logx)2
f(x)=x1logt1+tdtf(1x)=1x1logt1+tdtPut t=1zdt=1z2dzf(1x)=x1log1z1+1z(1z2)dz=x1logzz(1+z)dz=x1logtt(1+t)dtf(x)+f(1x)=x1logttdt=(logx)22

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