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Question

0f(x+1x).lnxxdx

A
Is equal to zero
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B
Is equal to one
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C
Is equal to 12
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D
Can not be evaluated
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Solution

The correct option is D Is equal to zero

Let: lnx=tx=et
1xdx=dt

As "x" varies from 0 to "lnx [t]" varies to .
Now,
0f(x+1x).lnxxdx

f(et+et).tdt=F(t)

Now,
Using properties of definite integral:
Here we can see above function is an odd function i.e F(t)=F(t)
therefore on integrating from to sum of area of odd function is zero.
f(et+et).tdt=0

Thus,
0f(x+1x).lnxxdx=0
Hence, correct option is "A"


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