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Question

f(x)=f(x),f(0)=1, then dxf(x)+f(x)

A
log(e2x+1)+C
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B
log(ex+ex)+C
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C
tan1(ex)+C
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D
None
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Solution

The correct option is B tan1(ex)+C
Given that f(x)=f(x)
f(x)f(x)dx=dx
lnf(x)=x+C
when x=0, f(0)=1
Therefore, C=0
f(x)=ex
now, dxf(x)+f(x)=dxex+ex
=exdx1+e2x
=tan1ex+c
Ans: C

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