A function f:R→R+ satisfies f(x+y)=f(x)f(y)∀x,y∈R. If f(0)=1 and f′(0)=2, then the value of loge3∫0[f(x)e−x]dx is
(where [⋅] is represents the greatest integer function)
A
loge(32)
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B
loge(92)
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C
loge(94)
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D
loge3
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Solution
The correct option is Bloge(92) f(x+y)=f(x)f(y)∀x,y∈R From the above equation, we can conclude that f(x)=ax,(a≠0,1) where a is some constant. ⇒f′(x)=axlna ⇒2=lna(∵f′(0)=2) ⇒a=e2 ⇒f(x)=ax=(e2)x=e2x