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Question

A function f:RR+ satisfies f(x+y)=f(x)f(y) x,yR. If f(0)=1 and f(0)=2, then the value of loge30[f(x)ex]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 B loge(92)
f(x+y)=f(x)f(y) x,yR
From the above equation, we can conclude that
f(x)=ax, (a0,1) where a is some constant.
f(x)=axlna
2=lna (f(0)=2)
a=e2
f(x)=ax=(e2)x=e2x

Now,
I=loge30[f(x)ex]dx =loge30[ex]dx =loge2loge11 dx+loge3loge22 dx =0+loge2+2(loge3loge2) =loge(92)

Alternate method :
f(x+y)=f(x)f(y)
Differentiating w.r.t x
f(x+y)(1+dydx)=f(x)f(y)+f(x)f(y)dydx
Put x=0 and y=x
f(0+x)(1+dxdx)=f(0)f(x)+f(0)f(x)dxdx
2f(x)=2f(x)+f(x)
f(x)=2f(x)
f(x)f(x)dx=2dx
loge|f(x)|=2x+c

Now, put x=0
loge|f(0)|=2(0)+c
loge1=c c=0
loge|f(x)|=2x
f(x)=e2x

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