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Question

Let f:R+R be a function which satifies :
f(x)f(y)=f(xy)+2(1x+1y+1) for x,y>0, then f(x) can be :

A
2x+1x
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B
(x+1x)
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C
5x+22x
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D
2(2x+1)x
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Solution

The correct option is A 2x+1x
Given function is f:R+R
Also given f(x)f(y)=f(xy)+2(1x+1y+1) for x,y>0
Put x=y=1 in the above function, we get
[f(1)]2=f(1)+6
[f(1)]2f(1)6=0
[f(1)+2)][f(1)3]=0
The root of the equation is f(1)=2 or f(1)=3
Since, the function domain is R+, so the f(1)=3 is accepted.
Now, put the y=1 in the given function, we get
f(x)f(1)=f(x)+2(1x+1+1)
3f(x)=f(x)+2(1x+2)
2f(x)=2(1x+2)
f(x)=2+1x=2x+1x

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