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Question

Let f(x) be defined xR and continuous.
Let f(x+y)f(xy)=4xyx,yR and f(0)=0, then f(x)<2x+3 will have solution :

A
(1,3)
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B
(15,1+5)
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C
(15,1)(3,1+5)
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D
(3,1+5)
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Solution

The correct option is A (1,3)
Put x=y in the given equation, we get
f(2x)f(0)=4x2f(2x)=4x2f(x)=x2
f(x)<2x+3x2<2x+3x22x3<0
(x3)(x+1)<0x(1,3)


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