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Question

If for all x,y the function f is defined by; f(x)+f(y)+f(x)f(y)=1 and f(x)>0.When f(x) is differentiable f(x)=,

A
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C
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D
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Solution

The correct option is C 0
Here, f(x)+f(y)+f(x)f(y)=1 .....(i)
Substitute x=y=0, we get
2f(0)+{f(0)}2=1{f(0)}2+2f(0)1=0
f(0)=2±4+42=12 and 1+2
As f(0)>0f(0)=21 [neglecting f(0)=12 as f(0) is positive]
Again, putting y=x is Eq, (i), 2f(x)+{f(x)}2=1
On differentiating w.r.t. x, 2f(x)+2f(x)f(x)=0
2f(x){1+f(x)}=0f(x)=0 because f(x)>0
Thus, f(x)=0 when f(x)>0

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