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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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B
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C
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D
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Solution

## The correct option is C 0Here, f(x)+f(y)+f(x)⋅f(y)=1 .....(i)Substitute x=y=0, we get2f(0)+{f(0)}2=1⇒{f(0)}2+2f(0)−1=0f(0)=−2±√4+42=−1−√2 and −1+√2As f(0)>0⇒f(0)=√2−1 [neglecting f(0)=−1−√2 as f(0) is positive]Again, putting y=x is Eq, (i), 2f(x)+{f(x)}2=1On differentiating w.r.t. x, 2f′(x)+2f(x)f′(x)=02f′(x){1+f(x)}=0⇒f′(x)=0 because f(x)>0Thus, f′(x)=0 when f(x)>0

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