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Question: Answered & Verified by Expert
Let f:RR be a differentiable function with f0=1 and satisfying the equation

fx+y=fxf'y+f'xfy for all x, yR.

Then, the value of logef4 is 
MathematicsDifferential EquationsJEE Main
Solution:
2378 Upvotes Verified Answer
The correct answer is: 2
Px,y: fx+y=fxf'y+f'xfy  x, yR

P0,0: f0=f0f'0+f'0f0

1=2f'0

f'0=12

Px,0: fx=fx.f'0+f'x.f0

   fx=12fx+f'x

   f'x=12fx

   fx=e12x

   lnf4=2

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