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JEE Advanced · Mathematics · 22. Functions

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 

  1. A 9
  2. B 7
  3. C 2
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 2

Step-by-step Solution

Detailed explanation

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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