Let `f: R->R`
be a
differentiable function with `f(0)=1`
and
satisfying the equation `f(x+y)=f(x)f^(prime)(y)+f^(prime)(x)f(y)`
for all `x ,\ y in R`
. Then, the
value of `(log)_e(f(4))`
is _______
Text Solution
Verified by Experts
`f(x+y)=f(x)cdotf'(y)+f'(x)cdotf(y)" ...(i)"` `"Putting "y=0` `f(x)=f(x)f'(0)+f'(x)cdotf(0)=f(x)f'(0)+f'(x)" ...(2)"` To find f'(0), in (1), put x=y=0. `therefore" "f(0)=2f(0)cdotf'(0)` `thereforef" "'(0)=(1)/(2)` So, from (2). we get `f'(x)=(f(x))/(2)` `rArr" "int(f'(x))/(f(x))dx=int (1)/(2)dx` `rArr" "log_(e)f(x)=(x//2)+c` `rArr" "log_(e)f(x)=x//2" (as f(0) = 1)"` `rArr" "log_(e)(f(4))=2`
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