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anil kumar Grade: 12

suppose a fuctin f satisfies the equation f(x+y)=f(x) f(y) for all x and y.f(x)=1+xg(x) where limit of g(x) as x tends to 0 is T,where T is apositive integer.if nth derivative of f(x)=k f(x) then what will be the value k,provide sol plz

7 years ago

Answers : (2)

Askiitians_Expert Yagyadutt
askIITians Faculty
74 Points

hello anil !


This question is very easy...just it rotates the condition...


See first you need to find that what is the f(x)  taking help of   f(x+y)=f(x).f(y)


Now it is clear the  f(x) = e^x    because   f(x+y) = e^(x+y) = e^x.e^y = f(x).f(y) get f(x) = e^x --------(1)


Now   f(x) = 1 + xg(x)


e^x = 1 +xg(x)  =>   e^x -1 = xg(x)  =>   g(x) = ( e^x -1)/x 


Lim x-->0  g(x)  = lim x-->0  (e^x-1)/x  =  1   = T

So T = 1


now if you derivate  f(x) for n will always ...get e^x

it...derivate of f(x) is always  f(x)


Hence k = 1


I hope it is clear....




7 years ago
vikas askiitian expert
510 Points

f(x+y) =f(x) . f(y)

 function which satisfies this relation is f(x) = ecx = cex          (c is a constant)

g(x) =(f(x)-1)/x

 lim x->0 g(x) = lim x->0 (f(x)-1)/x

                     =lim x->0 (cex -1)/x

     using L holpital rule

    at x = 0 numerator must be zero so value of c is 1...

                    =lim x->0 cex =T

                   c =1=T

 now f(x) =ex

       f1(x)=ex or kex where k=1 is constant....

so value of k is 1...

7 years ago
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