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

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)

Hence...you 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 times...you will always ...get e^x

it...derivate of f(x) is always  f(x)

Hence k = 1

I hope it is clear....

Regards

Yagya

askiitians_expert

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

 function which satisfies this relation is f(x) = e^cx = ce^x          (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 (ce^x -1)/x

     using L holpital rule

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

                    =lim x->0 ce^x =T

                   c =1=T

 now f(x) =e^x

       f¹(x)=e^x or ke^x where k=1 is constant....

so value of k is 1...