Find The Functions Quest :

1 . Does there exist a function " f " defined on the set of positive real numbers which takes only positive real values , and satisfies ,

i > f ( x f ( y ) ) = y f ( x )

ii > f ( x ) tends to zero as " x " approaches infinity ?

2 . Find all differentiable functions f : ( 0 , ∞ ) → ( 0 , ∞ ) for which there exists a real number " a " > 0 such that ,

f ' ( ax ) = xf ( x )

Last One Is One Of My Sir ' s Assignment Problems , I Really Enjoyed Solving It .

6 Answers

39
Dr.House ·

1)

IMO 1983 longlist

u still want the solution ricky?

1
bindaas ·

the second one ...i think f(x) has to be exponential in x
i.e x^(something)...trying to get the answer

341
Hari Shankar ·

f(x)f(xa)=x. .......1

Write a/x for x. We have f(x)f(xa)=xa.......2

If g(x)=f(x)f(xa), then we see that g(x)=f(x)f(xa)x2af(x)f(xa)=0

so that f(x)f(xa)=k ( a constant).....3

From 2 and 3, we get f(x)f(x)=xc where c = ka

This is easily solved (by integrating w.r.t x on both sides) and plugging back, we see that any
function of the formf(x)=txs with t,s>0 is a solution.

The corresponding a is obtained as a=[1s]t2s

341
Hari Shankar ·

I need to add that when s=1, we have t=1 and a can be any +ve real number.

6
AKHIL ·

@theprophet
if u wrote a/x for x then why didnt u replace the same on LHS??

66
kaymant ·

@AKHIL
Probably u didn't notice that the locations of the prime (') changed.

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