6@theprophet
if u wrote a/x for x then why didnt u replace the same on LHS??
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 .
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6 Answers
1the second one ...i think f(x) has to be exponential in x
i.e x^(something)...trying to get the answer
341f(x) f'\left(\frac{a}{x} \right) = x. .......1
Write a/x for x. We have f'(x) f\left(\frac{a}{x} \right) = \frac{a}{x}.......2
If g(x) = f(x) f\left(\frac{a}{x} \right), then we see that g'(x) = f'(x) f\left(\frac{a}{x} \right) - \frac{a}{x^2} f(x)f'\left(\frac{a}{x} \right) = 0
so that f(x) f\left(\frac{a}{x} \right) = k ( a constant).....3
From 2 and 3, we get \frac{f'(x)}{f(x)} = \frac{c}{x} 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) = tx^s with t,s>0 is a solution.
The corresponding a is obtained as a =\sqrt [1-s] {t^2s}
341I need to add that when s=1, we have t=1 and a can be any +ve real number.
6
66@AKHIL
Probably u didn't notice that the locations of the prime (') changed.
