Find The Functions Quest :

JEE Advanced Papers › Calculus questions › 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 , âˆž ) â†’ ( 0 , âˆž ) 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 .

#Mathematics #Calculus

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6 Answers

39

Dr.House ·2010-10-12 14:08:52

1)

IMO 1983 longlist

u still want the solution ricky?

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bindaas ·2010-10-13 00:09:52

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

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341

Hari Shankar ·2010-10-13 09:19:28

f(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}

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341

Hari Shankar ·2010-10-14 01:47:09

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

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AKHIL ·2010-10-14 03:24:50

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

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66

kaymant ·2010-10-14 04:04:48

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

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