39i do not know wat u have done sir , but i think i can contradict your answer:
from given condition when u put x=y then u get f`(1)=1
but in your function we do not get f`(1)=1
so your answer must be wrong
if f'(x/y).f(y/x) = \frac{x^{2}+y^{2}}{xy} for all x,y belongs to positive R and f(1)= 1 then f2(x) is?????????
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5 Answers
341Does it look like \frac{(x^4+4)e^{\left(\tan^{-1} \frac{x^2}{2} - \tan^{-1}\frac{1}{2}\right)}}{5}
39
1@ hassasin
f(1) = 1
f'(1)=2 .
putting x=1 in sir's solution we get 1.
341Ok, I will outline the steps:
Put y=1, and obtain f'(x) f \left(\frac{1}{x} \right) = \frac{x^2+1}{x}...........1
Put x = 1 and obtain f(x) f' \left(\frac{1}{x} \right) = \frac{x^2+1}{x}...................2
Now let g(x) = f(x) f \left(\frac{1}{x} \right)...................3
So that g'(x) = f'(x) f \left(\frac{1}{x} \right) - \frac{1}{x^2} f(x) f' \left(\frac{1}{x} \right) = x - \frac{1}{x^3}................4
Note by putting x =y that f'(1) =2
Now integrate between limits 1 and x to obtain
g(x) = f(x) f \left(\frac{1}{x} \right) = \frac{x^2}{2} + \frac{2}{x^2} - \frac{1}{2}...............5
From 1 and 5, we have
\frac{f'(x)}{f(x)} = \frac{2x(x^2+1)}{x^4-x^2+4}
From here again by integrating between limits 1 and x, you will get f(x) and hence f2(x)