Algebra - functions

341

Hari Shankar ·2010-01-17 01:48:55

Does it look like \frac{(x^4+4)e^{\left(\tan^{-1} \frac{x^2}{2} - \tan^{-1}\frac{1}{2}\right)}}{5}

1

dunno the ans....

39

i 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

1

@ hassasin

f(1) = 1

f^'(1)=2 .

putting x=1 in sir's solution we get 1.

341

Hari Shankar ·2010-01-17 07:02:21

Ok, 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)

functions