Hey guys!

2f(x)-3f(1/x)=x^2, x not = 0, find f(2). options are

1. -7/4, 2.5/2, 3.-1, 4.none

Please solve it asap...i'm unable to solve it........thanks

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- Aug 13th 2008, 08:07 AMsapanjotSolve this function
Hey guys!

2f(x)-3f(1/x)=x^2, x not = 0, find f(2). options are

1. -7/4, 2.5/2, 3.-1, 4.none

Please solve it asap...i'm unable to solve it........thanks - Aug 13th 2008, 08:33 AMchoovuck
plug in x=2 and x=1/2, you obtain system of two equations, easy to solve. answer is f(2)=-7/4.

- Aug 13th 2008, 09:17 AMsapanjotthanks......n another prob...
thanks...tht helped a lot.......

there is another question:

f(x)=cos(log x) then find f(x)f(4)-1/2 ( f(x/4)+f(4x)). Options are:

1,-1,0,1 &-1 both - Aug 13th 2008, 02:43 PMPlato
To work this you will need to know these facts.

$\displaystyle f\left( {\frac{x}{4}} \right) = \cos \left( {\log (x) - \log (4)} \right)\,\& \,f\left( {4x} \right) = \cos \left( {\log (x) + \log (4)} \right)$

$\displaystyle \cos \left( {A - B} \right) + \cos \left( {A + B} \right) = 2\cos (A)\cos (B)$

You can put all that together, after all it is your problem. - Aug 14th 2008, 06:11 AMsapanjotthanks...
Thanks mr.plato.....i knew it's my prob.....u didn't need to tell me tht...i was expecting only what u send-"a hint".......

nyways.....there is another question.....again i would appreciate a hint....

f is a real valued function given by $\displaystyle f(x)=27x^3+(1/x^3)$, and a,b are roots of $\displaystyle 3x+(1/x)=12$. Then

1. f(a) not equal to f(b), 2.f(a)=10, 3.f(b)=-10, 4. none of these. - Aug 14th 2008, 07:45 AMPlato
Well O.K., here is just a hint:

$\displaystyle 27x^3 + \frac{1}{{x^3 }} = \left( {3x + \frac{1}{x}} \right)\left( {9x^2 - 3 + \frac{1}{{x^2 }}} \right)$