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
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.
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.