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Let `f: R^+vecR`be a function which satisfies `f(x)dotf(y)=f(x y)+2(1/x+1/y+1)`for `x , y > 0.`Then find `f(x)dot` |
Answer» We have `f(x)xxf(y)=f(xy)+2((1)/(x)+(1)/(y)+1) " …(1)" ` To get `f(x)` we put `y=1`, ` :. f(x)xxf(1)=f(x)+2((1)/(x)+2)` `=f(x)+2((2x+1)/(x))` `impliesf(x)(f(1)-1)=(2(2x+1))/(x)` `implies f(x)=(2(2x+1))/(x(f(1)-1)) " ...(2)" ` Now, we need the value of f(1). Put `x=1` and `y=1` in (1), we get `(f(1))^(2)-f(1)-6=0` `implies f(1)=3 " or "f(1)= -2` For `f(1)=3,f(x)=(2x+1)/(x)` and for `f(1)= -2, f(x)=(2(2x+1))/(-3x)` |
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