Exercice 2



  • Determiner toutes les fonctions f:]0,+[]0,+[f:]0,+\infty[ \rightarrow ]0,+\infty [ telles que
    f(x)f(y)=f(y)f(xf(y))+1xyf(x)\cdot f(y)=f(y)f(xf(y))+\dfrac{1}{xy}
    pour tout x,y>0x,y \gt0



  • Since no one posted a solution , here's one :

    Let P(x,y)P(x,y) be the assertion f(x)f(y)=f(y)f(xf(y))+1xyf(x)f(y)=f(y)f(xf(y))+\frac 1{xy}

    P(1,x)P(1,x) :f(f(x))=f(1)1xf(x)f(f(x))=f(1)-\frac 1{xf(x)}

    P(f(x),y)P(f(x),y) :f(f(x))=f(f(x)f(y))+1f(x)f(y)yf(f(x))=f(f(x)f(y))+\frac 1{f(x)f(y)y}

    Subtracting, we get f(f(x)f(y))=f(1)1xf(x)1f(x)f(y)yf(f(x)f(y))=f(1)-\frac 1{xf(x)}-\frac 1{f(x)f(y)y}
    Swapping x,yx,y and subtracting, we get :

    1xf(x)+1f(x)f(y)y=1yf(y)+1f(x)f(y)x\frac 1{xf(x)}+\frac 1{f(x)f(y)y}=\frac 1{yf(y)}+\frac 1{f(x)f(y)x}

    Which is x(f(x)1)=y(f(y)1)x(f(x)-1)=y(f(y)-1) and so f(x)=1+axf(x)=1+\frac ax for some aa

    Plugging this back in original equation, we get a=1a=1


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