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Question Number 677 by prakash jain last updated on 22/Feb/15
$$\mathrm{Find}\:\mathrm{all}\:{f}:\:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left({x}^{\mathrm{2}} +{yf}\left({x}\right)\right)={xf}\left({x}+{y}\right) \\ $$
Commented by 123456 last updated on 22/Feb/15
![suposing f(x)=ax+b f[x^2 +yf(x)]=xf(x+y) f[x^2 +y(ax+b)]=x[a(x+y)+b] f(x^2 +axy+by)=x(ax+ay)+bx a(x^2 +axy+by)+b=ax^2 +axy+bx ax^2 +a^2 xy+aby+b=ax^2 +axy+by if b=0⇒ax^2 +a^2 xy=ax^2 +axy⇔axy=0∨a=1,f(x)=x if a=0⇒b=by⇔y=1∨b=0,f(x)=0](Q679.png)
$${suposing}\:{f}\left({x}\right)={ax}+{b} \\ $$$${f}\left[{x}^{\mathrm{2}} +{yf}\left({x}\right)\right]={xf}\left({x}+{y}\right) \\ $$$${f}\left[{x}^{\mathrm{2}} +{y}\left({ax}+{b}\right)\right]={x}\left[{a}\left({x}+{y}\right)+{b}\right] \\ $$$${f}\left({x}^{\mathrm{2}} +{axy}+{by}\right)={x}\left({ax}+{ay}\right)+{bx} \\ $$$${a}\left({x}^{\mathrm{2}} +{axy}+{by}\right)+{b}={ax}^{\mathrm{2}} +{axy}+{bx} \\ $$$${ax}^{\mathrm{2}} +{a}^{\mathrm{2}} {xy}+{aby}+{b}={ax}^{\mathrm{2}} +{axy}+{by} \\ $$$${if}\:{b}=\mathrm{0}\Rightarrow{ax}^{\mathrm{2}} +{a}^{\mathrm{2}} {xy}={ax}^{\mathrm{2}} +{axy}\Leftrightarrow{axy}=\mathrm{0}\vee{a}=\mathrm{1},{f}\left({x}\right)={x} \\ $$$${if}\:{a}=\mathrm{0}\Rightarrow{b}={by}\Leftrightarrow{y}=\mathrm{1}\vee{b}=\mathrm{0},{f}\left({x}\right)=\mathrm{0} \\ $$