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Question Number 1125 by 123456 last updated on 18/Jun/15
f:R→R g:R→R f(x+y)=f(x)+g(y) g(xy)=f(x)g(y) f(x)=? g(x)=?
$${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{g}\left({y}\right) \\ $$$${g}\left({xy}\right)={f}\left({x}\right){g}\left({y}\right) \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$
Commented by 123456 last updated on 18/Jun/15
f(0)=f(0)+g(0)⇒g(0)=0 f(x)=f(x)+g(0)⇒g(0)=0 g(0)=f(0)g(0)⇒f(0)=1∨g(0)=0 g(0)=f(x)g(0)⇒f(x)=1∨g(0)=0 g(1)=f(1)g(1)⇒f(1)=1∨g(1)=0 g(x)=f(1)g(x)⇒f(1)=1∨g(x)=0 g(−1)=f(1)g(−1)⇒f(1)=1∨g(−1)=0
$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{0}\right)+{g}\left(\mathrm{0}\right)\Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)={f}\left({x}\right)+{g}\left(\mathrm{0}\right)\Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${g}\left(\mathrm{0}\right)={f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1}\vee{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${g}\left(\mathrm{0}\right)={f}\left({x}\right){g}\left(\mathrm{0}\right)\Rightarrow{f}\left({x}\right)=\mathrm{1}\vee{g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${g}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right){g}\left(\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1}\vee{g}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${g}\left({x}\right)={f}\left(\mathrm{1}\right){g}\left({x}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1}\vee{g}\left({x}\right)=\mathrm{0} \\ $$$${g}\left(−\mathrm{1}\right)={f}\left(\mathrm{1}\right){g}\left(−\mathrm{1}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1}\vee{g}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$
Answered by prakash jain last updated on 20/Jun/15
g(x)=f(x)g(1) g(1)=1/k f(x)=kg(x) g(xy)=kg(x)g(y) g(x)=x, k=1 f(x)=kg(x)=x
$${g}\left({x}\right)={f}\left({x}\right){g}\left(\mathrm{1}\right) \\ $$$${g}\left(\mathrm{1}\right)=\mathrm{1}/{k} \\ $$$${f}\left({x}\right)={kg}\left({x}\right) \\ $$$${g}\left({xy}\right)={kg}\left({x}\right){g}\left({y}\right) \\ $$$${g}\left({x}\right)={x},\:{k}=\mathrm{1} \\ $$$${f}\left({x}\right)={kg}\left({x}\right)={x} \\ $$
Commented by prakash jain last updated on 20/Jun/15
g(xy)=((g(x)g(y))/(g(1)))⇒g(x)=x^r , g(1)=1 f(x)=x^r f(x+y)=f(x)+g(y) (x+y)^r =x^r +y^r ⇒r=1 g(x)=f(x)=x
$${g}\left({xy}\right)=\frac{{g}\left({x}\right){g}\left({y}\right)}{{g}\left(\mathrm{1}\right)}\Rightarrow{g}\left({x}\right)={x}^{{r}} ,\:{g}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left({x}\right)={x}^{{r}} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{g}\left({y}\right) \\ $$$$\left({x}+{y}\right)^{{r}} ={x}^{{r}} +{y}^{{r}} \Rightarrow{r}=\mathrm{1} \\ $$$${g}\left({x}\right)={f}\left({x}\right)={x} \\ $$

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