Question Number 950 by 123456 last updated on 06/May/15
![f:R→R,continuous such that ∀x∈R,f(x)f[f(x)]=1,f(2004)=2003,f(1999)=?](https://www.tinkutara.com/question/Q950.png)
$${f}:\mathbb{R}\rightarrow\mathbb{R},\mathrm{continuous}\:\mathrm{such}\:\mathrm{that}\:\forall{x}\in\mathbb{R},{f}\left({x}\right){f}\left[{f}\left({x}\right)\right]=\mathrm{1},{f}\left(\mathrm{2004}\right)=\mathrm{2003},{f}\left(\mathrm{1999}\right)=? \\ $$
Commented by 123456 last updated on 06/May/15
![f(2004)f[f(2004)]=2003f(2003)=1⇔f(2003)=(1/(2003)) f(2003)f[f(2003)]=(1/(2003))f((1/(2003)))=1⇔f((1/(2003)))=2003=f(2004) f((1/(2003)))f[f((1/(2003)))]=2003f(2003)=1](https://www.tinkutara.com/question/Q951.png)
$${f}\left(\mathrm{2004}\right){f}\left[{f}\left(\mathrm{2004}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1}\Leftrightarrow{f}\left(\mathrm{2003}\right)=\frac{\mathrm{1}}{\mathrm{2003}} \\ $$$${f}\left(\mathrm{2003}\right){f}\left[{f}\left(\mathrm{2003}\right)\right]=\frac{\mathrm{1}}{\mathrm{2003}}{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{1}\Leftrightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{2003}={f}\left(\mathrm{2004}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right){f}\left[{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1} \\ $$
Commented by 123456 last updated on 06/May/15
![f(x)f[f(x)]=1 f[f(x)]f{f[f(x)]}=1 f(x)=f{f[f(x)]}∨f[f(x)]=0 ∃x,f(x)=0⇒f(x)f[f(x)]=1⇒0f(0)=1 f(x)=f{f[f(x)]} f(2004)=f{f[f(2004)]}=2003 f(x)=y f(x)f[f(x)]=1 yf(y)=1 f(y)=(1/y)](https://www.tinkutara.com/question/Q952.png)
$${f}\left({x}\right){f}\left[{f}\left({x}\right)\right]=\mathrm{1} \\ $$$${f}\left[{f}\left({x}\right)\right]{f}\left\{{f}\left[{f}\left({x}\right)\right]\right\}=\mathrm{1} \\ $$$${f}\left({x}\right)={f}\left\{{f}\left[{f}\left({x}\right)\right]\right\}\vee{f}\left[{f}\left({x}\right)\right]=\mathrm{0} \\ $$$$\exists{x},{f}\left({x}\right)=\mathrm{0}\Rightarrow{f}\left({x}\right){f}\left[{f}\left({x}\right)\right]=\mathrm{1}\Rightarrow\mathrm{0}{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${f}\left({x}\right)={f}\left\{{f}\left[{f}\left({x}\right)\right]\right\} \\ $$$${f}\left(\mathrm{2004}\right)={f}\left\{{f}\left[{f}\left(\mathrm{2004}\right)\right]\right\}=\mathrm{2003} \\ $$$${f}\left({x}\right)={y} \\ $$$${f}\left({x}\right){f}\left[{f}\left({x}\right)\right]=\mathrm{1} \\ $$$${yf}\left({y}\right)=\mathrm{1} \\ $$$${f}\left({y}\right)=\frac{\mathrm{1}}{{y}} \\ $$
Answered by prakash jain last updated on 06/May/15
$${f}\left(\mathrm{2003}\right)=\frac{\mathrm{1}}{\mathrm{2003}}\:\mathrm{and}\:{f}\left(\mathrm{2004}\right)=\mathrm{2003} \\ $$$$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{continous}: \\ $$$$\exists{x}\:\mathrm{such}\:\mathrm{that}\:{f}\left({x}\right)=\mathrm{1999} \\ $$$$\mathrm{Since}\:{f}\left({f}\left({x}\right)\right)=\frac{\mathrm{1}}{{f}\left({x}\right)} \\ $$$${put}\:{f}\left({x}\right)=\mathrm{1999} \\ $$$${f}\left(\mathrm{1999}\right)=\frac{\mathrm{1}}{\mathrm{1999}} \\ $$