Question Number 1088 by 112358 last updated on 10/Jun/15 $${Solve}\:{the}\:{following}\:{integral} \\ $$$${equation}\:{for}\:{f}\left({x}\right): \\ $$$$\int_{\mathrm{0}} ^{\:{x}} {f}\left({t}\right){dt}=\mathrm{3}{f}\left({x}\right)+{k} \\ $$$${where}\:{k}\:{is}\:{a}\:{constant}.\: \\ $$ Answered by prakash jain last…
Question Number 132135 by benjo_mathlover last updated on 11/Feb/21 Answered by Olaf last updated on 11/Feb/21 $${f}\left({x}\right)+{f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)\:=\:\mathrm{1}+{x}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{x}\:=\:\frac{{u}−\mathrm{1}}{{u}} \\ $$$$\left(\mathrm{1}\right)\::\:{f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\frac{{u}−\mathrm{1}}{{u}}−\mathrm{1}}{\frac{{u}−\mathrm{1}}{{u}}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}} \\ $$$${f}\left(\frac{{u}−\mathrm{1}}{{u}}\right)+{f}\left(\frac{\mathrm{1}}{\mathrm{1}−{u}}\right)\:=\:\mathrm{1}+\frac{{u}−\mathrm{1}}{{u}}\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\::…
Question Number 950 by 123456 last updated on 06/May/15 $${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}\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}} \\ $$$${f}\left(\mathrm{2003}\right){f}\left[{f}\left(\mathrm{2003}\right)\right]=\frac{\mathrm{1}}{\mathrm{2003}}{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{1}\Leftrightarrow{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)=\mathrm{2003}={f}\left(\mathrm{2004}\right) \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right){f}\left[{f}\left(\frac{\mathrm{1}}{\mathrm{2003}}\right)\right]=\mathrm{2003}{f}\left(\mathrm{2003}\right)=\mathrm{1} \\…
Question Number 926 by 123456 last updated on 26/Apr/15 $${a}\left({n},{m}\right)=\begin{cases}{{m}}&{{n}\leqslant\mathrm{0}}\\{{na}\left({n}+{m}−\mathrm{1},{m}\right)+{m}}&{{n}>\mathrm{0}\vee{m}\leqslant\mathrm{0}}\\{{na}\left({n}−{m},{m}−\mathrm{1}\right)+{m}+{m}^{\mathrm{2}} }&{{n}>\mathrm{0}\vee{m}>\mathrm{0}}\end{cases} \\ $$$${a}\left(\mathrm{5},\mathrm{5}\right)=??? \\ $$$${a}\left(\mathrm{10},\mathrm{10}\right)=?? \\ $$ Commented by 123456 last updated on 27/Apr/15 $${a}\left(\mathrm{0},{m}\right)={m}…
Question Number 66462 by mathmax by abdo last updated on 15/Aug/19 $$\left.\mathrm{1}\right){simplify}\:{S}_{{n}} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{{k}} \left({x}\right){cos}\left({kx}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:{A}_{{n}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{cos}^{{k}}…
Question Number 874 by 123456 last updated on 08/Apr/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({x}\right)={g}\left({x}\right)−{f}\left(−{x}\right) \\ $$$${g}\left({x}\right)={f}\left({x}\right)+{g}\left(−{x}\right) \\ $$$${f}\left({x}\right)=? \\ $$$${g}\left({x}\right)=? \\ $$$${f}\left({x}\right)+{g}\left({x}\right)=? \\ $$$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\overset{?} {=}\boldsymbol{{f}}\left(−\boldsymbol{{x}}\right)…
Question Number 863 by 123456 last updated on 30/Mar/15 $${f}\left({xy}\right)={f}\left[{xf}\left({y}\right)\right] \\ $$$${f}\left({x}\right)=? \\ $$ Answered by prakash jain last updated on 30/Mar/15 $${f}\left({y}\right)={y} \\ $$…
Question Number 830 by prakash jain last updated on 20/Mar/15 $${f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{such}\:\mathrm{that} \\ $$$${f}\left({x}^{\mathrm{2}} \right)={f}\left({x}\right){f}\left({x}−\mathrm{1}\right) \\ $$$$\mathrm{Find}\:{f}\left({x}\right)\:\mathrm{such}\:\mathrm{that}\:{f}\left({x}\right)\neq\mathrm{0}. \\ $$ Commented by 123456 last updated on 23/Mar/15…
Question Number 821 by 123456 last updated on 17/Mar/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({xg}\right)={f}\left({x}\right){g}\left({y}\right) \\ $$$${g}\left({xy}\right)={f}\left({x}\right)+{g}\left({y}\right) \\ $$$$\frac{{d}\left({fg}\right)}{{dx}}=? \\ $$ Answered by prakash jain last…
Question Number 818 by 123456 last updated on 17/Mar/15 $${f}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${g}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${f}\left({xy}\right)={f}\left({x}\right){g}\left({x}\right)+{f}\left({y}\right){g}\left({y}\right) \\ $$$${g}\left({xy}\right)={f}\left({x}\right){g}\left({y}\right)+{f}\left({y}\right){g}\left({x}\right) \\ $$$$\frac{{d}\left({fg}\right)}{{dx}}=? \\ $$ Commented by prakash jain last…