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Question Number 1821 by 123456 last updated on 05/Oct/15
f(uv)=f(u)f(v)−f(u+v)  f(0)=?  f(1)=?  f(x)=?
$${f}\left({uv}\right)={f}\left({u}\right){f}\left({v}\right)−{f}\left({u}+{v}\right) \\ $$$${f}\left(\mathrm{0}\right)=? \\ $$$${f}\left(\mathrm{1}\right)=? \\ $$$${f}\left({x}\right)=? \\ $$
Answered by Rasheed Soomro last updated on 10/Oct/15
f(uv)=f(u)f(v)−f(u+v)  Let u=0 and v=0  f(0.0)=f(0)f(0)−f(0+0)  f(0)=[f(0)]^2 −f(0)  [f(0)]^2 −2f(0)=0  f(0)[f(0)−2]=0  f(0)=0  Or f(0)=2...................(i)  −−−−−−−−−−−−−−−−−−−  Let u=1  and v=0  f(1.0)=f(1)f(0)−f(1+0)  f(0)=f(1)[f(0)−1]  f(1)=((f(0))/(f(0)−1))...........................(ii)  When f(0)=0,f(1)=(0/(0−1))=0  When f(0)=2,f(1)=(2/(2−1))=2  f(1)=0   Or f(1)=2...........(iii)  −−−−−−−−−−−−−−−−−−  Let u=x and v=0  f(x.0)=f(x)f(0)−f(x+0)  f(0)=f(x)[f(0)−1]  f(x)=((f(0))/(f(0)−1)).................................(iv)  When f(0)=0,f(x)=(0/(0−1))=0  When f(0)=2,f(x)=(2/(2−1))=2  f(x)=0  Or f(x)=2........................(v)  −−−−−−−−−−−−−−−−−−−−−  (i) , (iii) and  (v) are answers.  (v) is general definition of the function.
$${f}\left({uv}\right)={f}\left({u}\right){f}\left({v}\right)−{f}\left({u}+{v}\right) \\ $$$${Let}\:{u}=\mathrm{0}\:{and}\:{v}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}.\mathrm{0}\right)={f}\left(\mathrm{0}\right){f}\left(\mathrm{0}\right)−{f}\left(\mathrm{0}+\mathrm{0}\right) \\ $$$${f}\left(\mathrm{0}\right)=\left[{f}\left(\mathrm{0}\right)\right]^{\mathrm{2}} −{f}\left(\mathrm{0}\right) \\ $$$$\left[{f}\left(\mathrm{0}\right)\right]^{\mathrm{2}} −\mathrm{2}{f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)\left[{f}\left(\mathrm{0}\right)−\mathrm{2}\right]=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\:{Or}\:{f}\left(\mathrm{0}\right)=\mathrm{2}……………….\left({i}\right) \\ $$$$−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{u}=\mathrm{1}\:\:{and}\:{v}=\mathrm{0} \\ $$$${f}\left(\mathrm{1}.\mathrm{0}\right)={f}\left(\mathrm{1}\right){f}\left(\mathrm{0}\right)−{f}\left(\mathrm{1}+\mathrm{0}\right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left(\mathrm{1}\right)\left[{f}\left(\mathrm{0}\right)−\mathrm{1}\right] \\ $$$${f}\left(\mathrm{1}\right)=\frac{{f}\left(\mathrm{0}\right)}{{f}\left(\mathrm{0}\right)−\mathrm{1}}………………………\left({ii}\right) \\ $$$${When}\:{f}\left(\mathrm{0}\right)=\mathrm{0},{f}\left(\mathrm{1}\right)=\frac{\mathrm{0}}{\mathrm{0}−\mathrm{1}}=\mathrm{0} \\ $$$${When}\:{f}\left(\mathrm{0}\right)=\mathrm{2},{f}\left(\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{2}−\mathrm{1}}=\mathrm{2} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0}\:\:\:{Or}\:{f}\left(\mathrm{1}\right)=\mathrm{2}………..\left({iii}\right) \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:{u}={x}\:{and}\:{v}=\mathrm{0} \\ $$$${f}\left({x}.\mathrm{0}\right)={f}\left({x}\right){f}\left(\mathrm{0}\right)−{f}\left({x}+\mathrm{0}\right) \\ $$$${f}\left(\mathrm{0}\right)={f}\left({x}\right)\left[{f}\left(\mathrm{0}\right)−\mathrm{1}\right] \\ $$$${f}\left({x}\right)=\frac{{f}\left(\mathrm{0}\right)}{{f}\left(\mathrm{0}\right)−\mathrm{1}}……………………………\left({iv}\right) \\ $$$${When}\:{f}\left(\mathrm{0}\right)=\mathrm{0},{f}\left({x}\right)=\frac{\mathrm{0}}{\mathrm{0}−\mathrm{1}}=\mathrm{0} \\ $$$${When}\:{f}\left(\mathrm{0}\right)=\mathrm{2},{f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{2}−\mathrm{1}}=\mathrm{2} \\ $$$${f}\left({x}\right)=\mathrm{0}\:\:{Or}\:{f}\left({x}\right)=\mathrm{2}……………………\left({v}\right) \\ $$$$−−−−−−−−−−−−−−−−−−−−− \\ $$$$\left({i}\right)\:,\:\left({iii}\right)\:{and}\:\:\left({v}\right)\:{are}\:{answers}. \\ $$$$\left({v}\right)\:{is}\:{general}\:{definition}\:{of}\:{the}\:{function}. \\ $$

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