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u-R-R-v-R-R-u-x-v-x-x-2-v-x-u-x-v-x-x-3-v-x-h-x-u-x-v-x-




Question Number 560 by 123456 last updated on 26/Jan/15
u:R→R  v:R→R   { ((u(x)v(x)=x^2 −v(−x))),((u(−x)v(x)=x^3 +v(−x))) :}  h(x)=u(x)v(x)=?
$${u}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$${v}:\mathbb{R}\rightarrow\mathbb{R} \\ $$$$\begin{cases}{{u}\left({x}\right){v}\left({x}\right)={x}^{\mathrm{2}} −{v}\left(−{x}\right)}\\{{u}\left(−{x}\right){v}\left({x}\right)={x}^{\mathrm{3}} +{v}\left(−{x}\right)}\end{cases} \\ $$$${h}\left({x}\right)={u}\left({x}\right){v}\left({x}\right)=? \\ $$
Answered by prakash jain last updated on 26/Jan/15
u(x)v(x)=x^2 −v(−x)     ...(i)  u(−x)v(x)=x^3 +v(−x)  ...(ii)  add (i) and (ii)  v(x)[u(x)+u(−x)]=x^2 +x^3 ⇒v(x)=((x^2 +x^3 )/(u(x)+u(−x)))  u(−x)v(−x)=x^2 −v(x) replace x by −x in (i)  u(−x)((x^2 −x^3 )/(u(x)+u(−x)))=x^2 −((x^2 +x^3 )/(u(x)+u(−x)))  u(x)=a  u(−x)=b  ((b(x^2 −x^3 ))/(a+b))+((x^2 +x^3 )/(a+b))=x^2   bx^2 −bx^3 +x^2 +x^3 =ax^2 +bx^2   bx^3 =x^2 +x^3 −ax^2   b=((x^2 +x^3 −ax^2 )/x^3 )=((1+x−a)/x)  1+x−a=xb  1+x−u(x)=xu(−x)  xu(−x)+u(x)−x−1=0       ...(iii)  −xu(x)+u(−x)+x−1=0    ...(iv) replace x by −x in (iii)  multiply (iv) by x  −x^2 u(x)+xu(−x)+x^2 −x=0      ...(v)  add (iii) and (v)  u(x)+x^2 u(x)−x^2 −1=0  u(x)=1, u(−x)=1  v(x)=((x^2 +x^3 )/2), v(−x)=((x^2 −x^3 )/2)  Check condition 1  u(x)v(x)=((x^2 +x^3 )/2)=x^2 −((x^2 −x^3 )/2)=((x^2 +x^3 )/2)  Check condition 1  u(−x)v(x)=((x^2 +x^3 )/2)=x^3 +((x^2 −x^3 )/2)=((x^2 +x^3 )/2)  Results  u(x)=1  v(x)=((x^2 +x^3 )/2)  u(x)v(x)=((x^2 +x^3 )/2)
$${u}\left({x}\right){v}\left({x}\right)={x}^{\mathrm{2}} −{v}\left(−{x}\right)\:\:\:\:\:…\left(\mathrm{i}\right) \\ $$$${u}\left(−{x}\right){v}\left({x}\right)={x}^{\mathrm{3}} +{v}\left(−{x}\right)\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{add}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$${v}\left({x}\right)\left[{u}\left({x}\right)+{u}\left(−{x}\right)\right]={x}^{\mathrm{2}} +{x}^{\mathrm{3}} \Rightarrow{v}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{{u}\left({x}\right)+{u}\left(−{x}\right)} \\ $$$${u}\left(−{x}\right){v}\left(−{x}\right)={x}^{\mathrm{2}} −{v}\left({x}\right)\:\mathrm{replace}\:{x}\:\mathrm{by}\:−{x}\:\mathrm{in}\:\left(\mathrm{i}\right) \\ $$$${u}\left(−{x}\right)\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{{u}\left({x}\right)+{u}\left(−{x}\right)}={x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{{u}\left({x}\right)+{u}\left(−{x}\right)} \\ $$$${u}\left({x}\right)={a} \\ $$$${u}\left(−{x}\right)={b} \\ $$$$\frac{{b}\left({x}^{\mathrm{2}} −{x}^{\mathrm{3}} \right)}{{a}+{b}}+\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{{a}+{b}}={x}^{\mathrm{2}} \\ $$$${bx}^{\mathrm{2}} −{bx}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}^{\mathrm{3}} ={ax}^{\mathrm{2}} +{bx}^{\mathrm{2}} \\ $$$${bx}^{\mathrm{3}} ={x}^{\mathrm{2}} +{x}^{\mathrm{3}} −{ax}^{\mathrm{2}} \\ $$$${b}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} −{ax}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\frac{\mathrm{1}+{x}−{a}}{{x}} \\ $$$$\mathrm{1}+{x}−{a}={xb} \\ $$$$\mathrm{1}+{x}−{u}\left({x}\right)={xu}\left(−{x}\right) \\ $$$${xu}\left(−{x}\right)+{u}\left({x}\right)−{x}−\mathrm{1}=\mathrm{0}\:\:\:\:\:\:\:…\left(\mathrm{iii}\right) \\ $$$$−{xu}\left({x}\right)+{u}\left(−{x}\right)+{x}−\mathrm{1}=\mathrm{0}\:\:\:\:…\left(\mathrm{iv}\right)\:\mathrm{replace}\:{x}\:\mathrm{by}\:−{x}\:\mathrm{in}\:\left(\mathrm{iii}\right) \\ $$$$\mathrm{multiply}\:\left(\mathrm{iv}\right)\:\mathrm{by}\:{x} \\ $$$$−{x}^{\mathrm{2}} {u}\left({x}\right)+{xu}\left(−{x}\right)+{x}^{\mathrm{2}} −{x}=\mathrm{0}\:\:\:\:\:\:…\left(\mathrm{v}\right) \\ $$$$\mathrm{add}\:\left(\mathrm{iii}\right)\:\mathrm{and}\:\left(\mathrm{v}\right) \\ $$$${u}\left({x}\right)+{x}^{\mathrm{2}} {u}\left({x}\right)−{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${u}\left({x}\right)=\mathrm{1},\:{u}\left(−{x}\right)=\mathrm{1} \\ $$$${v}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}},\:{v}\left(−{x}\right)=\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{\mathrm{2}} \\ $$$${Check}\:{condition}\:\mathrm{1} \\ $$$${u}\left({x}\right){v}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}}={x}^{\mathrm{2}} −\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{\mathrm{2}}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}} \\ $$$${Check}\:{condition}\:\mathrm{1} \\ $$$${u}\left(−{x}\right){v}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}}={x}^{\mathrm{3}} +\frac{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{\mathrm{2}}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}} \\ $$$$\mathrm{Results} \\ $$$${u}\left({x}\right)=\mathrm{1} \\ $$$${v}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}} \\ $$$${u}\left({x}\right){v}\left({x}\right)=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{3}} }{\mathrm{2}} \\ $$

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