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Question Number 187988 by cortano12 last updated on 24/Feb/23
 find function f(x) and g(x)   such that  { ((f(2x−1)+g(1−x)=x+1)),((f((x/(x+1)))+2g((1/(2x+2)))=3)) :}
$$\:\mathrm{find}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{1}}\\{\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)+\mathrm{2g}\left(\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}\right)=\mathrm{3}}\end{cases} \\ $$
Answered by MathGuy last updated on 24/Feb/23
Answer :− do x→x+1 in eq.1 & x→x−1 in eq.2  you will get  f(2x+1)+g(−x)=x+2 as new eq.1. call it eq.3  &  f(((x−1)/x))+2g((1/(2x)))=3 as new eq.2. call it eq.4  do x→−x in eq.3 & x→(1/(2x)) in eq.4  you will get  f(1−2x)+g(x)=2−x  ; call it eq.5  &  f(1−2x)+2g(x)=3  ; call it eq.6    doing [eq.6−eq.5] gives    g(x)=x+1    put value of g(x) in eq.5  you will get  f(1−2x)=1−2x  now putting x→((1−x)/2) , gives  f(x)=x    so we have f(x)=x & g(x)=x+1
$${Answer}\::−\:{do}\:{x}\rightarrow{x}+\mathrm{1}\:{in}\:{eq}.\mathrm{1}\:\&\:{x}\rightarrow{x}−\mathrm{1}\:{in}\:{eq}.\mathrm{2} \\ $$$${you}\:{will}\:{get} \\ $$$${f}\left(\mathrm{2}{x}+\mathrm{1}\right)+{g}\left(−{x}\right)={x}+\mathrm{2}\:{as}\:{new}\:{eq}.\mathrm{1}.\:{call}\:{it}\:{eq}.\mathrm{3} \\ $$$$\& \\ $$$${f}\left(\frac{{x}−\mathrm{1}}{{x}}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}}{\mathrm{2}{x}}\right)=\mathrm{3}\:{as}\:{new}\:{eq}.\mathrm{2}.\:{call}\:{it}\:{eq}.\mathrm{4} \\ $$$${do}\:{x}\rightarrow−{x}\:{in}\:{eq}.\mathrm{3}\:\&\:{x}\rightarrow\frac{\mathrm{1}}{\mathrm{2}{x}}\:{in}\:{eq}.\mathrm{4} \\ $$$${you}\:{will}\:{get} \\ $$$${f}\left(\mathrm{1}−\mathrm{2}{x}\right)+{g}\left({x}\right)=\mathrm{2}−{x}\:\:;\:{call}\:{it}\:{eq}.\mathrm{5} \\ $$$$\& \\ $$$${f}\left(\mathrm{1}−\mathrm{2}{x}\right)+\mathrm{2}{g}\left({x}\right)=\mathrm{3}\:\:;\:{call}\:{it}\:{eq}.\mathrm{6} \\ $$$$ \\ $$$${doing}\:\left[{eq}.\mathrm{6}−{eq}.\mathrm{5}\right]\:{gives} \\ $$$$ \\ $$$${g}\left({x}\right)={x}+\mathrm{1} \\ $$$$ \\ $$$${put}\:{value}\:{of}\:{g}\left({x}\right)\:{in}\:{eq}.\mathrm{5} \\ $$$${you}\:{will}\:{get} \\ $$$${f}\left(\mathrm{1}−\mathrm{2}{x}\right)=\mathrm{1}−\mathrm{2}{x} \\ $$$${now}\:{putting}\:{x}\rightarrow\frac{\mathrm{1}−{x}}{\mathrm{2}}\:,\:{gives} \\ $$$${f}\left({x}\right)={x} \\ $$$$ \\ $$$${so}\:{we}\:{have}\:{f}\left({x}\right)={x}\:\&\:{g}\left({x}\right)={x}+\mathrm{1} \\ $$
Answered by Rasheed.Sindhi last updated on 24/Feb/23
 such that  { ((f(2x−1)+g(1−x)=x+1...(i))),((f((x/(x+1)))+2g((1/(2x+2)))=3....(ii))) :}  (i):_(−)   2x−1=y⇒x=((y+1)/2)  (i)⇒f(y)+g(1−((y+1)/2))=((y+1)/2)+1             f(y)+g(((1−y)/2))=((y+3)/2)....(iii)  (ii):_(−)   (x/(x+1))=y  x−yx=y  x=(y/(1−y))  (ii)⇒f(y)+2g((1/(2((y/(1−y)))+2)))=3       f(y)+2g((1/((2y+2−2y)/(1−y))))=3       f(y)+2g(((1−y)/2))=3......(iv)  (iv)−(iii):g(((1−y)/2))=3−((y+3)/2)=((3−y)/2)  Let ((1−y)/2)=x            g(x)=((1−y+2)/2)=((1−y)/2)+1=x+1  2(iii)−(iv):  (vi):      f(y)+2g(((1−y)/2))=3  2(iii):  2f(y)+2g(((1−y)/2))=y+3    2(iii)−(iv):  f(y)=y+3−3                                f(y)=y                                f(x)=x
$$\:\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{1}…\left({i}\right)}\\{\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)+\mathrm{2g}\left(\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}\right)=\mathrm{3}….\left({ii}\right)}\end{cases} \\ $$$$\underset{−} {\left({i}\right):} \\ $$$$\mathrm{2}{x}−\mathrm{1}={y}\Rightarrow{x}=\frac{{y}+\mathrm{1}}{\mathrm{2}} \\ $$$$\left({i}\right)\Rightarrow{f}\left({y}\right)+{g}\left(\mathrm{1}−\frac{{y}+\mathrm{1}}{\mathrm{2}}\right)=\frac{{y}+\mathrm{1}}{\mathrm{2}}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{f}\left({y}\right)+{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\frac{{y}+\mathrm{3}}{\mathrm{2}}….\left({iii}\right) \\ $$$$\underset{−} {\left({ii}\right):} \\ $$$$\frac{{x}}{{x}+\mathrm{1}}={y} \\ $$$${x}−{yx}={y} \\ $$$${x}=\frac{{y}}{\mathrm{1}−{y}} \\ $$$$\left({ii}\right)\Rightarrow{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}}{\mathrm{2}\left(\frac{{y}}{\mathrm{1}−{y}}\right)+\mathrm{2}}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}}{\frac{\mathrm{2}{y}+\mathrm{2}−\mathrm{2}{y}}{\mathrm{1}−{y}}}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\mathrm{3}……\left({iv}\right) \\ $$$$\left({iv}\right)−\left({iii}\right):{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\mathrm{3}−\frac{{y}+\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{3}−{y}}{\mathrm{2}} \\ $$$${Let}\:\frac{\mathrm{1}−{y}}{\mathrm{2}}={x} \\ $$$$\:\:\:\:\:\:\:\:\:\:{g}\left({x}\right)=\frac{\mathrm{1}−{y}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{1}−{y}}{\mathrm{2}}+\mathrm{1}={x}+\mathrm{1} \\ $$$$\mathrm{2}\left({iii}\right)−\left({iv}\right): \\ $$$$\left({vi}\right):\:\:\:\:\:\:{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)=\mathrm{3} \\ $$$$\mathrm{2}\left({iii}\right):\:\:\mathrm{2}{f}\left({y}\right)+\mathrm{2}{g}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)={y}+\mathrm{3} \\ $$$$\:\:\mathrm{2}\left({iii}\right)−\left({iv}\right):\:\:{f}\left({y}\right)={y}+\mathrm{3}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({y}\right)={y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)={x} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Feb/23
 find function f(x) and g(x)   such that  { ((f(2x−1)+g(1−x)=x+1...(i))),((f((x/(x+1)))+2g((1/(2x+2)))=3.....(ii))) :}   (i):  1−x=y⇒ x=1−y         (i)⇒f( 2(1−y)−1 )+g(y)=1−y+1        f(1−2y)+g(y)=2−y.....(iii)  (ii): (1/(2x+2))=y⇒2xy+2y=1⇒x=((1−2y)/(2y))        (ii)⇒f((((1−2y)/(2y))/(((1−2y)/(2y))+1)))+2g(y)=3  (((1−2y)/(2y))/(((1−2y)/(2y))+1))=1−2y          f(1−2y)+2g(y)=3......(iv)  (iv)−(iii):   g(y)=3−(2−y)=y+1                          g(x)=x+1  2(iii)−(iv):      2(iii)⇒ 2f(1−2y)+2g(y)=4−2y        (iv)⇒    f(1−2y)+2g(y)=3                        f(1−2y)=4−2y−3                        f(1−2y)=1−2y                        f(x)=x
$$\:\mathrm{find}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\:\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{f}\left(\mathrm{2x}−\mathrm{1}\right)+\mathrm{g}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{x}+\mathrm{1}…\left({i}\right)}\\{\mathrm{f}\left(\frac{\mathrm{x}}{\mathrm{x}+\mathrm{1}}\right)+\mathrm{2g}\left(\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}\right)=\mathrm{3}…..\left({ii}\right)}\end{cases}\: \\ $$$$\left(\mathrm{i}\right):\:\:\mathrm{1}−\mathrm{x}=\mathrm{y}\Rightarrow\:\mathrm{x}=\mathrm{1}−\mathrm{y} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{i}\right)\Rightarrow\mathrm{f}\left(\:\mathrm{2}\left(\mathrm{1}−\mathrm{y}\right)−\mathrm{1}\:\right)+\mathrm{g}\left(\mathrm{y}\right)=\mathrm{1}−\mathrm{y}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{g}\left(\mathrm{y}\right)=\mathrm{2}−\mathrm{y}…..\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{ii}\right):\:\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{2}}=\mathrm{y}\Rightarrow\mathrm{2xy}+\mathrm{2y}=\mathrm{1}\Rightarrow\mathrm{x}=\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}} \\ $$$$\:\:\:\:\:\:\left(\mathrm{ii}\right)\Rightarrow\mathrm{f}\left(\frac{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}}{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}+\mathrm{1}}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{3} \\ $$$$\frac{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}}{\frac{\mathrm{1}−\mathrm{2y}}{\mathrm{2y}}+\mathrm{1}}=\mathrm{1}−\mathrm{2y} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{3}……\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iv}\right)−\left(\mathrm{iii}\right):\:\:\:\mathrm{g}\left(\mathrm{y}\right)=\mathrm{3}−\left(\mathrm{2}−\mathrm{y}\right)=\mathrm{y}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{2}\left(\mathrm{iii}\right)−\left(\mathrm{iv}\right): \\ $$$$\:\:\:\:\mathrm{2}\left(\mathrm{iii}\right)\Rightarrow\:\mathrm{2f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{4}−\mathrm{2y} \\ $$$$\:\:\:\:\:\:\left(\mathrm{iv}\right)\Rightarrow\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)+\mathrm{2g}\left(\mathrm{y}\right)=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)=\mathrm{4}−\mathrm{2y}−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{1}−\mathrm{2y}\right)=\mathrm{1}−\mathrm{2y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x} \\ $$

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