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Question Number 4844 by 123456 last updated on 17/Mar/16
y(x)=f(x)+g(x)+h(x) y(x)=x^2 +sin x+x(1−x) f(x) is even g(x) is odd f(0)g(0)−h(0)=?
$${y}\left({x}\right)={f}\left({x}\right)+{g}\left({x}\right)+{h}\left({x}\right) \\ $$$${y}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{sin}\:{x}+{x}\left(\mathrm{1}−{x}\right) \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{even} \\ $$$${g}\left({x}\right)\:\mathrm{is}\:\mathrm{odd} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)−{h}\left(\mathrm{0}\right)=? \\ $$
Commented by Rasheed Soomro last updated on 17/Mar/16
y(x)=f(x)+g(x)+h(x) y(x)=x^2 +sin x+x(1−x)⇒y(0)=0 f(x) is even & g(x) is odd f(0)g(0)−h(0)=? .................. y(−x)=f(−x)+g(−x)+h(−x) =f(x)−g(x)+h(−x) y(x)+y(−x)=2f(x)+h(x)+h(−x) y(x)−y(−x)=2g(x)+h(x)−h(−x) y(0)+y(0)=2f(0)+2h(0)⇒y(0)=f(0)+h(0) ⇒f(0)+h(0)=0⇒h(0)=−f(0)......A 0=2g(0)⇒g(0)=0...........................B f(0)g(0)−h(0)=f(0)g(0)+f(0) =f(0)[ g(0)+1 ] =f(0)[0+1]=f(0) Continue
$${y}\left({x}\right)={f}\left({x}\right)+{g}\left({x}\right)+{h}\left({x}\right) \\ $$$${y}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{sin}\:{x}+{x}\left(\mathrm{1}−{x}\right)\Rightarrow{y}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)\:\mathrm{is}\:\mathrm{even}\:\:\&\:{g}\left({x}\right)\:\mathrm{is}\:\mathrm{odd} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)−{h}\left(\mathrm{0}\right)=? \\ $$$$……………… \\ $$$${y}\left(−{x}\right)={f}\left(−{x}\right)+{g}\left(−{x}\right)+{h}\left(−{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left({x}\right)−{g}\left({x}\right)+{h}\left(−{x}\right) \\ $$$${y}\left({x}\right)+{y}\left(−{x}\right)=\mathrm{2}{f}\left({x}\right)+{h}\left({x}\right)+{h}\left(−{x}\right) \\ $$$${y}\left({x}\right)−{y}\left(−{x}\right)=\mathrm{2}{g}\left({x}\right)+{h}\left({x}\right)−{h}\left(−{x}\right) \\ $$$${y}\left(\mathrm{0}\right)+{y}\left(\mathrm{0}\right)=\mathrm{2}{f}\left(\mathrm{0}\right)+\mathrm{2}{h}\left(\mathrm{0}\right)\Rightarrow{y}\left(\mathrm{0}\right)={f}\left(\mathrm{0}\right)+{h}\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\Rightarrow{f}\left(\mathrm{0}\right)+{h}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{h}\left(\mathrm{0}\right)=−{f}\left(\mathrm{0}\right)……{A} \\ $$$$\mathrm{0}=\mathrm{2}{g}\left(\mathrm{0}\right)\Rightarrow{g}\left(\mathrm{0}\right)=\mathrm{0}………………………{B} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)−{h}\left(\mathrm{0}\right)={f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)+{f}\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left(\mathrm{0}\right)\left[\:{g}\left(\mathrm{0}\right)+\mathrm{1}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={f}\left(\mathrm{0}\right)\left[\mathrm{0}+\mathrm{1}\right]={f}\left(\mathrm{0}\right) \\ $$$${Continue} \\ $$
Answered by prakash jain last updated on 18/Mar/16
y(x)=x+sin x assume f(x)=5+x^2 (even) g(x)=x (odd) h(x)=sin x−5−x^2 y(x)=f(x)+g(x)+h(x)=sin x+x f(0)=5 g(0)=0 h(0)=−5 f(0)g(0)−h(0)=5 There is no unique answer to given question. 5 was taken as example and can be replaced by any value.
$${y}\left({x}\right)={x}+\mathrm{sin}\:{x} \\ $$$${assume} \\ $$$${f}\left({x}\right)=\mathrm{5}+{x}^{\mathrm{2}} \:\:\left({even}\right) \\ $$$${g}\left({x}\right)={x}\:\:\left({odd}\right) \\ $$$${h}\left({x}\right)=\mathrm{sin}\:{x}−\mathrm{5}−{x}^{\mathrm{2}} \\ $$$${y}\left({x}\right)={f}\left({x}\right)+{g}\left({x}\right)+{h}\left({x}\right)=\mathrm{sin}\:{x}+{x} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{5} \\ $$$${g}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${h}\left(\mathrm{0}\right)=−\mathrm{5} \\ $$$${f}\left(\mathrm{0}\right){g}\left(\mathrm{0}\right)−{h}\left(\mathrm{0}\right)=\mathrm{5} \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{unique}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{given}\:\mathrm{question}. \\ $$$$\mathrm{5}\:\mathrm{was}\:\mathrm{taken}\:\mathrm{as}\:\mathrm{example}\:\mathrm{and}\:\mathrm{can}\:\mathrm{be}\:\mathrm{replaced} \\ $$$$\mathrm{by}\:\mathrm{any}\:\mathrm{value}. \\ $$

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