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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(x)=f(x)+g(x)+h(x)y(x)=x2+sinx+x(1x)f(x)iseveng(x)isoddf(0)g(0)h(0)=?
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(x)=f(x)+g(x)+h(x)y(x)=x2+sinx+x(1x)y(0)=0f(x)iseven&g(x)isoddf(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)=0h(0)=f(0)A0=2g(0)g(0)=0Bf(0)g(0)h(0)=f(0)g(0)+f(0)=f(0)[g(0)+1]=f(0)[0+1]=f(0)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(x)=x+sinxassumef(x)=5+x2(even)g(x)=x(odd)h(x)=sinx5x2y(x)=f(x)+g(x)+h(x)=sinx+xf(0)=5g(0)=0h(0)=5f(0)g(0)h(0)=5Thereisnouniqueanswertogivenquestion.5wastakenasexampleandcanbereplacedbyanyvalue.

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