- #1

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- Homework Statement
- See attached questions (highlighted in Red)

- Relevant Equations
- Even and odd number concept

I am refreshing on this; of course i may need your insight where necessary...I intend to attempt the highlighted...this is a relatively new area to me...

For part (a),

We shall let ##f(x)=\dfrac{1}{x(2-x)}##, let ##g(x)## be the even function and ##h(x)## be the odd function. It follows that,

##f(x)=g(x)+ h(x)##

##g(x)=\dfrac{1}{2}\left[\dfrac{1}{2x-x^2}+\dfrac{1}{-2x-x^2}\right]##

##g(x) = \dfrac{1}{2}\left[\dfrac{2x+x^2-2x+x^2}{(2x-x^2)(2x+x^2}\right]##

##g(x)=\dfrac{1}{2}\left[\dfrac{2x^2}{(x^2(2-x)(2+x)}\right]##

##g(x)=\left[\dfrac{1}{(2-x)(2+x}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}-\dfrac{1}{-x(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}+\dfrac{1}{x(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1(2+x)+1(2-x)}{x(2-x)(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{2+x+2-x)}{x(2-x)(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{4}{x(2-x)(2+x)}\right]##

##h(x)=\left[\dfrac{2}{x(2-x)(2+x)}\right]##

therefore,

##\dfrac{1}{x(2-x)}=\dfrac{1}{(2-x)(2+x)}+\dfrac{2}{x(2-x)(2+x)} ##

Bingo! I will attempt the others later...

For part (a),

We shall let ##f(x)=\dfrac{1}{x(2-x)}##, let ##g(x)## be the even function and ##h(x)## be the odd function. It follows that,

##f(x)=g(x)+ h(x)##

##g(x)=\dfrac{1}{2}\left[\dfrac{1}{2x-x^2}+\dfrac{1}{-2x-x^2}\right]##

##g(x) = \dfrac{1}{2}\left[\dfrac{2x+x^2-2x+x^2}{(2x-x^2)(2x+x^2}\right]##

##g(x)=\dfrac{1}{2}\left[\dfrac{2x^2}{(x^2(2-x)(2+x)}\right]##

##g(x)=\left[\dfrac{1}{(2-x)(2+x}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}-\dfrac{1}{-x(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1}{x(2-x)}+\dfrac{1}{x(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{1(2+x)+1(2-x)}{x(2-x)(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{2+x+2-x)}{x(2-x)(2+x)}\right]##

##h(x)=\dfrac{1}{2}\left[\dfrac{4}{x(2-x)(2+x)}\right]##

##h(x)=\left[\dfrac{2}{x(2-x)(2+x)}\right]##

therefore,

##\dfrac{1}{x(2-x)}=\dfrac{1}{(2-x)(2+x)}+\dfrac{2}{x(2-x)(2+x)} ##

Bingo! I will attempt the others later...