Question

If $\frac{x^2+3}{x^4(x+3)}=\frac{A}{(x+3)}+\frac{B}{x}+\frac{C}{x^2}+\frac{D}{x^3}+\frac{E}{x^4}$ then
Match the following with I, II, III, IV, V

1) A	a) $\frac{4}{9}$
2) C	b) $\frac{4}{27}$
3) B	c) $-\frac{1}{3}$
4) E	d) $-\frac{4}{27}$
5) D	e) 1

• A. $1-b,2-a,3-d,4-e,5-c$
• B. $1-b,2-a,3-e,4-d,5-c$
• C. $1-d,2-b,3-e,4-a,5-c$
• D. $1-a,2-b,3-c,4-d,5-e$

Answer 1

Answer: (A)

$1-b,2-a,3-d,4-e,5-c$

$\frac{x^2+3}{x^4(x+3)}=\frac{x^2}{x^4(x+3)}+\frac{3}{x^4(x+3)}$

$=1/x^2(x+3)+3/x^4(x+3)$

$=\frac{1}{3}[\frac{1}{x^2}-\frac{1}{x}(x+3\left)\right]+[\frac{1}{x^4}-\frac{1}{x^3(x+3)}]$

$=\frac{1}{3x^2}+\frac{1}{x^4}-{\frac{1}{3x(x+3)}}^{-}\left[\frac{1}{x^3(x+3)}\right]--\left(1\right)$

$\frac{1}{x^3(x+3)}=\frac{1}{3}\left[\frac{(x+3)-x}{x^3(x+3)}\right]=\frac{1}{3}[\frac{1}{x^3}-\frac{1}{x^2(x+3)}]--\left(2\right)$

and $\frac{1}{x^2(x+3)}=\frac{1}{3}[\frac{1}{x^2}-\frac{1}{x(x+3)}]--\left(3\right)$

and $\frac{1}{x(x+3)}=\frac{1}{3}[\frac{1}{x}-\frac{1}{x+3}]--\left(4\right)$

substituting the values/results (2), (3) (4) in 1

we get

$1/3x^2+1/x^4=\frac{1}{3}\left[\frac{1}{3}\right(\frac{1}{x}-\frac{1}{x+3}\left)\right]$

$-\frac{1}{3}[\frac{1}{x^3}-[\frac{1}{3}\left]\right(\frac{1}{x^2}-\frac{1}{x(x+3)}\frac{1}{3}(\frac{1}{x}-\frac{1}{x+3})\left)\right]$

we get

$\frac{4}{27(x+3)}+\frac{-4}{27\left(x\right)}+\frac{4}{9x^2}-\frac{1}{3x^3}+\frac{1}{x^4}$

compare.