Question

If $\frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2}$ then $B+D+E$ is equal to

• A. A + C
• B. A - C
• C. 2A + C
• D. 2A + 2C

Answer 1

Answer: (A)

Given,
$\frac{x+1}{x^4(x+2)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{D}{x^4}+\frac{E}{x+2}$
$\Rightarrow x+1=A{x}^3(x+2)+B{x}^2(x+2)+Cx(x+2)+D(x+2)+E{x}^4$
At $x=-1$,
$\Rightarrow -1+1=-A+B-C+D+E$
$\Rightarrow 0=B+D+E-A-C$
$\therefore B+D+E=A+C$