Question

Find the value of :
$\frac{1}{x+p}+\frac{1}{x+q}+\frac{1}{x+r}+\frac{px}{x^3+p{x}^2}+\frac{qx}{x^3+p{x}^2}+\frac{rx}{x^3+rx}$

• A. $\frac{1}{x+p+q+r}$
• B. $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$
• C. $\frac{3}{x}$
• D. $\frac{3}{p+q+r}$

Answer 1

The correct option is C

$\frac{3}{x}$

Given,
$\frac{1}{x+p}+\frac{1}{x+q}+\frac{1}{x+r}+\frac{px}{x^3+p{x}^2}$
$+\frac{qx}{x^3+q{x}^2}+\frac{rx}{x^3+r{x}^2}=\frac{1}{x+p}+\frac{1}{x+q}+\frac{1}{x+r}+\frac{px}{x^2(x+p)}$
$+\frac{qx}{x^2(x+q)}+\frac{rx}{x^2(x+r)}=\frac{1}{x+p}+\frac{p}{x(x+p)}+\frac{1}{x+q}+\frac{q}{x(x+q)}$
$+\frac{1}{x+r}+\frac{r}{x(x+r)}=\frac{(x+p)}{x(x+p)}+\frac{(x+q)}{x(x+q)}+\frac{(x+r)}{x(x+r)}=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}=\frac{3}{x}$