Question

In resolving $\frac{1}{{\left(1-2x\right)}^2\left(1-3x\right)}$ as sum of the partial fractions, one of the terms is

• A. $\frac{-3}{1-2x}$
• B. $\frac{-2}{1-2x}$
• C. $\frac{-6}{1-2x}$
• D. $\frac{-9}{1-2x}$

Answer 1

Answer: (C)

Let $\frac{1}{{\left(1-2x\right)}^2\left(1-3x\right)}=\frac{A}{1-2x}+\frac{B}{{\left(1-2x\right)}^2}+\frac{C}{1-3x}$
$\therefore 1=A\left(1-2x\right)\left(1-3x\right)+B\left(1-3x\right)+C{\left(1-2x\right)}^2$ ...(i)
Putting $x=\frac{1}{3}$ in (i), we get
$1=0+0+C{\left(1-\frac{2}{3}\right)}^2\Rightarrow 1=\frac{1}{9}C\Rightarrow C=9$
Putting $x=\frac{1}{2}$ in (i), we get
$\therefore 1=B\left(1-\frac{3}{2}\right)\Rightarrow -\frac{1}{2}B=1\Rightarrow B=-2$
Putting $x=0$ in (i), we get $1=A+B+C$
$A=1-B-C=1+2-9=-6$
$\therefore \frac{1}{{\left(1-2x\right)}^2\left(1-3x\right)}=\frac{-6}{1-2x}-\frac{2}{{\left(1-2x\right)}^2}+\frac{9}{1-3x}$
One of the terms is $\frac{-6}{1-2x}$