Question

Resolve $\frac{1-2x}{3+2x-x^2}$ into partial fractions.

• A. $-\frac{5}{4(3-x)}+\frac{3}{4(1+x)}$
• B. $\frac{5}{4(3-x)}+\frac{3}{4(1+x)}$
• C. $-\frac{5}{2(3-x)}+\frac{3}{4(1+x)}$
• D. $-\frac{5}{4(3-x)}+\frac{3}{2(1+x)}$

Answer 1

The correct option is A $-\frac{5}{4(3-x)}+\frac{3}{4(1+x)}$

We have
$\frac{1-2x}{3+2x-x^2}=\frac{(1-2x)}{(3-x)(1+x)}$
Let $\frac{1-2x}{3+2x-x^2}=\frac{1-2x}{(3-x)(1+x)}=\frac{A}{3-x}+\frac{B}{1+x}$ ...(i)
$\therefore A=\underset{x\to 3}{lim}(3-x)\left[\frac{(1-2x)}{(3-x)(1+x)}\right]$
$=\underset{x\to 3}{lim}\left[\frac{(1-2x)}{(1+x)}\right]$ $=\frac{1-6}{1+3}$ $=-\frac{5}{4}$
And $B=\underset{x\to -1}{lim}(1+x)\left[\frac{(1-2x)}{(3x-x)(1+x)}\right]$
$=\underset{x\to -1}{lim}\left[\frac{1-2x}{3-x}\right]$ $=\frac{1+2}{3+1}$ $=\frac{3}{4}$
Substituting the value of A and B in (i), we have
$\frac{1-2x}{3+2x-x^2}=-\frac{5}{4(3-x)}+\frac{3}{4(1+x)}$
Which are the required partial fractions.
Hence, option 'A' is correct.