Question

lf $\frac{x-4}{x^2-5x+6}$ can be expanded in ascending power of $x$, then the coefficient of $x^3$ is

• A. $-\frac{73}{648}$
• B. $\frac{73}{648}$
• C. $\frac{71}{648}$
• D. $\frac{-71}{648}$

Answer 1

Answer: (A)

$-\frac{73}{648}$

Given, $\frac{x-4}{x^2-5x+6}$$=\frac{x-4}{(x-3)(x-2)}$
$=\frac{2(x-3)-(x-2)}{(x-3)(x-2)}$
$=\frac{2}{x-2}-\frac{1}{x-3}$
$=2(x-2{)}^{-1}-(x-3{)}^{-1}$
$=-(1-\frac{x}{2}{)}^{-1}+\frac{1}{3}(1-\frac{x}{3}{)}^{-1}$
$=\frac{1}{3}(1-\frac{x}{3}{)}^{-1}-(1-\frac{x}{2}{)}^{-1}$
$=\frac{1}{3}(1+\frac{x}{3}+\frac{x^2}{9}+\frac{x^3}{27}...\infty )-(1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}...\infty )$
Hence coefficient of $x^3$ will be
$=\frac{1}{3}\frac{x^3}{27}-\frac{x^3}{8}$
$=(\frac{1}{81}-\frac{1}{8})x^3$
$=\frac{8-81}{648}{x}^3$
$=\frac{-73x^3}{648}$
Hence, coefficient is $\frac{-73}{648}$