Question

If $\frac{x-4}{x^2-5x+6}$ can be expanded in ascending powers 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)

$\frac{-73}{648}$

Let $f\left(x\right)=\frac{x-4}{x^2-5x+6}$

So, $f^{{}^{\prime }}\left(x\right)=\frac{(x^2-5x+6)-(2x-5)(x-4)}{(x^2-5x+6{)}^2}=-\frac{x^2-8x+14}{(x^2-5x+6{)}^2}$

$f^{{}^{\prime \prime }}\left(x\right)=\frac{(x^2-5x+6{)}^2(-(2x-8))+(x^2-8x+14\left)\right(2(x^2-5x+6)(2x-5))}{(x^2-5x+6{)}^4}$

$=\frac{2x^5-34x^4+216x^3-656x^2+964x-552}{(x^2-5x+6{)}^4}$

Also,

$f^{{}^{\prime \prime \prime }}\left(x\right)=\frac{(x^2-5x+6{)}^4(10x^4-136x^3+648x^2-1312x+964)}{(x^2-5x+6{)}^8}$

$-\frac{(2x^5-34x^4+216x^3-656x^2+964x-552)\left(4\right(x^2-5x+6{)}^3(2x-5))}{(x^2-5x+6{)}^8}$

According to Tailor's expansion in maclaurin's form,

$f\left(x\right)={\sum }_{r=0}^{\infty }\frac{f^r\left(0\right)}{r!}{x}^r$

So, Coefficient of $x^3=\frac{f^{{}^{\prime \prime \prime }}\left(0\right)}{3!}=\frac{6^4\times 964+552\times 4\times 6^3\times (-5)}{6^8\times 3!}=\frac{964-92\times 20}{6^5}=\frac{-73}{648}$