Question

The coefficient of $x^{-5}$ in binomial expansion of ${\left(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right)}^{10}$, where $x\ne 0,1$ is :

• A. $1$
• B. $4$
• C. $-4$
• D. $-1$

Answer 1

Answer: (A)

$1$

${\left(\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{2}}+1}-\frac{x-1}{x-x^{\frac{1}{2}}}\right)}^{10}$

$\frac{x+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}$ $=\frac{(x^{\frac{1}{3}}{)}^3+1}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}$ $=\frac{(x^{\frac{1}{3}}+1)}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}$ $=\frac{(x^{\frac{1}{3}}+1)(x^{\frac{2}{3}}-x^{\frac{1}{3}}+1)}{x^{\frac{2}{3}}-x^{\frac{1}{3}}+1}$ $=x^{\frac{1}{3}}+1$

$\frac{x-1}{x-x^{\frac{1}{2}}}=\frac{(x^{\frac{1}{2}}-1)(x^{\frac{1}{2}}+1)}{x^{\frac{1}{2}}(x^{\frac{1}{2}}-1)}$ $=\frac{x^{\frac{1}{2}}+1}{x^{\frac{1}{2}}}$ $=1+x^{-\frac{1}{2}}$

$\therefore$ Expression $={[x^{\frac{1}{3}}+1-1-x^{-\frac{1}{2}}]}^{10}$ $={[x^{\frac{1}{3}}-x^{-\frac{1}{2}}]}^{10}$

General term $={}^{10}{C}_r\left(x^{\frac{1}{3}}{)}^r\right(-1{)}^{10-r}(x^{-\frac{1}{2}}{)}^{10-r}$

For coefficient of $x^{-5}$,

$\frac{r}{3}-\frac{1}{2}(10-r)=-5$

$\frac{r}{3}-5+\frac{r}{2}=-5$

$\frac{r}{3}+\frac{r}{2}=0$

$\Rightarrow r=0$

Coefficient $={}^{10}{C}_r(-1{)}^{10-r}{|}_{r=0}$
$={}^{10}{C}_0(-1{)}^{10}$
$=1$

Hence, option $B$ is correct.