Question

The term independent of $x$ in expansion of $(\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{x-1}{x-x^{1/2}}{)}^{10}$ is

• A. 4
• B. 120
• C. 210
• D. 310

Answer 1

Answer: (C)

$[\frac{x+1}{x^{2/3}-x^{1/3}+1}-\frac{(x-1)}{x-x^{1/2}}{]}^{10}$
=$[\frac{(x^{1/3}{)}^3+1^3}{x^{2/3}-x^{1/3}+1}-\frac{\{(\sqrt{x}{)}^2\}}{\sqrt{x}(\sqrt{x}-1)}{]}^{10}$
=$[\frac{(x^{1/3}+1)(x^{2/3}+1-x^{1/3})}{x^{2/3-x^{1/3}+1}}-\frac{\{(\sqrt{x}{)}^2-1\}}{\sqrt{x}(\sqrt{x}-1)}{]}^{10}$
$=[(x^{1/3}+1)-\frac{(\sqrt{x}+1)}{\sqrt{x}}{]}^{10}=(x^{1/3}-x^{-1/2}{)}^{10}$
$\therefore$The general term is
$T_{r+1}={}^{10}{C}_r(x^{1/3}{)}^{10-r}(-x^{-1/2}{)}^r=10C_r(-1{)}^r{x}^{\frac{10-r}{3}-\frac{r}{2}}$
For independent of x , put
$\frac{10-r}{3}-\frac{r}{2}=0$
$\Rightarrow 20-2r-3r=0$
$\Rightarrow 20=5r\Rightarrow 20-2r-3r=0$
$\therefore T_5={}^{10}{C}_4=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}=210$