Question

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

• A. 310
• B. 120
• C. 4
• D. 210

Answer 1

The correct option is D 210

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