A) \[8064\]
B) \[210\]
C) \[-546\]
D) \[5040\]
Correct Answer: B
Solution :
Given expansion is \[{{\left( \frac{x+1}{{{x}^{2/3}}-{{x}^{1/3}}+1}-\frac{x-1}{x-{{x}^{1/2}}} \right)}^{10}}\] \[={{\left[ \frac{{{({{x}^{1/3}})}^{3}}+{{1}^{3}}}{{{x}^{2/3}}-{{x}^{1/3}}+1}-\frac{\{{{(\sqrt{x})}^{2}}-1\}}{\sqrt{x}(\sqrt{x-1})} \right]}^{10}}\] \[={{\left[ \frac{({{x}^{1/3}}+1)({{x}^{2/3}}+1-{{x}^{1/3}})}{{{x}^{2/3}}-{{x}^{1/3}}+1}-\frac{\{{{(\sqrt{x})}^{2}}-1\}}{\sqrt{x}(\sqrt{x}-1)} \right]}^{10}}\] \[={{\left[ ({{x}^{1/3}}+1)-\frac{(\sqrt{x}+1)}{\sqrt{x}} \right]}^{10}}={{({{x}^{1/3}}-{{x}^{-1/2}})}^{10}}\] \[\therefore \] The general term is \[{{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{({{x}^{1/3}})}^{10-r}}{{(-x)}^{-t/2}}\] \[{{=}^{10}}{{C}_{r}}{{(-1)}^{r}}{{\times }^{\left( \frac{10-r}{3}-\frac{r}{2} \right)}}\] For the term independent of x, put \[\frac{10-r}{3}-\frac{r}{2}=0\] \[\Rightarrow \] \[20-2r-3r=0\] \[\Rightarrow \] \[20=5r\Rightarrow r=4\] \[\therefore \] \[{{T}_{5}}{{=}^{10}}{{C}_{4}}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}=210\]
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