A) \[\frac{5}{4}\]
B) \[\frac{7}{4}\]
C) \[\frac{9}{4}\]
D) \[45\]
Correct Answer: A
Solution :
General term, \[{{T}_{r+1}}{{=}^{10}}{{C}_{r}}{{\left( \sqrt{\frac{x}{3}} \right)}^{10-r}}.{{\left( \frac{3}{2{{x}^{2}}} \right)}^{r}}\] \[{{=}^{10}}{{C}_{r}}{{3}^{r-\frac{10-r}{2}}}{{\left( \frac{1}{2} \right)}^{r}}.{{x}^{\frac{10-r}{2}-2r}}\] \[{{=}^{10}}{{C}_{r}}{{3}^{r-\frac{3r-10}{2}}}{{\left( \frac{1}{2} \right)}^{r}}.{{x}^{\frac{10-5r}{2}}}\] For the term independent of x, put \[\frac{10-5r}{2}=0\] \[\Rightarrow \] \[5r=10\] \[\Rightarrow \] \[r=2\] \[\therefore \] The term independent of x, \[{{T}_{3}}{{=}^{10}}{{C}_{2}}\,{{3}^{\frac{6-10}{2}}}{{\left( \frac{1}{2} \right)}^{2}}\] \[=\frac{10\times 9}{2\times {{3}^{2}}\times 4}=\frac{5}{4}\]
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