A) \[405/226\]
B) \[504/289\]
C) \[450/263\]
D) None of the above
Correct Answer: D
Solution :
The general term in \[{{\left( \frac{x}{2}-\frac{3}{{{x}^{2}}} \right)}^{10}}\]is \[{{T}_{r+1}}={{(-1)}^{r-10}}{{C}_{r}}{{\left( \frac{x}{2} \right)}^{10-r}}{{\left( \frac{3}{{{x}^{2}}} \right)}^{r}}\] \[={{(-1)}^{r\,\,}}\,{{\,}^{10}}{{C}_{r}}.\frac{{{3}^{r}}}{{{2}^{10-r}}}.{{x}^{10-r-2r}}\] For coefficient of \[{{x}^{4}},\] we have to take \[10-3r=4\] \[\Rightarrow \] \[3r=6\,\,\,\,\,\Rightarrow \,\,\,\,r=2\] \[\therefore \] Coefficient of \[{{x}^{4}}\] in \[{{\left( \frac{x}{2}-\frac{3}{{{x}^{2}}} \right)}^{10}}\] \[={{(-1)}^{2}}.\,{{\,}^{10}}{{C}_{2}}.\frac{{{3}^{2}}}{{{2}^{8}}}=\frac{9\times 45}{256}=\frac{405}{256}\]
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