A) \[\frac{405}{256}\]
B) \[\frac{504}{259}\]
C) \[\frac{450}{263}\]
D) None of these
Correct Answer: A
Solution :
The general term in \[{{\left( \frac{x}{2}-\frac{3}{{{x}^{2}}} \right)}^{10}}\]is \[{{T}_{r+1}}={{\left( -1 \right)}^{r}}\,{{\,}^{10}}{{C}_{r}}{{\left( \frac{3}{{{x}^{2}}} \right)}^{r}}{{\left( \frac{x}{2} \right)}^{10-r}}\] \[={{(-1)}^{r}}\,{{\,}^{10}}{{C}_{r}}\frac{{{3}^{r}}}{{{2}^{10-r}}}.{{x}^{10-3r}}\] For coefficient of \[{{x}^{4}},\] we take \[10-3r=4\] \[\Rightarrow \] \[r=2\] \[\therefore \] 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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