The coefficient of x

(a) $\frac{405}{256}$

Suppose $x^{4}$ occurs at the $(r+1)$ th term in the given expansion.

Then, we have

$T_{r+1}={ }^{10} C_{r}\left(\frac{x}{2}\right)^{10-r}\left(\frac{-3}{2 x^{2}}\right)^{r}$

$=(-1)^{r}{ }^{10} C_{r} \frac{3^{r}}{2^{10-r}} x^{10-r-2 r}$

For this term to contain $x^{4}$, we must have :

$10-3 r=4$

$10-3 r=4$

$\Rightarrow r=2$

$\therefore$ Required coefficient $={ }^{10} C_{2} \frac{3^{2}}{2^{8}}=\frac{10 \times 9 \times 9}{2 \times 2^{8}}=\frac{405}{256}$