In the expansion of left( frac{x}{2} - frac{3 | vedclass

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(A) The general term in the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$ is given by $T_{r+1} = {}^{10}C_r \left( \frac{x}{2} \right

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$\frac{405}{256}$

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(A) The general term in the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} ight)^{10}$ is given by $T_{r+1} = {}^{10}C_r \left( \frac{x}{2} ight)^{10-r} \left( -\frac{3}{x^2} ight)^r$. Simplifying this,we get $T_{r+1} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-r-2r} = {}^{10}C_r (-1)^r \frac{3^r}{2^{10-r}} x^{10-3r}$. To find the coefficient of $x^4$,we set the exponent of $x$ equal to $4$: $10 - 3r = 4$ $\Rightarrow 3r = 6$ $\Rightarrow r = 2$. Substituting $r = 2$ into the expression for the term: $T_{2+1} = {}^{10}C_2 (-1)^2 \frac{3^2}{2^{10-2}} x^4 = \frac{10 	imes 9}{2 	imes 1} 	imes 1 	imes \frac{9}{2^8} x^4$. $T_3 = 45 	imes \frac{9}{256} x^4 = \frac{405}{256} x^4$. Thus,the coefficient of $x^4$ is $\frac{405}{256}$.

In the expansion of $\left( \frac{x}{2} - \frac{3}{x^2} \right)^{10}$,the coefficient of $x^4$ is

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