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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is given by:$T_{r+1} = {}^6C_r \left(\frac{3}{2} x^

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$\frac{20}{27}$

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(C) The general term $T_{r+1}$ in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}ight)^6$ is given by:$T_{r+1} = {}^6C_r \left(\frac{3}{2} x^2ight)^{6-r} \left(-\frac{1}{3x}ight)^r = {}^6C_r \left(\frac{3}{2}ight)^{6-r} \left(-\frac{1}{3}ight)^r x^{12-2r-r} = {}^6C_r \left(\frac{3}{2}ight)^{6-r} \left(-\frac{1}{3}ight)^r x^{12-3r}$.For the term to be independent of $x$,the exponent of $x$ must be $0$:$12 - 3r = 0 \implies r = 4$.The $k^{	ext{th}}$ term corresponds to $r+1$,so $k = 4+1 = 5$.Now,we find the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}ight)^5$ at $x = \frac{2}{3}$.Let $T = \left(\frac{3}{2} x^2 - \frac{1}{3x}ight)^5$. The ratio of consecutive terms is $\left|\frac{T_{r+1}}{T_r}ight| = \left|\frac{n-r+1}{r} \cdot \frac{b}{a}ight|$,where $a = \frac{3}{2}x^2$ and $b = -\frac{1}{3x}$.$\left|\frac{T_{r+1}}{T_r}ight| = \left|\frac{5-r+1}{r} \cdot \frac{-1/3x}{3/2x^2}ight| = \left|\frac{6-r}{r} \cdot \left(-\frac{2}{9x^3}ight)ight| = \frac{6-r}{r} \cdot \frac{2}{9(2/3)^3} = \frac{6-r}{r} \cdot \frac{2}{9(8/27)} = \frac{6-r}{r} \cdot \frac{2}{8/3} = \frac{6-r}{r} \cdot \frac{3}{4}$.Setting $\frac{6-r}{r} \cdot \frac{3}{4} \ge 1 \implies 18 - 3r \ge 4r \implies 7r \le 18 \implies r \le 2.57$.Thus,$T_3$ is the greatest term $(r=2)$.$T_3 = {}^5C_2 \left(\frac{3}{2} x^2ight)^3 \left(-\frac{1}{3x}ight)^2 = 10 \cdot \frac{27}{8} x^6 \cdot \frac{1}{9x^2} = 10 \cdot \frac{3}{8} x^4 = \frac{30}{8} \left(\frac{2}{3}ight)^4 = \frac{15}{4} \cdot \frac{16}{81} = \frac{5 \cdot 4}{27} = \frac{20}{27}$.

If the $k^{\text{th}}$ term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is independent of $x$,then the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^k$ when $x = \frac{2}{3}$ is:

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