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(D) The general term $T_{r+1}$ in the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}$ is given by:$T_{r+1} = {}^{18}C_r (9x)^{18-r} \left(

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$1$

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(D) The general term $T_{r+1}$ in the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}$ is given by:$T_{r+1} = {}^{18}C_r (9x)^{18-r} \left( -\frac{1}{3\sqrt{x}} \right)^r$$T_{r+1} = {}^{18}C_r (9)^{18-r} (x)^{18-r} (-1)^r (3)^{-r} (x)^{-r/2}$$T_{r+1} = {}^{18}C_r (9)^{18-r} (-1)^r (3)^{-r} (x)^{18 - r - r/2}$For the term to be independent of $x$,the exponent of $x$ must be $0$:$18 - \frac{3r}{2} = 0$$36 - 3r = 0 \Rightarrow r = 12$Substituting $r = 12$ into the expression:$T_{13} = {}^{18}C_{12} (9)^{18-12} (-1)^{12} (3)^{-12}$$T_{13} = {}^{18}C_{12} (9)^6 (1) (3)^{-12}$Since $9 = 3^2$,we have $9^6 = (3^2)^6 = 3^{12}$.$T_{13} = {}^{18}C_{12} \cdot 3^{12} \cdot 3^{-12} = {}^{18}C_{12} \cdot 1$Comparing this with $\alpha \cdot {}^{18}C_{12}$,we get $\alpha = 1$.

The term independent of $x$ in the expansion of $\left( 9x - \frac{1}{3\sqrt{x}} \right)^{18}, x > 0$,is $\alpha$ times the corresponding binomial coefficient. Then $\alpha$ is:

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