A) \[{{e}^{2}}\]
B) \[e\]
C) \[\frac{e}{2}\]
D) \[2e\]
Correct Answer: B
Solution :
Let\[r+1=7\Rightarrow r=6\] Given expansion is \[{{\left( \frac{3}{\sqrt[3]{84}}+\sqrt{3}\ln x \right)}^{9}},x>0\] We have \[{{T}_{r+1}}={{\,}^{n}}{{C}_{r}}{{(x)}^{n-r}}{{a}^{r}}\]for \[{{(x+a)}^{n}}.\] \[\therefore \] According to the question \[729={{\,}^{9}}{{C}_{6}}{{\left( \frac{3}{\sqrt[3]{84}} \right)}^{3}}.{{(\sqrt{3}\ln \,x)}^{6}}\] \[\Rightarrow \] \[{{3}^{6}}=84\times \frac{{{3}^{3}}}{84}\times {{3}^{3}}\times (6\,\ln \,x)\] \[\Rightarrow \]\[{{(\ln \,x)}^{6}}=1\]\[\Rightarrow \]\[{{(\ln x)}^{6}}={{(\ln e)}^{6}}\] \[\Rightarrow \]\[x=e\]
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