A) \[{{e}^{2}}\]
B) e
C) \[\frac{e}{2}\]
D) 2e
Correct Answer: B
Solution :
[b] 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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