JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    If the 7th  term in binomial expansion of \[{{\left( \frac{3}{^{3}\sqrt{84}}+\sqrt{3}\operatorname{In}x \right)}^{9}},x>0,\] equal to  729, then \[x\]cab be :                                   JEE Main  Online Paper (Held On 22 April 2013)

    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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