JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer If the fourth term in the binomial expansion of\[{{\left( \frac{2}{x}+{{x}^{{{\log }_{8}}x}} \right)}^{6}}(x>0)\]is \[20\times {{8}^{7}},\]then a value of  x is :             [JEE Main 9-4-2019 Morning]

    A) \[8\]                             

    B) \[{{8}^{2}}\]

    C) \[{{8}^{-2}}\]                                  

    D) \[{{8}^{3}}\]

    Correct Answer: B

    Solution :

    \[{{T}_{4}}={{T}_{3+1}}=\left( \begin{matrix}    6  \\    3  \\ \end{matrix} \right){{\left( \frac{2}{x} \right)}^{3}}.{{\left( {{x}^{{{\log }_{8}}x}} \right)}^{3}}\]           \[20\times {{8}^{7}}=\frac{160}{{{x}^{3}}}.{{x}^{3{{\log }_{8}}x}}\]           \[{{8}^{6}}={{x}^{{{\log }_{2}}x}}-3\]           \[{{2}^{18}}={{x}^{{{\log }_{2}}x-3}}\]           \[\Rightarrow 18=(lo{{g}_{2}}x-3)(lo{{g}_{2}}x)\]             Let\[{{\log }_{2}}x=t\]             \[\Rightarrow {{t}^{2}}-3t-18=0\]\[\Rightarrow (t-6)(t+3)=0\]             \[\Rightarrow t=6,-3\]             \[{{\log }_{2}}x=6\Rightarrow x={{2}^{6}}={{8}^{2}}\]             \[{{\log }_{2}}x=-3\Rightarrow x={{2}^{-3}}={{8}^{-1}}\]

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