BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer The ninth term in the expansion of\[{{[{{3}^{{{\log }_{3}}\sqrt{{{25}^{x-1}}+7}}}+{{3}^{-\frac{1}{8}{{\log }_{3}}({{5}^{x-1}}+1)}}]}^{10}}\]is equal  to 180, then \[x\]is equal to

    A)  1              

    B)         2                            

    C)         3                            

    D)         None of these

    Correct Answer: A

    Solution :

                    We have, \[{{[{{3}^{{{\log }_{3}}\sqrt{{{25}^{x-1}}+7}}}+{{3}^{-\frac{1}{8}{{\log }_{3}}{{5}^{x-1}}+1}}]}^{10}}\] \[={{[\sqrt{{{25}^{x-1}}+7}+{{({{5}^{x-1}}+1)}^{-1/8}}]}^{10}}\]                                                 \[(\because \,{{a}^{{{\log }_{a}}x}}=x)\]                 Here, \[{{T}_{9}}=180\]                 \[\Rightarrow \]\[{{\,}^{10}}{{C}_{8}}{{(\sqrt{{{25}^{x-1}}+7})}^{10-8}}{{\{{{({{5}^{x-1}}+1)}^{-1/8}}\}}^{8}}=180\]                 \[\Rightarrow \]\[{{\,}^{10}}{{C}_{8}}({{25}^{x-1}}+7).\frac{1}{({{5}^{x-1}}+1)}=180\]                 \[\Rightarrow \]               \[\frac{{{25}^{x-1}}+7}{{{5}^{x-1}}+1}=\frac{180}{45}=4\] Let \[y={{5}^{x-1}},\]then above equation becomes                                 \[\frac{{{y}^{2}}+7}{y+1}=4\]                 \[\Rightarrow \]               \[{{y}^{2}}+7-4y-4=0\]                 \[\Rightarrow \]               \[{{y}^{2}}-4y+3=0\]                 \[\Rightarrow \]               \[(y-3)(y-1)+0\]                 \[\Rightarrow \]               \[y=3,\,y=1\]                 If \[y=3\]                 \[\Rightarrow \]  \[{{5}^{x-1}}=3\Rightarrow {{5}^{x}}=15\Rightarrow x={{\log }_{5}}15\]                 If \[y=1\] \[\Rightarrow \]               \[{{5}^{x-1}}=1\Rightarrow {{5}^{x}}=5\Rightarrow x=1\]

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