JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    Let \[{{a}_{1}},\text{ }{{a}_{2}},{{a}_{3}},.....\text{ }{{a}_{10}}\] be in G.P. with \[{{a}_{i}}>0\] for \[i=1,\text{ }2,\text{ }......10\] and S be the set of pairs (r, k), r, k e N (the set of natural numbers) for which
    \[\left| \begin{matrix}    {{\log }_{e}}\,{{a}_{1}}{{\,}^{r}}{{a}_{2}}^{k} & {{\log }_{e}}\,{{a}_{2}}{{\,}^{r}}{{a}_{3}}^{k} & {{\log }_{e}}\,{{a}_{3}}{{\,}^{r}}{{a}_{4}}^{k}  \\    {{\log }_{e}}\,{{a}_{4}}{{\,}^{r}}{{a}_{5}}^{k} & {{\log }_{e}}\,{{a}_{5}}{{\,}^{r}}{{a}_{6}}^{k} & {{\log }_{e}}\,{{a}_{6}}{{\,}^{r}}{{a}_{7}}^{k}  \\    {{\log }_{e}}\,{{a}_{7}}{{\,}^{r}}{{a}_{8}}^{k} & {{\log }_{e}}\,{{a}_{8}}{{\,}^{r}}{{a}_{9}}^{k} & {{\log }_{e}}\,{{a}_{9}}{{\,}^{r}}{{a}_{10}}^{k}  \\ \end{matrix} \right|=0\]
    Then the number of elements in S, is -
      [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) 10                                            

    B) 4

    C) 2                     

    D)                  infinitely many

    Correct Answer: D

    Solution :

    \[{{a}^{r}}_{1}{{a}^{k}}_{2}={{a}^{r}}{{(aR)}^{k}}\,=\,{{a}^{r+k}}\,{{R}^{k}}\] \[{{a}^{r}}_{2}{{a}^{k}}_{3}=\,\,{{(aR)}^{r}}{{(a{{R}^{2}})}^{k}}={{a}^{r+k}}={{R}^{r+2k}}\] \[{{a}^{r}}_{3}{{a}^{k}}_{1}={{(a{{R}^{2}})}^{r}}{{(a{{R}^{3}})}^{k}}={{a}^{r+k}}\,{{R}^{2r+3k}}\] \[\Rightarrow \,\,\,\left| \begin{matrix}    In\,({{a}^{r+k}}\,{{R}^{k}}) & In\,({{a}^{r+k}}\,{{R}^{r+2k}}) & In\,({{a}^{r+k}}\,{{R}^{2r+3k}})  \\    In\,({{a}^{r+k}}\,{{R}^{3r+4k}}) & In\,({{a}^{r+k}}\,{{R}^{4r+5k}}\,) & In\,({{a}^{r+k}}\,{{R}^{5r+6k}})  \\    In\,({{a}^{r+k}}\,{{R}^{6r+7k}}) & In\,({{a}^{r+k}}\,{{R}^{7r+8k}}) & In\,({{a}^{r+k}}\,{{R}^{8r+9k}})  \\ \end{matrix} \right|\] = 0 \[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},\text{ }{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\] \[\Rightarrow \,\,\,\left| \begin{matrix}    In\,({{a}^{r+k}}\,{{R}^{k}}) & In\,({{a}^{r+k}}\,{{R}^{r+k}}) & In\,({{a}^{r+k}}\,{{R}^{2r+3k}})  \\    In\,\left( \frac{{{R}^{3r+4k}}}{{{R}^{k}}} \right) & In\,\left( \frac{{{R}^{4r+5k}}}{{{R}^{r+2k}}} \right) & In\,\left( \frac{{{R}^{5r+6k}}}{{{R}^{2r+3k}}} \right)  \\    In\,\left( \frac{{{R}^{6r+7k}}}{{{R}^{3r+4k}}} \right) & In\,\left( \frac{{{R}^{7r+8k}}}{{{R}^{4r+5k}}} \right) & In\,\left( \frac{{{R}^{8r+9k}}}{{{R}^{5r+6k}}} \right)  \\ \end{matrix} \right|\], = 0 \[\Rightarrow \,\,\,\left| \begin{matrix}    In\,({{a}^{r+k}}\,{{R}^{k}}) & In\,({{a}^{r+k}}\,{{R}^{r+k}}) & In\,({{a}^{r+k}}\,{{R}^{2r+3k}})  \\    In\,\left( {{R}^{3r+3k}} \right) & In\,\left( {{R}^{3r+3k}} \right) & In\,\left( {{R}^{3r+3k}} \right)  \\    In\,\left( {{R}^{3r+3k}} \right) & In\,\left( {{R}^{3r+3k}} \right) & In\,\left( {{R}^{3r+3k}} \right)  \\ \end{matrix} \right|\] = 0


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