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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    ) If m is the A.M. of two distinct real numbers \[l\] and \[n\left( l,n>1 \right)\]and \[{{G}_{1}},{{G}_{2}}\]and \[{{G}_{3}}\] are three geometric means between \[l\] and n, then \[G_{1}^{4}+2G_{2}^{4}+G_{3}^{4}\]equals. [JEE Main Solved Paper-2015 ]

    A) \[4lm{{n}^{2}}\]                               

    B) \[4{{l}^{2}}{{m}^{2}}{{n}^{2}}\]

    C) \[4{{l}^{2}}mn\]                               

    D) \[4l{{m}^{2}}n\]

    Correct Answer: D

    Solution :

    \[m=\frac{l+n}{2}\] \[{{L}_{1}}{{G}_{1}}{{G}_{2}}{{G}_{3}}n\]are in G.P. \[n=l{{(r)}^{4}}\] \[{{\left( \frac{n}{l} \right)}^{{}^{1}/{}_{4}}}=r\] \[{{G}_{1}}=l{{\left( \frac{n}{l} \right)}^{{}^{1}/{}_{4}}}\] \[{{G}_{2}}=l{{\left( \frac{n}{l} \right)}^{{}^{2}/{}_{4}}}\] \[{{G}_{3}}=l{{\left( \frac{n}{l} \right)}^{{}^{3}/{}_{4}}}\] \[G_{1}^{4}+2G_{2}^{4}+G_{3}^{4}\] \[={{l}^{4}}\left[ \frac{n}{l}+2{{\left( \frac{n}{l} \right)}^{2}}+{{\left( \frac{n}{l} \right)}^{3}} \right]\] \[={{l}^{4}}\left[ \frac{n{{l}^{2}}+2{{n}^{2}}l+{{n}^{3}}}{{{l}^{3}}} \right]\] \[=l\left[ n\left( {{l}^{2}}+2nl+{{n}^{2}} \right) \right]\] \[=\ln {{\left( l+n \right)}^{2}}=4{{m}^{2}}\ln \]


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