JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If \[\alpha \] and \[\beta \] are the roots of the quadratic equation, \[{{x}^{2}}+x\sin \theta -2\sin \theta =0,\theta \in \left( 0,\frac{\pi }{2} \right),\] then \[\frac{{{\alpha }^{12}}+{{\beta }^{12}}}{\left( {{\alpha }^{-12}}+{{\beta }^{-12}} \right){{\left( \alpha -\beta  \right)}^{24}}}\]is equal to : [JEE Main 10-4-2019 Morning]

    A) \[\frac{{{2}^{6}}}{{{\left( \sin \theta +8 \right)}^{12}}}\]                     

    B) \[\frac{{{2}^{12}}}{{{\left( \sin \theta -8 \right)}^{6}}}\]

    C) \[\frac{{{2}^{12}}}{{{\left( \sin \theta -4 \right)}^{12}}}\]                     

    D) \[\frac{{{2}^{12}}}{{{\left( \sin \theta +8 \right)}^{12}}}\]

    Correct Answer: D

    Solution :

    \[\frac{{{\alpha }^{12}}+{{\beta }^{12}}}{\left( \frac{1}{{{\alpha }^{12}}}+\frac{1}{{{\beta }^{12}}} \right){{\left( \alpha -\beta  \right)}^{24}}}=\frac{{{\left( \alpha \beta  \right)}^{12}}}{{{\left( \alpha -\beta  \right)}^{24}}}\]           \[=\frac{{{\left( \alpha \beta  \right)}^{12}}}{{{\left[ {{\left( \alpha +\beta  \right)}^{2}}-4\alpha \beta  \right]}^{12}}}={{\left[ \frac{\alpha \beta }{{{\left( \alpha +\beta  \right)}^{2}}-4\alpha \beta } \right]}^{12}}\]           \[={{\left( \frac{-2\sin \theta }{{{\sin }^{2}}\theta +8\sin \theta } \right)}^{12}}=\frac{{{2}^{12}}}{{{\left( \sin \theta +8 \right)}^{12}}}\]


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