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

  • question_answer All the pairs (x, y) that satisfy the inequality \[2\sqrt{{{\sin }^{2}}x-2\sin x+5}.\]\[\frac{1}{{{4}^{{{\sin }^{2}}y}}}\le 1\]also satisfy the equation. [JEE Main 10-4-2019 Morning]

    A) \[\sin x=|\sin y|\]

    B) \[\sin x=2\sin y\]

    C) \[2|\sin x|=3\sin y\]   

    D) \[2\sin x=\sin y\]

    Correct Answer: A

    Solution :

     \[{{2}^{\sqrt{{{\sin }^{2}}x-2\sin x+5}}}{{.4}^{-{{\sin }^{2}}y}}\le 1\]           \[\Rightarrow \]\[{{2}^{\sqrt{{{(\sin x-1)}^{2}}+4}}}\le {{2}^{2{{\sin }^{2}}y}}\] \[\Rightarrow \]\[\sqrt{{{(sinx-1)}^{2}}+4}\le 2{{\sin }^{2}}y\] \[\Rightarrow \]\[{{2}^{\sqrt{{{(sinx-1)}^{2}}+4}}}\le {{2}^{2{{\sin }^{2}}y}}\] \[\Rightarrow \]\[\sqrt{{{(sinx-1)}^{2}}+4}\le 2{{\sin }^{2}}y\] \[\Rightarrow \]\[\sin x=1\]and\[|\sin y|=1\]

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