JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    The number of \[x\in [0,2\pi ]\]for which\[\left| \sqrt{2{{\sin }^{4}}x+18{{\cos }^{2}}x}-\sqrt{2{{\cos }^{4}}x+18{{\sin }^{2}}x} \right|=1\]   JEE Main Online Paper (Held On 09 April 2016)

    A) 6                                             

    B) 4

    C) 8             

    D) 2

    Correct Answer: C

    Solution :

                                    \[2\sqrt{2{{\cos }^{4}}x+18{{\sin }^{2}}x}\]                 \[2({{\sin }^{4}}x-{{\cos }^{2}}x)+18({{\cos }^{2}}x-{{\sin }^{2}}x)=1+2\] \[\sqrt{2{{\cos }^{4}}x+18{{\sin }^{2}}x}\]       \[\Rightarrow \]\[16({{\cos }^{2}}x-{{\sin }^{2}}x)=1+2\sqrt{2{{\cos }^{4}}x+18{{\sin }^{2}}x}\] \[\Rightarrow \]\[16\cos 2x-1=2\sqrt{2{{\left( \frac{1+\cos 2x}{2} \right)}^{2}}+9(1-\cos 2x)}\] \[\Rightarrow \]\[256{{\cos }^{2}}2x+1-32\cos 2x=4\] \[\Rightarrow \]\[\left( \frac{1+2\cos 2x+{{\cos }^{2}}2x}{2}+9(1-\cos 2x) \right)\] \[\Rightarrow \]\[256{{\cos }^{2}}2x+1-32\cos 2x=2(19-16\cos 2x+{{\cos }^{2}}2x)\] \[\Rightarrow \]\[254{{\cos }^{2}}2x=37\] \[\Rightarrow \]\[\cos 2x=\pm \sqrt{\frac{34}{254}}\in [-1,1]\]clearly 8 solutions                                

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