JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    For any \[\theta \,\,\,\,\in \,\,\,\,\left( \frac{\pi }{4},\,\,\frac{\pi }{2} \right)\], the expression \[3{{\left( sin\,\theta -cos\,\theta  \right)}^{4}}+6{{\left( sin\,\theta +cos\,\theta  \right)}^{2}}+4\,si{{n}^{6}}\theta \]equals: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) \[13-4\,co{{s}^{4}}\theta +2\,si{{n}^{2}}\theta \,co{{s}^{2}}\theta \]

    B) \[13-4\,co{{s}^{2}}\theta +6\,si{{n}^{2}}\theta \,co{{s}^{2}}\theta \]

    C) \[13-4\text{ }cos{{\,}^{2}}\,\theta +6\,\,co{{s}^{4}}\,\theta \]

    D) \[13-4\text{ }cos{{\,}^{6}}\,\theta \]

    Correct Answer: D

    Solution :

      \[3{{\left( sin\,\theta -cos\,\theta  \right)}^{4}}+6{{\left( sin\,\theta +cos\,\theta  \right)}^{2}}+\,\,4\,si{{n}^{6}}\theta \]   \[3{{(1-2\,sin\,\theta \,\,cos\,\theta )}^{2}}+6(1+2\,sin\,\theta \text{ }cos\,\theta )+4\,{{(1-{{\cos }^{2}}\,\theta )}^{3}}\] \[=\,\,3\,(1+4{{\sin }^{2}}\theta \,{{\cos }^{2}}\,{{\theta }^{2}}-4\,sin\,\theta \,cos\,\theta )+6+\] \[12\,sin\,\theta \,\,cos\,\theta +4(1-co{{s}^{6}}\,\theta -3\,co{{s}^{2}}\theta +3\,co{{s}^{4}}\theta )\]\[=13\text{ }+12\,si{{n}^{2}}\theta \,co{{s}^{2}}\theta -4\,co{{s}^{6}}\theta -12\,co{{s}^{2}}\theta \] \[+\,\,12\,\cos {{\,}^{4}}\theta \] \[=13+12(1-co{{s}^{2}}\theta )\,co{{s}^{2}}\theta -12\,co{{s}^{2}}\theta \,\,+12\,cos{{\,}^{4}}\theta \] \[=\text{ }13-4\,{{\cos }^{6}}\,\theta \]


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