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JEE Main & Advanced Sample Paper JEE Main Sample Paper-40

  • question_answer
    The value of \[\underset{x\to \pi /6}{\mathop{\lim }}\,{{(4-3\sin x-2{{\cos }^{2}}x)}^{\frac{1}{2\sin x-1}}}\]is

    A) 1                     

    B) e      

    C) Öe                               

    D) \[{{e}^{-1/2}}\]

    Correct Answer: D

    Solution :

     \[\underset{x\to \pi /6}{\mathop{\lim }}\,{{\left( 4-3\sin x-2{{\cos }^{2}}x \right)}^{1/(2\sin x-1)}}\] \[\underset{x\to \pi /6}{\mathop{\lim }}\,{{\left[ \frac{\left\{ 1+(2\sin x-1)(\sin x-1) \right\}}{{}^{1}/{}_{(2\sin x-1)(\sin x-1)}} \right]}^{\sin x-1}}={{e}^{-1/2}}\]


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