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JEE Main & Advanced Mathematics Limits, Continuity and Differentiability Question Bank Self Evaluation Test - Continuity and Differentiability

  • question_answer
    The set of points of discontinuity of the function\[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(2\,\sin \,x)}^{2}}^{n}}{{{3}^{n}}-{{(2\cos \,x)}^{2n}}}\] is given by

    A) R

    B) \[\left\{ n\pi \pm \frac{\pi }{3},n\in I \right\}\]

    C) \[\left\{ n\pi \pm \frac{\pi }{6},n\in I \right\}\]

    D) None of these

    Correct Answer: C

    Solution :

    [c] We have, \[f(x)=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(2\,\sin \,x)}^{2n}}}{-{{(2\cos \,x)}^{2n}}}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{(2\,\,\sin \,\,x)}^{2n}}}{{{(\sqrt{3})}^{2n}}-{{(2\,\cos \,x)}^{2n}}}\] \[f(x)\] is discontinuous when \[{{(\sqrt{3})}^{2n}}-{{(2\,\,\cos \,\,x)}^{2n}}=0\] i.e. \[\cos \,\,x=\pm \frac{\sqrt{3}}{2}\Rightarrow x=n\pi \pm \frac{\pi }{6}(n\in I)\]


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