JEE Main & Advanced Sample Paper JEE Main Sample Paper-5

  • question_answer
    The set of points of continuity of the function \[f(x)=\sqrt{\frac{1}{2}-{{\cos }^{2}}x,}\] is

    A)  \[\left\{ x:\frac{\pi }{4}+2n\,\pi \le x\le \frac{3\pi }{4}+2n\pi ;n\in l \right\}\]

    B)  \[\left\{ x:\frac{5\pi }{4}+2n\,\pi \le x\le \frac{7\pi }{4}+2n\,\pi ;n\in l \right\}\]

    C)  \[\left\{ x:\frac{\pi }{4}+2n\,\pi \le x\le \frac{3\pi }{4}+2n\,\pi  \right\}\cup \]\[\left\{ x:\frac{5\pi }{4}+2n\,\pi \le x\le \frac{7\pi }{4}+2n\,\pi  \right\}\]

    D)  None of the above

    Correct Answer: C

    Solution :

     The given function \[f(x)=\sqrt{\frac{1}{2}-{{\cos }^{2}}x}\]is continuous at all the points where\[\frac{1}{2}-{{\cos }^{2}}x\ge 0\] \[\Rightarrow \]\[|\cos x|\le \frac{1}{\sqrt{2}}\]\[\Rightarrow \]\[\frac{\pi }{4}+2n\,\pi \le x\le \frac{3\pi }{4}+2n\pi \] or\[\frac{5\pi }{4}+2n\,\pi \le x\le \frac{7\pi }{4}+2n\,\pi ,n\in l\]

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