JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    Range of the function \[f(x)={{\cos }^{-1}}\left( \frac{1}{{{e}^{x}}+{{e}^{-x}}} \right)\] is

    A) \[(0,\,\pi )\]        

    B)        \[\left[ \frac{\pi }{3},\frac{\pi }{2} \right)\]           

    C) \[\left[ \frac{\pi }{6},\frac{\pi }{2} \right)\]           

    D)        \[\left[ \frac{\pi }{2},\frac{2\pi }{3} \right)\]

    Correct Answer: B

    Solution :

    [b] \[f(x)={{\cos }^{-1}}\left( \frac{1}{{{e}^{x}}+{{e}^{-x}}} \right)\] \[\frac{1}{{{e}^{x}}+{{e}^{-x}}}\] is an even function For  \[x\ge 0,\,\,\,2\le {{e}^{x}}+{{e}^{-x}}<\infty \] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{{{e}^{x}}+{{e}^{-x}}}\in \left( 0,\frac{1}{2} \right]\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\cos }^{-1}}\left( \frac{1}{{{e}^{x}}+{{e}^{-x}}} \right)\,\in \left[ \frac{\pi }{3},\frac{\pi }{2} \right)\]


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