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JEE Main & Advanced Mathematics Relations Question Bank Mock Test - Relations and Functions

  • question_answer
    The range of \[f(x)=si{{n}^{-1}}(\sqrt{{{x}^{2}}+x+1})\] is

    A) \[\left( 0,\frac{\pi }{2} \right]\]     

    B) \[\left( 0,\frac{\pi }{3} \right]\]

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

    D) \[\left[ \frac{\pi }{6},\frac{\pi }{3} \right]\]

    Correct Answer: C

    Solution :

    [c] for the function to get defined, \[0\le {{x}^{2}}+x+1\le 1,\] but \[{{x}^{2}}+x+1\ge \frac{3}{4}\] Or  \[\frac{\sqrt{3}}{2}\le \sqrt{{{x}^{2}}+x+1}\le 1\] Or  \[\frac{\pi }{3}\le {{\sin }^{-1}}(\sqrt{{{x}^{2}}+x+1})\le \frac{\pi }{2}\]


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