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JEE Main & Advanced Sample Paper JEE Main Sample Paper-6

  • question_answer
    Let f: (-1,1) \[\to \] B, be a function defined by \[f(x)={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right),\] then f is both one-to-one and onto when B is the interval

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

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

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

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

    Correct Answer: A

    Solution :

    Since, \[x\in (-1,1)\] \[\Rightarrow \]\[{{\tan }^{-1}}x\in \left( -\frac{\pi }{4},\frac{\pi }{4} \right)\]\[\Rightarrow \]\[2{{\tan }^{-1}}x\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right)\] and \[f(x)={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)=2{{\tan }^{-1}}x,({{x}^{2}}<1)\] So, \[f(x)\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right)\] \[\therefore \]Function is one-to-one onto.


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