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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    Let \[f:(-1,1)\to B\], be a function defined by \[f(x)={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}},\] then f is both one- one and onto when B is the interval [AIEEE 2005]

    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: B

    Solution :

               For ? 1< x< 1, \[{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}=2{{\tan }^{-1}}x\]                    Range of \[f(x)=\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\].                                                               \Co-domain of function = B \[=\left( -\frac{\pi }{2},\frac{\pi }{2} \right)\].


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