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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    Let \[f:R\to R\]be defined by \[f(x)=\frac{x}{1+{{x}^{2}}},x\in R.\]Then the range of f is      [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[(-1,1)-\{0\}\]                           

    B)               \[R-[-1,1]\]

    C)               \[\left[ -\frac{1}{2},\frac{1}{2} \right]\]                           

    D)               \[R-\left[ -\frac{1}{2},\frac{1}{2} \right]\]

    Correct Answer: C

    Solution :

    Here, f(x) is an odd function and f(0) = 0 Now, when \[x>0,f(x)=\frac{1}{x+\frac{1}{x}}\in \left( 0,\frac{1}{2} \right]\] When \[x<0,f(x)=\frac{1}{x+\frac{1}{x}}\in \left[ -\frac{1}{2},0 \right)\] \[\therefore \]Range of f(x) is \[\left[ -\frac{1}{2},\frac{1}{2} \right].\]


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