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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    Let X and Y be subsets of R, the set of all real numbers. The function \[f:X\to Y\]defined by \[f(x)={{x}^{2}}\] for \[x\in X\] is one-one but not onto if (Here \[{{R}^{+}}\] is the set of all positive real numbers)                 [EAMCET 2000]

    A)                    \[X=Y={{R}^{+}}\]

    B)            \[X=R,\ Y={{R}^{+}}\]

    C)            \[X={{R}^{+}},\ Y=R\]

    D)            \[X=Y=R\]

    Correct Answer: C

    Solution :

               \[f({{x}_{1}})=f({{x}_{2}})\Rightarrow x_{1}^{2}=x_{2}^{2}\Rightarrow {{x}_{1}}={{x}_{2}}\], [if \[X={{R}^{+}}]\]            Þ f is one-one. Since \[{{R}_{f}}={{R}^{+}}\subseteq R=Y\]; \ f is not onto.


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