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JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relation and Functions-II

  • question_answer
    Let \[f(x)={{x}^{2}}+3x-3,x>0.\] If n points \[{{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}}\] are so chosen on the x-axis such that (i) \[\frac{1}{n}\sum\limits_{i=1}^{n}{{{f}^{-1}}({{x}_{i}})}=f\left( \frac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}} \right)\] (ii) \[\sum\limits_{i=1}^{n}{{{f}^{-1}}}({{x}_{i}})=\sum\limits_{i=1}^{n}{{{x}_{i}}},\] where \[{{f}^{-1}}\] denotes the inverse of f. The value of \[\frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}=\]

    A) 1

    B) 2

    C) 3

    D) 4

    Correct Answer: A

    Solution :

    [a] \[\frac{{{f}^{-1}}({{x}_{1}})+{{f}^{-1}}({{x}_{2}})+...+{{f}^{-1}}({{x}_{n}})}{n}\] \[=f\left( \frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} \right)\] and \[\frac{{{f}^{-1}}({{x}_{1}})+{{f}^{-1}}({{x}_{2}})+...+{{f}^{-1}}({{x}_{n}})}{n}\] \[=f\left( \frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n} \right)\] \[\Rightarrow f(\bar{x})=\bar{x},\] Where \[\bar{x}=\frac{{{x}_{1}}+{{x}_{2}}+...+{{x}_{n}}}{n}\] \[\Rightarrow {{\bar{x}}^{2}}+3\bar{x}-3=\bar{x}\Rightarrow {{\bar{x}}^{2}}+2\bar{x}-3=0\] \[\Rightarrow \bar{x}=-3,1\Rightarrow \bar{x}=1\,\,as\,\,\bar{x}>0\]


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