JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    Let f be a function defined by \[f(x)={{(x-1)}^{2}}+1,(x\ge 1).\] Statement -1 : The set\[\{x:f(x)={{f}^{-1}}(x)\}=\{1,2\}.\] Statement - 2 : f is a bisection and \[{{f}^{-1}}(x)=1+\sqrt{x-1},x\ge 1.\]     AIEEE  Solved  Paper (Held On 11 May  2011)

    A)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

    B)  Statement-1 is true, Statement-2 is true; Statement-2 is NOT a correct explanation for Statement-1

    C)  Statement-1 is true, Statement-2 is false

    D)  Statement-1 is false, Statement-2 is true.

    Correct Answer: A

    Solution :

                    \[f(x)={{(x-1)}^{2}}+1,x\ge 1\] \[f:[1,\infty )\to [1,\infty )\]is a bijective function \[\Rightarrow \]\[y={{(x-1)}^{2}}+1\Rightarrow {{(x-1)}^{2}}=y-1\] \[\Rightarrow \]\[x=1\pm \sqrt{y-1}\Rightarrow {{f}^{-1}}(y)=1\pm \sqrt{y-1}\] \[\Rightarrow \]\[{{f}^{-1}}(x)=1+\sqrt{x-1}\{\therefore x\ge 1\}\] so statement-2 is correct Now \[f(x)={{f}^{-1}}(x)\Rightarrow f(x)=x\Rightarrow {{(x-1)}^{2}}+1=x\] \[\Rightarrow \]\[{{x}^{2}}-3x+2=0\Rightarrow x=1,2\] so statement-1 is correct

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