• # 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.

$f(x)={{(x-1)}^{2}}+1,x\ge 1$ $f:[1,\infty )\to [1,\infty )$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