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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    Let \[f(x)={{(x+1)}^{2}}-1,\ \ (x\ge -1)\]. Then the set \[S=\{x:f(x)={{f}^{-1}}(x)\}\] is                           [IIT 1995]

    A) Empty

    B) {0, -1}

    C) {0, 1, -1}

    D) \[\left\{ 0,\ -1,\ \frac{-3+i\sqrt{3}}{2},\ \frac{-3-i\sqrt{3}}{2} \right\}\]

    Correct Answer: D

    Solution :

    • Let \[f(x)={{(x+1)}^{2}}-1,\,\,x\ge -1.\] Since \[f(x)={{f}^{-1}}(x)\]           
    • \[\therefore \,\,{{(x+1)}^{2}}-1=\sqrt{1+x}-1\]  \[\left( \because \,\,{{f}^{-1}}(x)=\sqrt{1+x}-1 \right)\]           
    • \[\Rightarrow \,\,{{(x+1)}^{4}}=1+x\,\,\Rightarrow \,\,(x+1)\,\,[{{(x+1)}^{3}}-1]=0\]           
    • \[\Rightarrow \,\,x=-1\] or \[{{(x+1)}^{3}}=1\,\Rightarrow x+1=1,\,\,\omega ,\,\,{{\omega }^{2}}\]           
    • \[\Rightarrow \,\,x=0,\,\,-1,\,\frac{-3+i\sqrt{3}}{2},\,\,\frac{-3-i\sqrt{3}}{2}.\]


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