JEE Main & Advanced Mathematics Quadratic Equations Question Bank Critical Thinking

  • question_answer 1) If \[x=\sqrt{1+\sqrt{1+\sqrt{1+.......\text{to infinity}}}},\]then x =

    A) \[\frac{1+\sqrt{5}}{2}\]

    B) \[\frac{1-\sqrt{5}}{2}\]

    C) \[\frac{1\pm \sqrt{5}}{2}\]

    D) None of these

    Correct Answer: A

    Solution :

    \[x=\sqrt{1+\sqrt{1+\sqrt{1+.....}}}\]to \[\infty \]\[\infty \] We have \[x=\sqrt{1+x}\] Þ \[{{x}^{2}}=1+x\,\,\,\,\Rightarrow {{x}^{2}}-x-1=0\] Þ \[x=\frac{1\pm \sqrt{1+4}}{2}=\frac{1\pm \sqrt{5}}{2}\] As\[x>0\], we get  \[x=\frac{1+\sqrt{5}}{2}\]

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