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  • If,  <img alt="\mathrm{y=x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%3Dx&plus;%5Csqrt%7Bx&plus;%5Csqrt%7Bx&plus

If,  \mathrm{y=x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}} }, then \mathrm{y(2)=}

Option: 1

4 or 1
 


Option: 2

 4 only
 


Option: 3

1 only
 


Option: 4

undefined


Answers (1)

best_answer

                          \mathrm{y=x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}}

\mathrm{\text { At } x=2}

                         \mathrm{y=2+\sqrt{2+\sqrt{2+\ldots \infty}}}

                         \mathrm{y=2+\sqrt{y}}                                           ....(1)

              \mathrm{(y-2)^2=y \quad \text { (Squaring on bath sides) }}

\mathrm{y^2-4 y+4=y}

\mathrm{\begin{array}{r} y^2-5 y+4=0 \\ (y-1)(y-4)=0 \end{array}}

                         \mathrm{y=1 \text { or } y=4}

But \mathrm{y=1} does not satisfy equation (1)

\therefore                    \mathrm{y(2)=4}

Posted by

shivangi.shekhar

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