​ If x = (√1 + (√1 + (√1...........))) then ​

Question

If x = \(\sqrt{1+\sqrt{1+\sqrt{1 \,+............}}}\) then

A) 1 < x < 2 

B) x = 1 

C) 0 < x < 1 

D) x is infinite

Answer 1

Correct option is (A) 1 < x < 2

We have \(x=\sqrt{1+\sqrt{1+\sqrt{1\,+......}}}\)

\(\Rightarrow\) \(x=\sqrt{1+x}\)

\(\Rightarrow x^2=1+x\)     (By squaring both sides)

\(\Rightarrow x^2-x-1=0\)

\(\Rightarrow x=\frac{-(-1)\pm\sqrt{(-1)^2-4\times1\times-1}}2\)

\(=\frac{1\pm\sqrt5}2\)

\(=\frac{1+\sqrt5}2\) or \(\frac{1-\sqrt5}2\)

\(\because\) \(x=\sqrt{1+\sqrt{1+\sqrt{1\,+......}}}\)

\(\Rightarrow\) x > 1

but \(\frac{1-\sqrt5}2<0<1\)

\(\therefore\) \(x\neq\frac{1-\sqrt5}2\)

Therefore, \(x=\frac{1+\sqrt5}2=\frac{1+2.236}2\)

\(=\frac{3.236}2\)

= 1.618 which is lie between 1 & 2.

\(\therefore\) \(1<x<2\)

Answer 2

Correct option is A) 1 < x < 2