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

1.	If x = \(\sqrt{1+\sqrt{1+\sqrt{1 \,+............}}}\) thenA) 1 < x < 2 B) x = 1 C) 0 < x < 1 D) x is infinite
Answer» 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. \(\therefore\) \(1<x<2\) Correct option is A) 1 < x < 2

If x = \(\sqrt{1+\sqrt{1+\sqrt{1 \,+............}}}\) thenA) 1 < x < 2 B) x = 1 C) 0 < x < 1 D) x is infinite

Answer»

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.

\(\therefore\) \(1<x<2\)

Correct option is A) 1 < x < 2