If x -1-1-1- infinity - then x -A- 1-5-2B- 1-5-21 -5-2D- None of these

The correct option is A x = √(1+√(1+√(1+......to ∞))) We have x = √(1+x) ⇒ x^2 = 1+ x ⇒ x^2 x 1 = 0 ⇒ x = 1±√(1+4)/2 = 1±√(5)/2 As x gt; 0, we get x=1 ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

If x =√1+√1+√...

Question

$I f x = \sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . . t o i n f i n i t y}}}, t h e n x =$

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None of these

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Solution

The correct option is A
$x = \sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . . t o \infty}}} W e h a v e x = \sqrt{1 + x} \Rightarrow x^{2} = 1 + x \Rightarrow x^{2} - x - 1 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} A s x > 0, w e g e t x = \frac{1 \pm \sqrt{5}}{2}$

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Similar questions

Q. If x =

\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . t o i n f i n i t y}}},

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Q. If

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, then
(a)

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$F i n d t h e i n t e g r a l v a l u e o f x s u c h t h a t \frac{3 x + 1}{x + 1} + \frac{x + 1}{3 x + 1} = \frac{5}{2}$

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(a) 1/2
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If x=√1+√1+√1+......to infinity, then x=

The correct option is A x=√1+√1+√1+......to∞We have x=√1+x⇒x2=1+x⇒x2−x−1=0⇒x=1±√1+42=1±√52Asx>0,we get x=1±√52

$I f x = \sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . . t o i n f i n i t y}}}, t h e n x =$

The correct option is A
$x = \sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . . t o \infty}}} W e h a v e x = \sqrt{1 + x} \Rightarrow x^{2} = 1 + x \Rightarrow x^{2} - x - 1 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} A s x > 0, w e g e t x = \frac{1 \pm \sqrt{5}}{2}$