If x - -1-1-1-to-infinity- then x -None of these

The correct option is D 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 0, We get x =1+√(5)/2 ...

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Byju's Answer

Standard XII

Mathematics

Quadratic Formula for Finding Roots

If x = √(1+√(...

Question

If x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . t o i n f i n i t y}}},$ then x =

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None of these

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Solution

The correct option is A
x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . .}}},$ to $\infty$
We have $x = \sqrt{1 + x}$
$\Rightarrow x^{2} = 1 + x \Rightarrow x^{2} - x - 1 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$
As $x > 0,$ We get x = $\frac{1 + \sqrt{5}}{2}$

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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±√52 As x>0, We get x =1+√52

If x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . . t o i n f i n i t y}}},$ then x =

The correct option is A
x = $\sqrt{1 + \sqrt{1 + \sqrt{1 + . . . . .}}},$ to $\infty$
We have $x = \sqrt{1 + x}$
$\Rightarrow x^{2} = 1 + x \Rightarrow x^{2} - x - 1 = 0$
$\Rightarrow x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$
As $x > 0,$ We get x = $\frac{1 + \sqrt{5}}{2}$