Question-98516

Question Number 98516 by bemath last updated on 14/Jun/20

Commented by Aziztisffola last updated on 14/Jun/20

Hi sir best draw! which app do you use ?) = \sqrt{x + \sqrt{x + \sqrt{x + \dots}}} \Leftrightarrow f^{2} (x) = x + f (x)

I need the best one .

x) - f (x) - x = 0

= {(- 1)}^{2} - 4 \times 1 \times (- x) = 1 + 4 x

\Rightarrow f (x) = \frac{1 + \sqrt{1 + 4 x}}{2} (x and f (x) > 0)

f^{'} x) = \frac{1}{2} \frac{4}{2 \sqrt{1 + 4 x}} = \frac{1}{\sqrt{1 + 4 x}}

^{'} (5) = \frac{1}{\sqrt{1 + 4 \times 5}} = \frac{1}{\sqrt{21}}

Commented by Quvonchbek last updated on 14/Jun/20

Answered by mr W last updated on 14/Jun/20

s a y c e n t e r o f b i g c i r c l e i s (k, h)

{(k - 4)}^{2} + {(h - 3)}^{2} = {(R - 5)}^{2} \dots (i)

{(k - 0)}^{2} + {(h - 3)}^{2} = {(R - 3)}^{2} \dots (i i)

{(k - 4)}^{2} + {(h - 0)}^{2} = {(R - 4)}^{2} \dots (i i i)

(i i i) - (i) :

6 h - 9 = 2 R - 9

\Rightarrow h = \frac{R}{3}

(i i) - (i) :

4 k - 8 = 2 R - 8

\Rightarrow k = \frac{R}{2}

p u t t h i s i n t o (i i) :

\frac{R^{2}}{4} + {(\frac{R}{3} - 3)}^{2} = {(R - 3)}^{2}

\Rightarrow R = \frac{144}{23}

Answered by john santu last updated on 14/Jun/20

Commented by john santu last updated on 14/Jun/20

{(R - 3)}^{2} = 4^{2} + {(R - 5)}^{2} - 2 \times 4 \times \cos α

{(R - 4)}^{2} = 3^{2} + {(R - 5)}^{2} - 2 \times 3 \times \sin α

[\sin^{2} α = 1 - \cos^{2} α]

{(\frac{R - 9}{5 (R - 5)})}^{2} + {(\frac{R - 8}{2 (R - 5)})}^{2} = 1

R = \frac{144}{23} .