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Question Number 37089 by behi83417@gmail.com last updated on 08/Jun/18

Commented by math khazana by abdo last updated on 09/Jun/18

$1) f i d d e f i n e d o n [0, + \infty [$ $f^{^{'}} (x) = \frac{\frac{1}{2 \sqrt{1 + x}} (1 + \sqrt{x}) - \sqrt{1 + x} \frac{1}{2 \sqrt{x}}}{{(1 + \sqrt{x})}^{2}}$ $= \frac{2 \sqrt{x} (1 + \sqrt{x}) - 2 (1 + x)}{{(1 + \sqrt{x})}^{2}} = \frac{2 \sqrt{x} + 2 x - 2 - 2 x}{{(1 + \sqrt{x})}^{2}}$ $= \frac{2 (\sqrt{x} - 2)}{{(1 + \sqrt{x})}^{2}} i f x ⩾ 4 f^{^{'}} ⩾ 0 a n d f i s i n c r e a z i n g$ $f ([4, + \infty [) = [f (4), f (+ \infty) [= [\frac{\sqrt{5}}{3}, 1 [$ $i f 0 ⩽ x ⩽ 4 f i s d e c r e a s i n g s o$ $f ([0, 4]) = [f (4), f (0)] = [\frac{\sqrt{5}}{3}, 1]$
Commented by math khazana by abdo last updated on 09/Jun/18
$2) let f(x)=y ⇒ x=f^(−1) (y) ⇒ (√(1+x)) =y(1+(√x)) ⇒ 1+x=y^2 (1+2(√x) +x) ⇒ 1+x =y^2 +2y^2 (√x) +y^2 x ⇒ 1+x −y^2 −y^2 x = 2y^2 (√x) ⇒ (1+x −y^2 −y^2 x)^2 =4y^4 x ⇒ (1+x −(1+x)y^2 )^2 =4 y^4 x ⇒ (1+x)^2 (1−y^2 )^2 =4y^4 x ⇒ (x^2 +2x +1)(1−y^2 )^2 −4y^4 x =0 ⇒ (1−y^2 )^2 x^2 +2(1−y^2 )^2 x +(1−y^2 )^2 −4y^4 x =0 ⇒ {1−y^2 }^2 x^2 +{ 2 −4y^2 +2y^4 −4y^4 }x +(1−y^2 )^2 =0 {1−y^2 }^2 x^2 +2{ 1 −2y^2 −y^4 }x +(1−y^2 )^2 =0 ⇒ (1−y^2 )^2 x^2 −2(y^4 +2y^2 −1)Δx +(1−y^2 )^2 =0 Δ^′ =(y^4 +2y^2 −1)^2 −(1−y^2 )^4 ....be continued...$
$2) l e t f (x) = y \Rightarrow x = f^{- 1} (y) \Rightarrow \sqrt{1 + x} = y (1 + \sqrt{x})$ $\Rightarrow 1 + x = y^{2} (1 + 2 \sqrt{x} + x) \Rightarrow$ $1 + x = y^{2} + 2 y^{2} \sqrt{x} + y^{2} x \Rightarrow$ $1 + x - y^{2} - y^{2} x = 2 y^{2} \sqrt{x} \Rightarrow$ ${(1 + x - y^{2} - y^{2} x)}^{2} = 4 y^{4} x \Rightarrow$ ${(1 + x - (1 + x) y^{2})}^{2} = 4 y^{4} x \Rightarrow$ ${(1 + x)}^{2} {(1 - y^{2})}^{2} = 4 y^{4} x \Rightarrow$ $(x^{2} + 2 x + 1) {(1 - y^{2})}^{2} - 4 y^{4} x = 0 \Rightarrow$ ${(1 - y^{2})}^{2} x^{2} + 2 {(1 - y^{2})}^{2} x + {(1 - y^{2})}^{2} - 4 y^{4} x = 0 \Rightarrow$ ${1 - y^{2}}^{2} x^{2} + {2 - 4 y^{2} + 2 y^{4} - 4 y^{4}} x + {(1 - y^{2})}^{2} = 0$ ${1 - y^{2}}^{2} x^{2} + 2 {1 - 2 y^{2} - y^{4}} x + {(1 - y^{2})}^{2} = 0 \Rightarrow$ ${(1 - y^{2})}^{2} x^{2} - 2 (y^{4} + 2 y^{2} - 1) Δ x + {(1 - y^{2})}^{2} = 0$ $Δ^{^{'}} = {(y^{4} + 2 y^{2} - 1)}^{2} - {(1 - y^{2})}^{4} . . . . b e c o n t i n u e d . . .$
Commented by math khazana by abdo last updated on 09/Jun/18
$3) let I = ∫_0 ^1 ((√(1+x))/(1+(√x))) dx changement (√x)=t give I = ∫_0 ^1 ((√(1+t^2 ))/(1+t)) 2t dt =2 ∫_0 ^1 (t/(1+t))(√(1+t^2 )) dt =2{ ∫_0 ^1 (√(1+t^2 )) dt −∫_0 ^1 ((√(1+t^2 ))/(1+t))dt } ∫_0 ^1 (√(1+t^2 )) dt =_(t=sh(u)) ∫_o ^(ln(1+(√2))) ch(u)ch(u)du =(1/2) ∫_0 ^(ln(1+(√2))) (1+ch(2u))du =(1/2)ln(1+(√2)) +(1/4) [sh(2u)]_0 ^(ln(1+(√2))) but sh(2u) =((e^(2u) +e^(−2u) )/2) ⇒ sh(2ln(1+(√2)) = (((1+(√2))^2 +(1+(√2))^(−2) )/2) ⇒ ∫_0 ^1 (√(1+t^2 ))dt = (1/2)ln(1+(√2)) +(1/8){ (1+(√2))^2 +(1+(√2))^(−2) } letfind ∫_0 ^1 ((√(1+t^2 ))/(1+t)) dt chang. t=sh(x) give ∫_0 ^1 ((√(1+t^2 ))/(1+t)) dt= ∫_0 ^(ln(1+(√2))) ((ch(x))/(1+sh(x))) ch(x)dx = ∫_0 ^(ln(1+(√2))) (1/(1+sh(x))) ((1+ch(2x))/2) dx =(1/2) ∫_0 ^(ln(1+(√2))) ((1 +((e^(2x) +e^(−2x) )/2))/(1+ ((e^x +e^(−x) )/2))) dx =(1/2) ∫_0 ^(ln(1+(√2))) ((2 +e^(2x) +e^(−2x) )/(2+e^x +e^(−x) )) dx =_(e^x =t) (1/2) ∫_1 ^(1+(√2)) ((2 +t^(2 ) +(1/t^2 ))/(2 +t +(1/t))) dt =(1/2) ∫_1 ^(1+(√2)) ((2t^2 +t^4 +1)/(2t +t^2 +1)) (1/t) dt =(1/2) ∫_1 ^(1 +(√2)) ((t^4 +2t^2 +1)/(t^3 +2t^2 +t)) dt =(1/2) ∫_1 ^(1 +(√2)) ((t(t^3 +2t^2 +t) −2t^3 +t^2 +2t^2 +1)/(t^3 +2t^2 +t)) dt =(1/2) ∫_1 ^(1+(√2)) t dt +(1/2) ∫_1 ^(1+(√2)) ((−2t^3 +3t^2 +1)/(t^3 +2t^2 +t))dt =(1/4){ (1+(√2))^2 −1} +(1/2) ∫_1 ^(1+(√2)) ((−2(t^3 +2t^2 +t)+4t^2 +2t +3t^(2 ) +1)/(t^3 +2t^2 +t)) =(1/4){2−2(√2)) −1((√2)) +(1/2)∫_1 ^(1+(√2)) ((7t^2 +2t +1)/(t(t^2 +2t +1)))dt let decompose F(t) = ((7t^2 +2t +1)/(t(t+1)^2 )) F(t) = (a/t) +(b/(t+1)) +(c/((t+1)^2 )) ....be continued...$
$3) l e t I = \int_{0}^{1} \frac{\sqrt{1 + x}}{1 + \sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t g i v e$ $I = \int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{1 + t} 2 t d t$ $= 2 \int_{0}^{1} \frac{t}{1 + t} \sqrt{1 + t^{2}} d t = 2 {\int_{0}^{1} \sqrt{1 + t^{2}} d t - \int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{1 + t} d t}$ $\int_{0}^{1} \sqrt{1 + t^{2}} d t =_{t = s h (u)} \int_{o}^{l n (1 + \sqrt{2})} c h (u) c h (u) d u$ $= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} (1 + c h (2 u)) d u$ $= \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{4} {[s h (2 u)]}_{0}^{l n (1 + \sqrt{2})} b u t$ $s h (2 u) = \frac{e^{2 u} + e^{- 2 u}}{2} \Rightarrow s h (2 l n (1 + \sqrt{2})$ $= \frac{{(1 + \sqrt{2})}^{2} + {(1 + \sqrt{2})}^{- 2}}{2} \Rightarrow$ $\int_{0}^{1} \sqrt{1 + t^{2}} d t = \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{8} {{(1 + \sqrt{2})}^{2} + {(1 + \sqrt{2})}^{- 2}}$ $l e t f i n d \int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{1 + t} d t c h a n g . t = s h (x) g i v e$ $\int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{1 + t} d t = \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (x)}{1 + s h (x)} c h (x) d x$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{1}{1 + s h (x)} \frac{1 + c h (2 x)}{2} d x$ $= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} \frac{1 + \frac{e^{2 x} + e^{- 2 x}}{2}}{1 + \frac{e^{x} + e^{- x}}{2}} d x$ $= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} \frac{2 + e^{2 x} + e^{- 2 x}}{2 + e^{x} + e^{- x}} d x$ $=_{e^{x} = t} \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{2 + t^{2} + \frac{1}{t^{2}}}{2 + t + \frac{1}{t}} d t$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{2 t^{2} + t^{4} + 1}{2 t + t^{2} + 1} \frac{1}{t} d t$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{t^{4} + 2 t^{2} + 1}{t^{3} + 2 t^{2} + t} d t$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{t (t^{3} + 2 t^{2} + t) - 2 t^{3} + t^{2} + 2 t^{2} + 1}{t^{3} + 2 t^{2} + t} d t$ $= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} t d t + \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{- 2 t^{3} + 3 t^{2} + 1}{t^{3} + 2 t^{2} + t} d t$ $= \frac{1}{4} {{(1 + \sqrt{2})}^{2} - 1} + \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{- 2 (t^{3} + 2 t^{2} + t) + 4 t^{2} + 2 t + 3 t^{2} + 1}{t^{3} + 2 t^{2} + t}$ $= \frac{1}{4} {2 - 2 \sqrt{2}) - 1 (\sqrt{2}) + \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{7 t^{2} + 2 t + 1}{t (t^{2} + 2 t + 1)} d t$ $l e t d e c o m p o s e F (t) = \frac{7 t^{2} + 2 t + 1}{t {(t + 1)}^{2}}$ $F (t) = \frac{a}{t} + \frac{b}{t + 1} + \frac{c}{{(t + 1)}^{2}} . . . . b e c o n t i n u e d . . .$
Commented by behi83417@gmail.com last updated on 09/Jun/18

Commented by prakash jain last updated on 09/Jun/18

$x = 4$ $f (x) = \frac{\sqrt{5}}{3}$ $x = \frac{1}{4} \Rightarrow f (x) = \frac{\sqrt{5}}{4}$ $f^{- 1} (x) f o r f (x) = \frac{\sqrt{1 + x}}{1 + \sqrt{x}}$ $is not defined for since f (x) is not$ $one to one .$
Commented by behi83417@gmail.com last updated on 09/Jun/18

$2) 2 y^{2} - 1 ⩾ 0 \Rightarrow y ⩾ \frac{\sqrt{2}}{2}, y \neq \pm 1, D_{f} = [0, + \infty)$ $f (0) = \lim_{x \to + \infty} f (x) = 1$ $\Rightarrow \frac{\sqrt{2}}{2} ⩽ y < 1 \Rightarrow R_{f} = [\frac{\sqrt{2}}{2}, 1) .$
Commented by prakash jain last updated on 09/Jun/18

$f (x) = \frac{\sqrt{5}}{3} what is f^{- 1} (x) ?$ $f (\frac{1}{4}) and f (4) both give$ $f (x) = \frac{\sqrt{5}}{3} so f^{- 1} (x) is multivalued$ $and hence f^{- 1} (x) not a function by definition .$
Commented by math khazana by abdo last updated on 09/Jun/18

$h e l l o s i r p r a k a s h w h y a r e y o u a b s e n t f r o m t h i s$ $p l a t f o r m . . .$
Commented by prakash jain last updated on 10/Jun/18

$I am not completely absent . I read$ $messages daily .$ $My office work load is high and I$ $am not getting time to work$ $problems .$
Commented by math khazana by abdo last updated on 10/Jun/18

$n e v e r m i n d s i r . .$
	Answered by ajfour last updated on 09/Jun/18

	$x^{2} {(1 + \sqrt{y})}^{2} = 1 + y$ $x^{2} (1 + y + 2 \sqrt{y}) = 1 + y$ ${(x^{2} - 1)}^{2} {(1 + y)}^{2} = 4 x^{4} y$ $l e t y = \cos 2 θ =$ $\Rightarrow 4 {(x^{2} - 1)}^{2} \cos^{4} θ = 4 x^{4} (2 \cos^{2} θ - 1)$ $\Rightarrow {(x^{2} - 1)}^{2} \cos^{4} θ - 8 x^{4} \cos^{2} θ + 4 x^{4} = 0$ $\cos^{2} θ = \frac{8 x^{4} \pm \sqrt{64 x^{8} - 16 x^{4} {(x^{2} - 1)}^{2}}}{2 {(x^{2} - 1)}^{2}}$ $\cos^{2} θ = \frac{8 x^{4} \pm \sqrt{48 x^{8} + 32 x^{6} - 16 x^{4}}}{2 {(x^{2} - 1)}^{2}}$ $\cos^{2} θ = \frac{8 x^{4} \pm 4 x^{2} \sqrt{3 x^{4} + 2 x^{2} - 1}}{2 {(x^{2} - 1)}^{2}}$ $= \frac{4 x^{4} \pm 2 x^{2} \sqrt{(3 x^{2} - 1) (x^{2} + 1)}}{{(x^{2} - 1)}^{2}}$ $4 x^{4} - (3 x^{4} + 2 x^{2} - 1) = x^{4} - 2 x^{2} + 1$ $= {(x^{2} - 1)}^{2}$ $\Rightarrow \cos^{2} θ = \frac{4 x^{4} \pm 2 x^{2} \sqrt{4 x^{4} - {(x^{2} - 1)}^{2}}}{{(x^{2} - 1)}^{2}}$ $y = \cos 2 θ = 2 \cos^{2} θ - 1$ $y = \frac{8 x^{4} - {(x^{2} - 1)}^{2} \pm 4 x^{2} \sqrt{4 x^{4} - {(x^{2} - 1)}^{2}}}{{(x^{2} - 1)}^{2}} .$
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