Math Question and Answers


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 39222 by behi83417@gmail.com last updated on 04/Jul/18

Commented by math khazana by abdo last updated on 04/Jul/18
$1) f is defined on [0,1[ f^′ (x)=((((−1)/(2(√(1−x))))(1−(√x)) −(√(1−x))(((−1)/(2(√x)))))/((1−(√x))^2 )) =((−2(√x)(1−(√x))+2(1−x))/(4(√x)(√(1−x))(1−(√x))^2 )) =((−2(√x) +2x +2−2x)/(4(√x)(√(1−x))(1−(√x))^2 )) = (2/(4(√x)(√(1−x))(1−(√x)))) >0 ⇒ f is increasing on[0,1[ f(o)= 1 and lim _(x→1^− ) f(x) =lim_(x→1^− ) ((√(1−x))/(1−(√x))) =lim_(x→1^− ) (((−1)/(2(√(1−x))))/(−(1/(2(√x))))) = lim_(x→1^− ) ((2(√x))/(2(√(1−x)))) =lim_(x→1^− ) ((√x)/(√(1−x))) = +∞ ⇒ f([0,1[)=[1,+∞[ . 2)f(x)=y ⇔ x=f^(−1) (y) ⇒ ((√(1−x))/(1−(√x)))=y ⇒ (√( ((1−x)/((1−(√x))^2 )))) =y ⇒ ((1−x)/(1+x−2(√x))) =y^2 let (√x)=t ⇒ ((1−t^2 )/(1+t^2 −2t))=y^2 ⇒ 1−t^2 =y^2 +y^2 t^2 −2y^2 t ⇒ (1+y^2 )t^2 −2y^2 t +y^2 −1=0 Δ^′ =y^4 −(y^2 +1)(y^2 −1)=y^4 −(y^4 −1)=1 t_1^ =((y^(2 ) +1)/(1+y^2 )) =1 and t_2 = ((y^2 −1)/(y^2 +1)) ⇒x =1 or x=(((y^2 −1)/(y^2 +1)))^2 f^(−1) is not constant ⇒ f^(−1) (x) = {((x^2 −1)/(x^2 +1))}^2 .$
$1) f i s d e f i n e d o n [0, 1 [$ $f^{^{'}} (x) = \frac{\frac{- 1}{2 \sqrt{1 - x}} (1 - \sqrt{x}) - \sqrt{1 - x} (\frac{- 1}{2 \sqrt{x}})}{{(1 - \sqrt{x})}^{2}}$ $= \frac{- 2 \sqrt{x} (1 - \sqrt{x}) + 2 (1 - x)}{4 \sqrt{x} \sqrt{1 - x} {(1 - \sqrt{x})}^{2}} = \frac{- 2 \sqrt{x} + 2 x + 2 - 2 x}{4 \sqrt{x} \sqrt{1 - x} {(1 - \sqrt{x})}^{2}}$ $= \frac{2}{4 \sqrt{x} \sqrt{1 - x} (1 - \sqrt{x})} > 0 \Rightarrow f i s i n c r e a s i n g o n [0, 1 [$ $f (o) = 1 a n d$ $l i m_{x \to 1^{-}} f (x) = {l i m}_{x \to 1^{-}} \frac{\sqrt{1 - x}}{1 - \sqrt{x}}$ $= {l i m}_{x \to 1^{-}} \frac{\frac{- 1}{2 \sqrt{1 - x}}}{- \frac{1}{2 \sqrt{x}}} = {l i m}_{x \to 1^{-}} \frac{2 \sqrt{x}}{2 \sqrt{1 - x}}$ $= {l i m}_{x \to 1^{-}} \frac{\sqrt{x}}{\sqrt{1 - x}} = + \infty \Rightarrow$ $f ([0, 1 [) = [1, + \infty [.$ $2) f (x) = y \Leftrightarrow x = f^{- 1} (y) \Rightarrow \frac{\sqrt{1 - x}}{1 - \sqrt{x}} = y \Rightarrow$ $\sqrt{\frac{1 - x}{{(1 - \sqrt{x})}^{2}}} = y \Rightarrow \frac{1 - x}{1 + x - 2 \sqrt{x}} = y^{2}$ $l e t \sqrt{x} = t \Rightarrow \frac{1 - t^{2}}{1 + t^{2} - 2 t} = y^{2} \Rightarrow$ $1 - t^{2} = y^{2} + y^{2} t^{2} - 2 y^{2} t \Rightarrow$ $(1 + y^{2}) t^{2} - 2 y^{2} t + y^{2} - 1 = 0$ $Δ^{^{'}} = y^{4} - (y^{2} + 1) (y^{2} - 1) = y^{4} - (y^{4} - 1) = 1$ $t_{1^{}} = \frac{y^{2} + 1}{1 + y^{2}} = 1 a n d t_{2} = \frac{y^{2} - 1}{y^{2} + 1}$ $\Rightarrow x = 1 o r x = {(\frac{y^{2} - 1}{y^{2} + 1})}^{2} f^{- 1} i s n o t c o n s t a n t \Rightarrow$ $f^{- 1} (x) = {\frac{x^{2} - 1}{x^{2} + 1}}^{2} .$
Commented by maxmathsup by imad last updated on 04/Jul/18

$3) l e t I = \int_{0}^{\frac{\sqrt{2}}{2}} \frac{\sqrt{1 - x}}{1 - \sqrt{x}} d x c h a n g e m e n t \sqrt{x} = t g i v e$ $I = \int_{0}^{\frac{1}{(^{4} \sqrt{2})}} \frac{\sqrt{1 - t^{2}}}{1 - t} 2 t d t$ $I = - 2 \int_{0}^{2^{- \frac{1}{4}}} \frac{1 - t - 1}{1 - t} \sqrt{1 - t^{2}} d t$ $= - 2 \int_{0}^{2^{- \frac{1}{4}}} \sqrt{1 - t^{2}} + 2 \int_{0}^{2^{- \frac{1}{4}}} \frac{\sqrt{1 - t^{2}}}{1 - t} d t$ $\int_{0}^{2^{- \frac{1}{4}}} \sqrt{1 - t^{2}} d t =_{t = s i n α} \int_{0}^{a r c s i n (2^{- \frac{1}{4}})} c o s α c o s α d α$ $= \frac{1}{2} \int_{0}^{a r c s i n (2^{- \frac{1}{4}})} (1 + c o s (2 α)) d α = \frac{1}{2} a r c s i n (2^{- \frac{1}{4}}) + \frac{1}{4} s i n (2 a r c s i n (2^{- \frac{1}{4}}))$ $\int_{0}^{2^{- \frac{1}{4}}} \frac{\sqrt{1 - t^{2}}}{1 - t} =_{t = s i n α} \int_{0}^{a r c s i n (2^{- \frac{1}{4}})} \frac{c o s α}{1 - s i n α} c o s α d α$ $= \int_{0}^{a r c s i n (2^{- \frac{1}{4}})} \frac{1 - {s i n}^{2} α}{1 - s i n α} d α = \int_{0}^{a r c s i n (2^{- \frac{1}{4}})} (1 + s i n α) d α$ $= a r c s i n (2^{- \frac{1}{4}}) + {[- c o s α]}_{0}^{a r s i n (2^{- \frac{1}{4}})}$ $= a r c s i n (2^{- \frac{1}{4}}) + 1 - c o s (a r c s i n (2^{- \frac{1}{4}})) s o t h e v a l u e o f I i s k n o w n .$
Commented by behi83417@gmail.com last updated on 04/Jul/18

$d e a r p r o . a b d o! t h a n k y o u f o r h a r d w o r k$ $g o d b l e s s y o u s i r . w h a t d o y o u t h i n k$ $You can't use 'macro parameter character #' in math mode$
Commented by math khazana by abdo last updated on 04/Jul/18

$f o r x = y w e g e t f^{2} (4 x) = \frac{1 - f (2 x)}{{(1 - f (x))}^{2}}$ $x = \frac{1}{8} \Rightarrow f^{2} (\frac{1}{2}) = {(\frac{\frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}})}^{2} = {(\frac{1}{\sqrt{2} - 1})}^{2} = \frac{1}{3 - 2 \sqrt{2}}$ $\frac{1 - f (2 x)}{{(1 - f (x))}^{2}} = \frac{1 - f (\frac{1}{4})}{{(1 - f (\frac{1}{8}))}^{2}} b u t f (\frac{1}{4}) = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$ $f (\frac{1}{8}) = \frac{\frac{\sqrt{7}}{\sqrt{8}}}{1 - \frac{1}{\sqrt{8}}} = \frac{\sqrt{7}}{2 \sqrt{2} - 1} \Rightarrow 1 - f (\frac{1}{8})$ $= 1 - \frac{\sqrt{7}}{2 \sqrt{2} - 1} = \frac{2 \sqrt{2} - 1 - \sqrt{7}}{2 \sqrt{2}} \Rightarrow {(1 - f (\frac{1}{8}))}^{2}$ $= {(\frac{2 \sqrt{2} - 1 - \sqrt{7}}{2 \sqrt{2}})}^{2} a n d i t s c l e a r t h a t$ $f^{2} (4 x) = \frac{1 - f (2 x)}{{(1 - f (x))}^{2}} i s n o t t r u e w i t h x = \frac{1}{8}$ $s o t b e e q u a l i t y i s n o t c o r r e c t .$
Commented by math khazana by abdo last updated on 04/Jul/18

$n e v e r m i n d s i r B e h i .$
	Answered by MJS last updated on 04/Jul/18

	$y = \frac{\sqrt{1 - x}}{1 - \sqrt{x}} \Rightarrow x \in [0; 1 [\land y \in [1; + \infty [$ $f^{- 1} (x) = {(\frac{x^{2} - 1}{x^{2} + 1})}^{2} with x \in [1; + \infty [\land y \in [0; 1 [$ $\int \frac{\sqrt{1 - x}}{1 - \sqrt{x}} d x = \int \frac{(1 + \sqrt{x}) \sqrt{1 - x}}{(1 + \sqrt{x}) (1 - \sqrt{x})} d x =$ $= \int \frac{\sqrt{1 - x} + \sqrt{x - x^{2}}}{1 - x} d x = \int \frac{d x}{\sqrt{1 - x}} + \int \sqrt{\frac{x}{1 - x}} d x =$ $[t = \sqrt{x} \to d x = 2 \sqrt{x} d t]$ $2 \int \frac{t}{\sqrt{1 - t^{2}}} d t + 2 \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t =$ $2 \int \frac{t}{\sqrt{1 - t^{2}}} d t = - 2 \sqrt{1 - t^{2}} = - 2 \sqrt{1 - x}$ $2 \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t =$ $[u = \arcsin t \to d t = \sqrt{1 - t^{2}} d u]$ $= 2 \int \sin^{2} u d u = u - \sin u \cos u =$ $= \arcsin t - t \sqrt{1 - t^{2}} = \arcsin \sqrt{x} - \sqrt{x} \sqrt{1 - x}$ $= \arcsin \sqrt{x} - (2 + \sqrt{x}) \sqrt{1 - x} + C$ $\underset{0}{\int^{\frac{\sqrt{2}}{2}}} \frac{\sqrt{1 - x}}{1 - \sqrt{x}} d x \approx 1.46146$ $but \underset{0}{\int^{1}} \frac{\sqrt{1 - x}}{1 - \sqrt{x}} d x = 2 + \frac{π}{2}$ $(4) is not true for the given function$
	Commented by behi83417@gmail.com last updated on 04/Jul/18

	$t h a n k y o u s o m u c h d e a r M J S!$ $g o d b l e s s y o u s i r .$ $You can't use 'macro parameter character #' in math mode$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum