If , x∈ [0 , 1] , ∣ (√(1−x^( 2) )) −ax−b ∣≤ (((√2) −1)/2) find the values of ( a , b ) a , b∈ R.


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Question Number 173829 by mnjuly1970 last updated on 19/Jul/22

$I f, x \in [0, 1]$ $, ∣ \sqrt{1 - x^{2}} - a x - b ∣⩽ \frac{\sqrt{2} - 1}{2}$ $f i n d t h e v a l u e s o f (a, b)$ $a, b \in R .$
Commented by kaivan.ahmadi last updated on 19/Jul/22

$w e c a n f i n d b, e a s i l y$ $l e t f (x) =∣ \sqrt{1 - x^{2}} - a x - b ∣$ $f (0) =∣ 1 - b ∣⩽ \frac{\sqrt{2} - 1}{2}$ $\Rightarrow \frac{1 - \sqrt{2}}{2} ⩽ 1 - b ⩽ \frac{\sqrt{2} - 1}{2} \Rightarrow$ $\Rightarrow \frac{1 - \sqrt{2}}{2} ⩽ b - 1 ⩽ \frac{\sqrt{2} - 1}{2}$ $\Rightarrow \frac{3 - \sqrt{2}}{2} ⩽ b ⩽ \frac{1 + \sqrt{2}}{2}$
	Answered by mahdipoor last updated on 19/Jul/22
	$f(x)=(√(1−x^2 ))−ax−b (df/dx)=((−x)/( (√(1−x^2 ))))−a=0 or ∄ critical point in [0,1] ⇒ 0,1, (if a<0) ((−a)/( (√(1+a^2 )))) ∣f∣≤(((√2)−1)/2)=c ⇒ −c≤f≤c { ((x=0 ⇒ −c≤1−b≤c)),((x=1 ⇒ −c≤a+b≤c)),((x=((−a)/( (√(1+a^2 )))) ⇒ −c≤(√(1+a^2 ))−b≤c)) :} ⇒^((I)) { ((1−c≤b≤1+c (I))),((1−2c≤a≤2c+1 )),((−2(√(c^2 +c))≤a≤2(√(c^2 +c)))) :} ⇒ { ((1−c≤b≤1+c)),((−2(√(c^2 −c))≤1−2c≤a≤2(√(c^2 +c))≤2c+1)) :} a<0 & { ((((3−(√2))/2)≤b≤(((√2)+1)/2))),((2−(√2)≤a≤(√2))) :} ⇒ ∄ if a≥0 : critical point in [0,1] ⇒ 0,1 { ((x=0 ⇒ −c≤1−b≤c ⇒1−c≤b≤1+c)),((x=1 ⇒ −c≤a+b≤c ⇒1−2c≤a≤2c+1 )) :} ⇒ a≥0 & { ((((3−(√2))/2)≤b≤(((√2)+1)/2))),((2−(√2)≤a≤(√2))) :} ⇒⇒ { ((((3−(√2))/2)≤b≤(((√2)+1)/2))),((2−(√2)≤a≤(√2))) :}$
	$f (x) = \sqrt{1 - x^{2}} - a x - b$ $\frac{d f}{d x} = \frac{- x}{\sqrt{1 - x^{2}}} - a = 0 o r ∄$ $c r i t i c a l p o i n t i n [0, 1] \Rightarrow 0, 1, (i f a < 0) \frac{- a}{\sqrt{1 + a^{2}}}$ $∣ f ∣⩽ \frac{\sqrt{2} - 1}{2} = c \Rightarrow - c ⩽ f ⩽ c$ ${\begin{cases} x = 0 \Rightarrow - c ⩽ 1 - b ⩽ c \\ x = 1 \Rightarrow - c ⩽ a + b ⩽ c \\ x = \frac{- a}{\sqrt{1 + a^{2}}} \Rightarrow - c ⩽ \sqrt{1 + a^{2}} - b ⩽ c \end{cases}$ $\overset{(I)}{\Rightarrow} {\begin{cases} 1 - c ⩽ b ⩽ 1 + c (I) \\ 1 - 2 c ⩽ a ⩽ 2 c + 1 \\ - 2 \sqrt{c^{2} + c} ⩽ a ⩽ 2 \sqrt{c^{2} + c} \end{cases}$ $\Rightarrow {\begin{cases} 1 - c ⩽ b ⩽ 1 + c \\ - 2 \sqrt{c^{2} - c} ⩽ 1 - 2 c ⩽ a ⩽ 2 \sqrt{c^{2} + c} ⩽ 2 c + 1 \end{cases}$ $a < 0 & {\begin{cases} \frac{3 - \sqrt{2}}{2} ⩽ b ⩽ \frac{\sqrt{2} + 1}{2} \\ 2 - \sqrt{2} ⩽ a ⩽ \sqrt{2} \end{cases} \Rightarrow ∄$ $i f a ⩾ 0 :$ $c r i t i c a l p o i n t i n [0, 1] \Rightarrow 0, 1$ ${\begin{cases} x = 0 \Rightarrow - c ⩽ 1 - b ⩽ c \Rightarrow 1 - c ⩽ b ⩽ 1 + c \\ x = 1 \Rightarrow - c ⩽ a + b ⩽ c \Rightarrow 1 - 2 c ⩽ a ⩽ 2 c + 1 \end{cases}$ $\Rightarrow a ⩾ 0 & {\begin{cases} \frac{3 - \sqrt{2}}{2} ⩽ b ⩽ \frac{\sqrt{2} + 1}{2} \\ 2 - \sqrt{2} ⩽ a ⩽ \sqrt{2} \end{cases}$ $\Rightarrow\Rightarrow {\begin{cases} \frac{3 - \sqrt{2}}{2} ⩽ b ⩽ \frac{\sqrt{2} + 1}{2} \\ 2 - \sqrt{2} ⩽ a ⩽ \sqrt{2} \end{cases}$
	Commented by mnjuly1970 last updated on 20/Jul/22

	$z e n d e h b a s h i d o s t a d$ $s e p a s . . .$
	Commented by mahdipoor last updated on 20/Jul/22

	$s h a r m a n d e m i k o n i d m a n o,$ $m a n o s t a d n i s t a m . . . :)$
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