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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
If , x∈ [0 , 1] , ∣ (√(1−x^( 2) )) −ax−b ∣≤ (((√2) −1)/2) find the values of ( a , b ) a , b∈ R.
If,x∈[0,1],∣1−x2−ax−b∣⩽2−12findthevaluesof(a,b)a,b∈R.
Commented by kaivan.ahmadi last updated on 19/Jul/22
we can find b, easily let f(x)=∣(√(1−x^2 ))−ax−b∣ f(0)=∣1−b∣≤(((√2)−1)/2) ⇒((1−(√2))/2)≤1−b≤(((√2)−1)/2)⇒ ⇒((1−(√2))/2)≤b−1≤(((√2)−1)/2) ⇒((3−(√2))/2)≤b≤((1+(√2))/2)
wecanfindb,easilyletf(x)=∣1−x2−ax−b∣f(0)=∣1−b∣⩽2−12⇒1−22⩽1−b⩽2−12⇒⇒1−22⩽b−1⩽2−12⇒3−22⩽b⩽1+22
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)=1−x2−ax−bdfdx=−x1−x2−a=0or∄criticalpointin[0,1]⇒0,1,(ifa<0)−a1+a2∣f∣⩽2−12=c⇒−c⩽f⩽c{x=0⇒−c⩽1−b⩽cx=1⇒−c⩽a+b⩽cx=−a1+a2⇒−c⩽1+a2−b⩽c⇒(I){1−c⩽b⩽1+c(I)1−2c⩽a⩽2c+1−2c2+c⩽a⩽2c2+c⇒{1−c⩽b⩽1+c−2c2−c⩽1−2c⩽a⩽2c2+c⩽2c+1a<0&{3−22⩽b⩽2+122−2⩽a⩽2⇒∄ifa⩾0:criticalpointin[0,1]⇒0,1{x=0⇒−c⩽1−b⩽c⇒1−c⩽b⩽1+cx=1⇒−c⩽a+b⩽c⇒1−2c⩽a⩽2c+1⇒a⩾0&{3−22⩽b⩽2+122−2⩽a⩽2⇒⇒{3−22⩽b⩽2+122−2⩽a⩽2
Commented by mnjuly1970 last updated on 20/Jul/22
zendeh bashid ostad sepas...
zendehbashidostadsepas…
Commented by mahdipoor last updated on 20/Jul/22
sharmande mikonid mano , man ostad nistam ... :)
sharmandemikonidmano,manostadnistam…:)

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