Menu Close

given-f-x-2-1-sin-x-2-1-cos-x-find-masimum-value-of-function-f-x-2-




Question Number 85130 by john santu last updated on 19/Mar/20
given f(x)= ((√2)+1)sin x +((√2)−1)cos x  find masimum value of function  [f(x)]^2
givenf(x)=(2+1)sinx+(21)cosxfindmasimumvalueoffunction[f(x)]2
Answered by Rio Michael last updated on 19/Mar/20
f(x) = asin x + bcos x can be expressed in  the form  Rsin(x + α) where  R>0 and    0^° <α<90°  if R is the maximum value of f(x) then  R^2  will be the  maximum value of [f(x)]^2   ⇒ ((√2) +1)sin x + ((√2) − 1)cos x = Rsin x cosα + Rcos x sinα  ⇒  R cosα = ((√2) + 1).......(i)        Rsinα = ((√2) −1)..........(ii)     R^2  = ((√2) + 1)^2  + ((√2) −1)^2       R^2  = 2 + 2(√2) + 1 + 2 −2(√(2 )) + 1 = 4 + 2 = 6  ⇒ maximum value of [f(x)]^2  = 6
f(x)=asinx+bcosxcanbeexpressedintheformRsin(x+α)whereR>0and0°<α<90°ifRisthemaximumvalueoff(x)thenR2willbethemaximumvalueof[f(x)]2(2+1)sinx+(21)cosx=Rsinxcosα+RcosxsinαRcosα=(2+1).(i)Rsinα=(21).(ii)R2=(2+1)2+(21)2R2=2+22+1+222+1=4+2=6maximumvalueof[f(x)]2=6
Commented by john santu last updated on 19/Mar/20
thank you
thankyou
Commented by Rio Michael last updated on 19/Mar/20
welcome sir
welcomesir
Answered by jagoll last updated on 19/Mar/20
if my method   f(x) = k cos (x−θ)   with k = (√(((√2)−1)^2 +((√2)+1)^2 ))  so max value of [f(x)]^2  equal to  k = 3−2(√2) + 3+2(√2) = 6
ifmymethodf(x)=kcos(xθ)withk=(21)2+(2+1)2somaxvalueof[f(x)]2equaltok=322+3+22=6
Commented by john santu last updated on 19/Mar/20
good. thank[you
good.thank[you

Leave a Reply

Your email address will not be published. Required fields are marked *