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](Q85130.png)
$$\mathrm{given}\:\mathrm{f}\left(\mathrm{x}\right)=\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\mathrm{sin}\:\mathrm{x}\:+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\mathrm{cos}\:\mathrm{x} \\ $$$$\mathrm{find}\:\mathrm{masimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{function} \\ $$$$\left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \\ $$
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](Q85134.png)
$$\mathrm{f}\left({x}\right)\:=\:{a}\mathrm{sin}\:{x}\:+\:\mathrm{bcos}\:{x}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{form}\:\:\mathrm{Rsin}\left({x}\:+\:\alpha\right)\:\mathrm{where}\:\:\mathrm{R}>\mathrm{0}\:\mathrm{and}\:\:\:\:\mathrm{0}^{°} <\alpha<\mathrm{90}° \\ $$$$\mathrm{if}\:\mathrm{R}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left({x}\right)\:\mathrm{then}\:\:{R}^{\mathrm{2}} \:\mathrm{will}\:\mathrm{be}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left[\mathrm{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\sqrt{\mathrm{2}}\:+\mathrm{1}\right)\mathrm{sin}\:{x}\:+\:\left(\sqrt{\mathrm{2}}\:−\:\mathrm{1}\right)\mathrm{cos}\:{x}\:=\:\mathrm{Rsin}\:{x}\:\mathrm{cos}\alpha\:+\:{R}\mathrm{cos}\:{x}\:\mathrm{sin}\alpha \\ $$$$\Rightarrow\:\:{R}\:\mathrm{cos}\alpha\:=\:\left(\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right).......\left({i}\right) \\ $$$$\:\:\:\:\:\:{R}\mathrm{sin}\alpha\:=\:\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)..........\left({ii}\right) \\ $$$$\:\:\:{R}^{\mathrm{2}} \:=\:\left(\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right)^{\mathrm{2}} \:+\:\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{R}^{\mathrm{2}} \:=\:\mathrm{2}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{1}\:+\:\mathrm{2}\:−\mathrm{2}\sqrt{\mathrm{2}\:}\:+\:\mathrm{1}\:=\:\mathrm{4}\:+\:\mathrm{2}\:=\:\mathrm{6} \\ $$$$\Rightarrow\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} \:=\:\mathrm{6} \\ $$
Commented by john santu last updated on 19/Mar/20
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by Rio Michael last updated on 19/Mar/20
$$\mathrm{welcome}\:\mathrm{sir} \\ $$
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](Q85139.png)
$$\mathrm{if}\:\mathrm{my}\:\mathrm{method}\: \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{k}\:\mathrm{cos}\:\left(\mathrm{x}−\theta\right)\: \\ $$$$\mathrm{with}\:\mathrm{k}\:=\:\sqrt{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{so}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of}\:\left[\mathrm{f}\left(\mathrm{x}\right)\right]^{\mathrm{2}} \:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{k}\:=\:\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\:+\:\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:=\:\mathrm{6} \\ $$
Commented by john santu last updated on 19/Mar/20
$$\mathrm{good}.\:\mathrm{thank}\left[\mathrm{you}\right. \\ $$