Question Number 170155 by cortano1 last updated on 17/May/22
$$\:\:{Given}\:{f}\left({x}\right)={x}\sqrt{\mathrm{1}−{x}+\sqrt{\mathrm{1}−{x}}} \\ $$$$\:{where}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1} \\ $$$$\:{find}\:{max}\:{f}\left({x}\right) \\ $$
Answered by Mathspace last updated on 18/May/22
![x∈[0,1] ⇒x=cosθ ⇒ f(x)=cosθ(√(1−cosθ+(√(1−cosθ)))) =cosθ(√(2sin^2 ((θ/2))+(√(2sin^2 (θ/2))))) =cosθ(√(2sin^2 ((θ/2))+(√2)sin((θ/2)))) =cosθ(√((√2)sin((θ/2))))(√((√2)sin((θ/2))+1)) ≤(^4 (√2))(√(1+(√2))) ....](https://www.tinkutara.com/question/Q170201.png)
$${x}\in\left[\mathrm{0},\mathrm{1}\right]\:\Rightarrow{x}={cos}\theta\:\Rightarrow \\ $$$${f}\left({x}\right)={cos}\theta\sqrt{\mathrm{1}−{cos}\theta+\sqrt{\mathrm{1}−{cos}\theta}} \\ $$$$={cos}\theta\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)+\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}} \\ $$$$={cos}\theta\sqrt{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)+\sqrt{\mathrm{2}}{sin}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$={cos}\theta\sqrt{\sqrt{\mathrm{2}}{sin}\left(\frac{\theta}{\mathrm{2}}\right)}\sqrt{\sqrt{\mathrm{2}}{sin}\left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{1}} \\ $$$$\leqslant\left(^{\mathrm{4}} \sqrt{\mathrm{2}}\right)\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}} \\ $$$$…. \\ $$
Commented by cortano1 last updated on 18/May/22
$${wrong} \\ $$