Задача 53469 Найти множество значений...
Условие
Решение
[m]\left\{\begin{matrix}4x-1-2x^2 ≥0 \\ cos(\frac{2\pi}{3}\cdot (\sqrt{4x-1-2x^2}-\frac{1}{2})) ≥ 0\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}2x^2-4x+1 ≤ 0 \\ -\frac{\pi}{2}+2\pi n ≤ \frac{2\pi}{3}\cdot (\sqrt{4x-1-2x^2}-\frac{1}{2}) ≤ \frac{\pi}{2}+2\pi n, n \in Z\end{matrix}\right.[/m]
2x^2-4x+1=0; D=(-4)^2-4*2*1=8
x_(1)=2-sqrt(2); x_(2)=2+sqrt(2).
[m]\left\{\begin{matrix}2-\sqrt{2}≤x ≤ 2+\sqrt{2} \\ -\frac{\pi}{2}+2\pi n ≤ \frac{2\pi}{3}\cdot (\sqrt{4x-1-2x^2}-\frac{1}{2}) ≤ \frac{\pi}{2}+2\pi n, n \in Z\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}2-\sqrt{2}≤x ≤ 2+\sqrt{2} \\ -\frac{3}{4}+3n ≤\sqrt{4x-1-2x^2}-\frac{1}{2} ≤ \frac{3}{4}+3n, n \in Z\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}2-\sqrt{2}≤x ≤ 2+\sqrt{2} \\ -\frac{3}{4}+\frac{1}{2}+3n ≤\sqrt{4x-1-2x^2} ≤ \frac{3}{4}+\frac{1}{2}+3n, n \in Z\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}2-\sqrt{2}≤x ≤ 2+\sqrt{2} \\ -\frac{1}{4}+3n ≤\sqrt{4x-1-2x^2} ≤ \frac{5}{4}+3n, n \in Z\end{matrix}\right.[/m]
Второе неравенство:
[m]-\frac{1}{4}+3n ≤\sqrt{4x-1-2x^2} ≤ \frac{5}{4}+3n, n \in Z[/m]
равносильно системе:
[m]\left\{\begin{matrix}\sqrt{4x-1-2x^2} ≤ \frac{5}{4}+3n, n \in Z \\\sqrt{4x-1-2x^2} \geq-\frac{1}{4}+3n, n \in Z\end{matrix}\right.[/m]
верно при n=0 для любых х
ОДЗ: [m][2-\sqrt{2};2+\sqrt{2}][/m]
При x ∈ [m][2-\sqrt{2};2+\sqrt{2}][/m]
[m]0\leq\sqrt{4x-1-2x^2}\leq 1[/m], так как квадратный трехчлен принимает наибольшее значение в вершине, т.е в точке x=1.
(4x-1-2x^2)=-2(x^2-2x+1)+1=-2*(x-1)^2+1 ≤ 1
Тогда
[m]\sqrt{4x-1-2x^2}-\frac{1}{2}\leq\frac{1}{2}[/m]
[m]cos(\frac{2\pi}{3}\cdot (\sqrt{4x-1-2x^2}-\frac{1}{2}))\leq cos\frac{2\pi}{3}\frac{1}{2}=cos\frac{\pi}{3}=\frac{1}{2}[/m]
О т в е т. [m] [0;\frac{\sqrt{2}}{2}] [/m]