Страница 289 номер 971, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(f(x) = 2\sin x + \sin 2x\) на отрезке \(\left[0;\ \frac{3}{2}\pi\right]\)
\(f'(x) = 2\cos x + 2\cos 2x\)
\(2\cos x + 2\cos 2x = 0\)
\(2\cos x + 4\cos^{2} x - 2 = 0\)
\(\cos x = z\)
\(2z^{2} + z - 1 = 0\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 8}}{4}\)
\(z_{1} = \dfrac{1}{2}, z_{2} = -1\)
\(x_{1} = \pm \dfrac{\pi}{3} + 2\pi n, x_{2} = \pi + 2\pi n, n \in Z.\)

Отрезку \(\left[0;\ \frac{3}{2}\pi\right]\) принадлежат только \(x = \frac{\pi}{3}\) (\(n = 0\)) и \(x = \pi\) (\(n = 0\));

\(f(0) = 0\)
\(f\left(\dfrac{3\pi}{2}\right) = -2\)
\(f\left(\dfrac{\pi}{3}\right) = \sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2}\)
\(f(\pi) = 0;\)

Наибольшее: \(f\left(\frac{\pi}{3}\right) = \frac{3\sqrt{3}}{2}\)

Наименьшее: \(f\left(\frac{3\pi}{2}\right) = -2\);

2) \(f(x) = 2\cos x + \sin 2x\) на отрезке \([0;\ \pi]\)
\(f'(x) = -2\sin x + 2\cos 2x\)
\(-2\sin x + 2(1 - 2\sin^{2} x) = 0\)
\(\sin x = z, -4z^{2} - 2z + 2 = 0\)
\(2z^{2} + z - 1 = 0\)
\(z_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 8}}{4}\)
\(z_{1} = \frac{1}{2}\), \(z_{2} = -1\);
\(x_{1} = (-1)^{n}\frac{\pi}{6} + \pi n\): \(\frac{\pi}{6}\) (\(n = 0\)), \(\frac{5\pi}{6}\) (\(n = 1\)); \(x_{2} = -\frac{\pi}{2} + 2\pi n\) – не принадлежит отрезку \([0;\ \pi]\);
\(f(0) = 2, f\left(\dfrac{\pi}{6}\right) = \sqrt{3} + \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2};\)
\(f\left(\dfrac{5\pi}{6}\right) = -\dfrac{3\sqrt{3}}{2}; f(\pi) = -2;\)

Наибольшее: \(f\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{2}\)

Наименьшее: \(f\left(\frac{5\pi}{6}\right) = -\frac{3\sqrt{3}}{2}\).