Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 1201

\[\boxed{\mathbf{1201}\mathbf{.}}\]

\[1)\ a\sin x + (1 + a)\cos x = \sqrt{5}\]

\[R = \sqrt{a^{2} + (1 + a)^{2}} =\]

\[= \sqrt{2a^{2} + 2a + 1}\]

\[R\left( \frac{a}{R}\sin x + \frac{1 + a}{R}\cos x \right) = \sqrt{5}\]

\[\cos 4 = \frac{a}{R};\ \ \sin 4 = \frac{1 + a}{R};\]

\[R \cdot \sin(x + 4) = \sqrt{5}.\]

\[Это\ уравнение\ имеет\ корни,\ \]

\[если\ \ - 1 \leq \frac{\sqrt{5}}{R} \leq 1;\ \ \]

\[так\ как\ R > 0 \rightarrow R \geq \sqrt{5}.\]

\[\sqrt{2a^{2} + 2a + 1} \geq \sqrt{5}\]

\[2a^{2} + 2a + 1 \geq 5\]

\[2a^{2} + 2a - 4 \geq 0\]

\[a^{2} + a - 2 \geq 0\]

\[D = 1 + 8 = 9\]

\[a_{1} = \frac{- 1 - 3}{2} = - 2;\ \ \]

\[a_{2} = \frac{- 1 + 3}{2} = 1.\]

\[(a + 2)(a - 1) \geq 0\]

\[a \leq - 2;\ \ a \geq 1.\]

\[Ответ:при\ a \leq - 2;\ \ a \geq 1.\]

\[2)\ a\cos x + (1 - a)\sin x = \sqrt{5}\]

\[R = \sqrt{a^{2} + (1 - a)^{2}} =\]

\[= \sqrt{2a^{2} - 2a + 1}\]

\[R\left( \frac{a}{R}\sin x + \frac{1 - a}{R}\sin x \right) = \sqrt{5}\]

\[\cos 4 = \frac{a}{R};\ \sin 4 = \frac{1 - a}{R}\]

\[R\sin{(x + 4)} = \sqrt{5}\]

\[Это\ уравнение\ имеет\ корни,\ \]

\[если\ \ - 1 \leq \frac{\sqrt{5}}{R} \leq 1;\ \ \]

\[так\ как\ R > 0 \rightarrow R \geq \sqrt{5}.\]

\[\sqrt{2a^{2} - 2a + 1} \geq \sqrt{5}\ \]

\[2a^{2} - 2a + 1 \geq 5\]

\[2a^{2} - 2a - 4 \geq 0\]

\[a^{2} - a - 2 \geq 0\]

\[D = 1 + 8 = 9\]

\[a_{1} = \frac{1 + 3}{2} = 2;\ \]

\[\ a_{2} = \frac{1 - 3}{2} = - 1\]

\[(a + 1)(a - 2) \geq 0\]

\[a \leq - 1;\ \ a \geq 2\]

\[Ответ:при\ a \leq - 1;\ \ a \geq 2.\]