Đề giữa kỳ 1 Toán 11 năm 2024 – 2025 trường THPT Việt Nam – Ba Lan – Hà Nội

$D = {\mathbb{R}}\backslash\left\{ k\pi;\dfrac{\pi}{6} + k\pi \middle| k \in {\mathbb{Z}} \right\}.$

$D = {\mathbb{R}}\backslash\left\{ \dfrac{\pi}{6} + k\pi \middle| k \in {\mathbb{Z}} \right\}.$

$D = {\mathbb{R}}\backslash\left\{ \dfrac{\pi}{2} + k\pi;\dfrac{\pi}{6} + k\pi \middle| k \in {\mathbb{Z}} \right\}.$

$D = {\mathbb{R}}\backslash\left\{ k\pi \middle| k \in {\mathbb{Z}} \right\}.$

$u_{3} = 5.$

$u_{3} = 3.$

$u_{3} = - 2.$

$u_{3} = 1.$

$u_{n} = 2 + 3n$.

$u_{n} = 5n$.

$u_{n} = 3 + 2n$.

$u_{n} = 1 + 4n$.

$u_{n} = \left( {- 1} \right)^{n + 1}\sin n,\,\forall n \in {\mathbb{N}}^{\ast}.$

$u_{n} = \dfrac{n}{n^{2} + 1},\,\forall n \in {\mathbb{N}}^{\ast}.$

$u_{n} = \left( {- 1} \right)^{2n}\left( {5^{n} + 1} \right),\,\forall n \in {\mathbb{N}}^{\ast}.$

$u_{n} = \dfrac{1}{n + 2},\,\forall n \in {\mathbb{N}}^{\ast}.$

$y = \cos 2x.$

$y = \tan^{2}x.$

$y = x^{3} + 1.$

$y = \sin 2x.$

$9$.

$40$.

$30$.

$35$.

$240^{\text{o}}$.

$270^{\text{o}}$.

$135^{\text{o}}$.

$72^{\text{o}}$.

$\left\{ \begin{array}{l} {u_{1}^{} = 2} \\ {u_{n + 1}^{} = u_{n}^{} - 3,\,\forall n \in {\mathbb{N}}^{*}} \end{array} \right.$.

$u_{n} = 1 + 3^{n}$.

$u_{n} = \left( {n + 3} \right)^{2} - n^{2}$.

$u_{n} = 2n + 5$.

$102m$.

$92,45m$.

$96,50m$.

$94,25m$.

$4,\, 6,\, 9,\, 13,\, 18.$

$4,\, 16,\, 32,\, 64,\, 128.$

$4,\, 5,\, 7,\, 10,\, 14.$

$4,\, 5,\, 6,\, 7,\, 8.$

$x \in \left\lbrack {- \dfrac{\pi}{2};\dfrac{\pi}{2}} \right\rbrack.$

$x \in \left( {- \dfrac{3\pi}{2}; - \dfrac{\pi}{2}} \right).$

$x \in \left( {- \dfrac{\pi}{2};\dfrac{\pi}{2}} \right).$

$x \in \left( {- \pi;0} \right).$

$\sin a.\sin b = \dfrac{1}{2}\left\lbrack {\cos\left( {a - b} \right) - \cos\left( {a + b} \right)} \right\rbrack.$

$\sin a.\cos b = \dfrac{1}{2}\left\lbrack {\sin\left( {a - b} \right) + \sin\left( {a + b} \right)} \right\rbrack.$

$\sin a.\cos b = \dfrac{1}{2}\left\lbrack {\sin\left( {a - b} \right) - \sin\left( {a + b} \right)} \right\rbrack.$

$\cos a.\cos b = \dfrac{1}{2}\left\lbrack {\cos\left( {a - b} \right) + \cos\left( {a + b} \right)} \right\rbrack.$