Bộ 10 đề thi cuối kì 1 Toán 11 Cánh diều có đáp án - Đề 8

\(\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b\).

\(\cos \left( {a - b} \right) = \cos a\cos b - \sin a\sin b.\)

\(\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b\).

\(\sin \left( {a + b} \right) = \sin a\cos b + \cos a\sin b.\)

\(D = \mathbb{R}\backslash \left\{ 0 \right\}.\)

\(D = \mathbb{R}\backslash \left\{ {k2\pi |k \in \mathbb{Z}} \right\}.\)

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

\(D = \mathbb{R}\backslash \left\{ {0;\pi } \right\}.\)

\(y = - \sin x.\)

\(y = \cos x - \sin x.\)

\[y = \cos x + {\sin ^2}x.\]

\(y = \cos x\sin x.\)

\[\left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right).\]

\[\left[ \begin{array}{l}x = \frac{{3\pi }}{4} + k2\pi \\x = \frac{{ - 3\pi }}{4} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right).\]

\[\left[ \begin{array}{l}x = \frac{{5\pi }}{4} + k2\pi \\x = \frac{{ - 5\pi }}{4} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right).\]

\[\left[ \begin{array}{l}x = \frac{{5\pi }}{4} + k2\pi \\x = \frac{{ - 3\pi }}{4} + k2\pi \end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right).\]

\({u_3} = - \frac{8}{3}.\)

\({u_3} = 2.\)

\({u_3} = - 2.\)

\({u_3} = \frac{8}{3}.\)

\({u_n} = 3{n^2} + 2017.\)

\({u_n} = 3n + 2008.\)

\({u_n} = {3^n}.\)

\({u_n} = {\left( { - 3} \right)^{n + 1}}.\)

\(25.\)

\(15.\)

\(12.\)

\(31.\)

\(1;\,\,2;\,\,4;\,\,8;...\)

\(3;\,\,{3^2};\,\,{3^3};\,\,{3^4};...\)

\(4;\,\,2;\,\,1;\,\,\frac{1}{2};\,\,\frac{1}{4};...\)

\(\frac{1}{\pi };\,\,\frac{1}{{{\pi ^2}}};\,\,\frac{1}{{{\pi ^4}}};\,\,\frac{1}{{{\pi ^6}}};...\)

\[ - 2;\,\,10;\,\,50;\,\, - 250.\]

\[ - 2;\,\,10;\,\, - 50;\,\,250.\]

\[ - 2;\,\, - 10;\,\, - 50;\,\, - 250.\]

\[ - 2;\,\,10;\,\,50;\,\,250.\]

\[\lim {u_n} = c\] (\(c\) là hằng số).

\[\lim {q^n} = 0\] (\[\left| q \right| > 1\]).

\[\lim \frac{1}{n} = 0.\]

\[\lim \frac{1}{{{n^k}}} = 0\] (\(k > 1\)).

\(\frac{3}{2}.\)

0.

\(\frac{6}{5}.\)

\( - 6.\)

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} = + \infty .\)

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{x} = - \infty .\)

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{{x^5}}} = + \infty .\)

\(\mathop {\lim }\limits_{x \to {0^ + }} \frac{1}{{\sqrt x }} = + \infty .\)

\(L = 1.\)

\(L = \frac{1}{2}.\)

\(L = - \frac{1}{2}.\)

\(L = - \frac{3}{4}.\)

Hàm số liên tục tại \(x = - 1\).

Hàm số liên tục tại \(x = 0\).

Hàm số liên tục tại \(x = 1\).

Hàm số liên tục tại \(x = \frac{1}{2}\).

\(y = {x^3} - x.\)

\(y = \cot x.\)

\(y = \frac{{2x - 1}}{{x - 1}}.\)

\(y = \sqrt {{x^2} - 1} .\)

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Đường thẳng \(AB\).

Đường thẳng \(AD\).

Đường thẳng \(AC\).

Đường thẳng \(SA\).

\(MN{\rm{//}}\left( {SBC} \right).\)

\(MN{\rm{//}}BD.\)

\(MN{\rm{//}}\left( {SAB} \right).\)

\(MN\) cắt \(BC.\)