Đề kiểm tra 1 tiết chương 1: Hàm số lượng giác và phương trình lượng giác - Đề số 3

\(M\left( {0;1} \right)\)

\(N\left( {\dfrac{\pi }{2};0} \right)\)

\(P\left( {1;0} \right)\)

\( - \dfrac{{5{\pi ^2}}}{{12}}\)           

\(\dfrac{{5{\pi ^2}}}{{144}}\)

\(\dfrac{{{\pi ^2}}}{{12}}\) 

\(y = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)\)

\(x = k\pi \left( {k \in Z} \right)\)

\(y = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\)

\(m =  - 3\)

\(m =  - 2\)      

\(m = 3\)

\(\dfrac{\pi }{3}\)

\(\dfrac{{4\pi }}{3}\)

\(\dfrac{{7\pi }}{3}\)

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x = \pi  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)    

\(\left[ \begin{array}{l}x = \alpha  + k2\pi \\x =  - \alpha  + k2\pi \end{array} \right.\left( {k \in Z} \right)\)      

\(\left[ \begin{array}{l}x = \alpha  + k\pi \\x =  - \alpha  + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\cos x \ne 1 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\)

\(\cos x \ne  - 1 \Leftrightarrow x \ne  - \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\)    

\(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{8} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{8} + k\pi ,k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{4} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\) 

\(\max y =  - 2,\min y = 4\)

\(\max y = 2,\min y = 4\)

\(\max y =  - 2,\min y = 3\)

$0$     

$1$     

$4$

\(\left( {\dfrac{\pi }{2} + k2\pi ;\dfrac{{3\pi }}{2} + k2\pi } \right)\)

\(\left( { - \pi  + k2\pi ;k2\pi } \right)\)

\(R\)

$m \le 6$

$m \le 24$.

$m \le 3$.

\(x = \dfrac{\pi }{2} + k2\pi \).

\(x = k2\pi \).

\(x = \dfrac{\pi }{6} + k2\pi \).

\(y = {x^2} - \sin x\)

\(y = {x^2} + \sin x\)

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

\(x = \dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)\)

\(x = k2\pi \left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\)

$x =  - \dfrac{\pi }{2} + k2\pi $.

$x = \dfrac{\pi }{2} + k\pi $.

$x = k\pi $.

\(x = \dfrac{\pi }{2} + k\pi \,\,\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \dfrac{\pi }{4} + k\pi \\x = \arctan \left( { - 2} \right) + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)\)

\(x =  - \dfrac{\pi }{4} + k\pi \,\,\left( {k \in Z} \right)\)

\(1\)

\(3\)

Vô nghiệm

\(\left[ \begin{array}{l}x =  - \dfrac{\pi }{6} + \dfrac{{k2\pi }}{9}\\x = \dfrac{{7\pi }}{6} + \dfrac{{k2\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

 \(\left[ \begin{array}{l}x =  - \dfrac{\pi }{9} + \dfrac{{k2\pi }}{9}\\x = \dfrac{{7\pi }}{9} + \dfrac{{k2\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\(\left[ \begin{array}{l}x =  - \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{9}\\x = \dfrac{{7\pi }}{{12}} + \dfrac{{k2\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\)

\(x = k\pi \,\,\left( {k \in \mathbb{Z}} \right)\)         

\(x = \dfrac{\pi }{6} + k2\pi \,\,\left( {k \in \mathbb{Z}} \right)\)    

Tất cả đều đúng.

\(\min y = \dfrac{{5 - 2\sqrt {22} }}{4}\); \(\max y = \dfrac{{5 + 2\sqrt {22} }}{4}\)

\(\min y = \dfrac{{5 - 2\sqrt {22} }}{{14}}\); \(\max y = \dfrac{{5 + 2\sqrt {22} }}{{14}}\)

\(\min y = \dfrac{{5 - 2\sqrt {22} }}{8}\); \(\max y = \dfrac{{5 + 2\sqrt {22} }}{8}\)

\(x =  \pm \dfrac{\pi }{6} + k\pi \).

\(x =  \pm \dfrac{\pi }{6} + k2\pi \).

\(x =  \pm \dfrac{\pi }{3} + k2\pi \).

\(\left\{ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\y =  - \dfrac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)          

\(\left\{ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k2\pi \\y = \dfrac{\pi }{3} - k2\pi \end{array} \right.\left( {k \in Z} \right)\)                      

\(\left\{ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\y = \dfrac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)

$\cot x = \sqrt 3 $.

$\cot 3x = \sqrt 3 $.

$\tan x = \sqrt 3 $.

$ - \dfrac{5}{2} \le m \le 0$.

$0 < m \le 1$.

$1 < m \le \dfrac{3}{2}$.