Bộ 8 Đề kiểm tra giữa học kì 1 Toán 11 có đáp án (Mới nhất) - Đề số 2

$x\ne \frac{\pi }{2}+k\pi ,\left(k\in \mathbb{Z}\right).$

$x\ne \frac{\pi }{2}+k2\pi ,\left(k\in \mathbb{Z}\right).$

$x\ne k\pi ,\left(k\in \mathbb{Z}\right).$

$x\ne k2\pi ,\left(k\in \mathbb{Z}\right).$

$D=ℝ\backslash \left\{\frac{\pi }{12}+k\pi ,k\in \mathbb{Z}\right\}.$

$D=ℝ\backslash \left\{\frac{\pi }{3}+\frac{k\pi }{2},k\in \mathbb{Z}\right\}.$

$D=ℝ\backslash \left\{\frac{\pi }{12}+\frac{k\pi }{2},k\in \mathbb{Z}\right\}.$

$D=ℝ\backslash \left\{\frac{\pi }{3}+k\pi ,k\in \mathbb{Z}\right\}.$

2

4

3

5

$\tan x=1\iff x=\frac{\pi }{4}+k\pi ,k\in \mathbb{Z}.$

$\tan x=1\iff x=\frac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.$

$\tan x=0\iff x=k2\pi ,k\in \mathbb{Z}.$

$\tan x=0\iff x=\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}.$

1

4

2

3

9

8

7

6

$D=ℝ\backslash \left\{\frac{k\pi }{2},k\in \mathbb{Z}\right\}.$

$D=ℝ\backslash \left\{\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}\right\}.$

$D=ℝ\backslash \left\{k2\pi ,k\in \mathbb{Z}\right\}.$

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

-5 và 2

-8 và -2

2 và 8

-5 và 3

T = [-1;1]

T = [0;1]

T = (-1;1)

T = [-2;2]

$\left[\begin{array}{l}x=\frac{\pi }{6}+k\pi \\ x=\frac{5\pi }{6}+k\pi \end{array}\right.(k\in {\rm Z}).$

$\left[\begin{array}{l}x=\frac{5\pi }{12}+k2\pi \\ x=\frac{13\pi }{12}+k2\pi \end{array}\right.(k\in {\rm Z}).$

$\left[\begin{array}{l}x=\frac{5\pi }{24}+k\pi \\ x=\frac{13\pi }{24}+k\pi \end{array}\right.(k\in {\rm Z}).$

$\left[\begin{array}{l}x=\frac{2\pi }{3}+k\pi \\ x=\frac{\pi }{3}+k\pi \end{array}\right.(k\in {\rm Z}).$

$x=k2\pi .$

$x=\frac{\pi }{2}+k2\pi .$

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

$x=k\pi .$

2

1

3

4

365

353

235

153

t = 6 (giờ)

t = 8 (giờ)

t = 10 (giờ)

t = 11 (giờ)

5 và -5

10 và 0

1 và -1

2 và -1

$\left[\begin{array}{l}x=\pm \frac{\pi }{3}+k2\pi \\ x=-\frac{\pi }{4}+k\pi \end{array}\right..$

$\left[\begin{array}{l}x=\pm \frac{\pi }{6}+k2\pi \\ x=-\frac{\pi }{6}+k\pi \end{array}\right..$

$\left[\begin{array}{l}x=\pm \frac{\pi }{3}+k\pi \\ x=-\frac{\pi }{4}+k2\pi \end{array}\right..$

$\left[\begin{array}{l}x=\pm \frac{\pi }{2}+k2\pi \\ x=\pm \frac{\pi }{4}+k\pi \end{array}\right..$

$P\left(-3;11\right)$

$F\left(-1;-3\right)$

$E\left(3;1\right)$

$Q\left(1;3\right)$

$OM=O{M}^{\text{'}}\mathit{}và\mathit{}(OM;O{M}^{\mathit{\text{'}}})=\phi$

$OM=O{M}^{\text{'}}\mathit{}và\mathit{}\widehat{\mathit{M}\mathit{O}{\mathit{M}}^{\mathit{\text{'}}}}=\phi$

$\overrightarrow{OM}=\overrightarrow{O{M}^{\text{'}}}\mathrm{và}\widehat{MO{M}^{\text{'}}}=\phi$

$\stackrel{}{\overrightarrow{OM}=\overrightarrow{O{M}^{\text{'}}}\mathrm{và}(OM;O{M}^{\text{'}})=\phi }$

${\left(x+2\right)}^2+{\left(y+5\right)}^2=4.$

${\left(x+4\right)}^2+{\left(y-1\right)}^2=4.$

${\left(x-2\right)}^2+{\left(y-5\right)}^2=4.$

${\left(x-1\right)}^2+{\left(y+3\right)}^2=4.$

$\left(ABCD\right)\cap \left(SAB\right)=AB.$

$\left(APQ\right)\cap \left(SBC\right)=EQ.$

$\left(SAB\right)\cap \left(SCD\right)=SE.$

$\left(SAD\right)\cap \left(ABQ\right)=AP.$