Bộ 24 Đề kiểm tra Giữa kì 2 Toán 11 có đáp án (Mới nhất) (Đề 3)

$\lim \frac{1}{n}=+\infty$

$\lim \left(-2n+1\right)=-\infty$

$\lim \frac{2-n}{3n^2}=-\infty$

$\lim \frac{-3}{-2n+1}=\frac{3}{2}$

$+\infty$.

$-\infty$.

-1.

4.

1.

0.

$+\infty$.

$-\infty$

2.

0.

$\frac{1}{2}$.

1

${\left(\frac{1}{3}\right)}^n$

${\left(\frac{4}{e}\right)}^n$

${\left(\frac{-5}{3}\right)}^n$

${\left(\frac{5}{3}\right)}^n$

18

7

11

-7

13

-13

30

-30

5

4

25

9

3

10

-3

2021

1

2020

2022

2

1

$+\infty$

0

$+\infty .$

$-\infty .$

0

1

$+\infty$

$-\infty$

2

-2

x = 2022

x = 2020

x = 2023

x = -2022

x = -2

x = 3

x = 1

x = 2

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{AD}}$

$2\overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{DC}}$

$\overrightarrow{\mathrm{AD}}+2\overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}$

$2\overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AD}}$

$\overrightarrow{\mathrm{BD}\text{'}}$

$\overrightarrow{\mathrm{BD}}$

$\overrightarrow{\mathrm{BA}\text{'}}$

$\overrightarrow{\mathrm{BC}\text{'}}$

$\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|.\cos \left(\overrightarrow{u},\overrightarrow{v}\right)$

$-\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|.\cos \left(\overrightarrow{u},\overrightarrow{v}\right)$

$\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|.\cot \left(\overrightarrow{u},\overrightarrow{v}\right)$

$-\left|\overrightarrow{u}\right|.\left|\overrightarrow{v}\right|.\cot \left(\overrightarrow{u},\overrightarrow{v}\right)$

${90}^0$

${60}^0$

${45}^0$

${120}^0$

$\frac{-2}{3}$

1

3

-2

S = 2

S = $\frac{3}{2}$

S = 1

S = $\frac{2}{3}$

$\frac{-1}{3}$

$-\infty$

2

1

0

$-\infty$

$+\infty$

2

0

$+\infty$

1

$-\infty$

$\frac{5}{4}$

$\frac{-5}{4}$

$\frac{1}{4}$

2

(-3;2)

$\left(-2;+\infty \right)$

$\left(-\infty ;3\right)$

(2;3)

m = 3

m = 1

m = $\frac{3}{4}$

m = $\frac{1}{2}$

$y=\frac{x-2}{x+2020}$

$y=\frac{2x+1}{x-25}$

$y=\frac{x+1}{x-2020}$

$y=\frac{1}{x^2-4}$

f(x) = tan x + 5

$f\left(x\right)=\frac{x^2+3}{5-x}$

$f\left(x\right)=\sqrt{x-6}$

$f\left(x\right)=\frac{x+5}{x^2+4}$

${90}^0$

${45}^0$

${60}^0$

${30}^0$

${60}^0$

${30}^0$

${90}^0$

${45}^0$

$\mathrm{cos\alpha }=\frac{3}{8}$

$\alpha =30°$

$\mathrm{cos\alpha }=\frac{1}{3}$

$\alpha =60°$

$\overrightarrow{\mathrm{AG}}=\frac{2}{3}\left(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AD}}\right)$

$\overrightarrow{\mathrm{AG}}=\frac{1}{4}\left(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AD}}\right)$

$\overrightarrow{\mathrm{OG}}=\frac{1}{4}\left(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}+\overrightarrow{\mathrm{OD}}\right)$

$\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}+\overrightarrow{\mathrm{GD}}=\overrightarrow{0}$

$\overrightarrow{\mathrm{EB}}+\overrightarrow{\mathrm{EC}}+\overrightarrow{\mathrm{ED}}=3\overrightarrow{\mathrm{EG}}$

$\overrightarrow{\mathrm{GA}}+\overrightarrow{\mathrm{GB}}+\overrightarrow{\mathrm{GC}}+\overrightarrow{\mathrm{GD}}=\overrightarrow{0}$

$\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{AD}}=3\overrightarrow{\mathrm{AG}}$

$2\overrightarrow{\mathrm{EF}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{DC}}$