Đề thi giữa kì 1 Toán 11 THPT Số 3 Tuy Phước (Bình Định) năm 2023-2024 (có đáp án)

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

$k2\pi$.

$\frac{3\pi }{4}+k2\pi$.

$-\frac{\pi }{4}+k2\pi$.

$sd\left(Ou,Ov\right)=60^{\circ }+k360^{\circ }\left(k\in \mathbb{Z}\right)$.

$sd\left(Ou,Ov\right)=-60^{\circ }+k360^{\circ }\left(k\in \mathbb{Z}\right)$

$sd\left(Ou,Ov\right)=60^{\circ }+k180^{\circ }\left(k\in \mathbb{Z}\right)$.

$sd\left(Ou,Ov\right)=-60^{\circ }+k180^{\circ }\left(k\in \mathbb{Z}\right)$.

$$\sin \alpha >0.$$

$$\cos \alpha >0.$$

$$\tan \alpha >0.$$

$$\cot \alpha >0.$$

Thứ $\mathrm{II}.$

Thứ $\mathrm{IV}.$

Thứ $\mathrm{II}$ hoặc $\mathrm{IV}.$

Thứ $\mathrm{I}$ hoặc $\mathrm{III}.$

$\tan \alpha >0;\cot \alpha >0.$

$\tan \alpha <0;\cot \alpha <0.$

$\tan \alpha >0;\cot \alpha <0.$

$\tan \alpha <0;\cot \alpha >0.$

$\cos \left[\frac{\pi }{4}+\left(2k+1\right)\pi \right]=-\frac{\sqrt{3}}{2}.$

$\cos \left[\frac{\pi }{4}+\left(2k+1\right)\pi \right]=-\frac{\sqrt{2}}{2}.$

$\cos \left[\frac{\pi }{4}+\left(2k+1\right)\pi \right]=-\frac{1}{2}.$

$\cos \left[\frac{\pi }{4}+\left(2k+1\right)\pi \right]=\frac{\sqrt{3}}{2}.$

$[-1;1]$.

$\mathbb{R}$.

$(0;+\infty )$.

$(-\infty ;0)$.

$$\mathrm{D}=\mathbb{R}\backslash \left\{\frac{\pi }{4}+k\pi \left(k\in \mathbb{Z}\right)\right\}.$$

$$\mathrm{D}=\varnothing .$$

$$\mathrm{D}=\mathbb{R}\backslash \left\{\frac{\pi }{8}+k\frac{\pi }{2}\left(k\in \mathbb{Z}\right)\right\}.$$

$$\mathrm{D}=\mathbb{R}.$$

$y=\cot 4x.$

$y=\frac{\sin x+1}{\cos x}.$

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

$y=\left|\cot x\right|.$

$\left(0;\frac{\pi }{4}\right)$.

$\left(\frac{\pi }{2};\pi \right)$.

$\left(\pi ;\frac{3\pi }{2}\right)$.

$\left(\frac{3\pi }{2};2\pi \right)$.

$[\begin{array}{l}x=\frac{2\pi }{3}+k2\pi \\ x=-\frac{2\pi }{3}+k2\pi \end{array}\left(k\in \mathbb{Z}\right).$

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

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

$[\begin{array}{l}x=\frac{2\pi }{3}+k2\pi \\ x=\frac{\pi }{3}+k2\pi \end{array}\left(k\in \mathbb{Z}\right).$

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

$$x=\pm \frac{\pi }{3}+k2\pi \left(k\in \mathbb{Z}\right).$$

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

$$x=\frac{\pi }{3}+k2\pi \left(k\in \mathbb{Z}\right).$$

$$x=-\frac{\pi }{3}+k2\pi \left(k\in \mathbb{Z}\right).$$

$$x=-\frac{\pi }{3}+k\pi \left(k\in \mathbb{Z}\right).$$

$$x=\frac{\pi }{3}+k2\pi \left(k\in \mathbb{Z}\right).$$

$$x=\frac{\pi }{6}+k\pi \left(k\in \mathbb{Z}\right).$$

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

$$x=\frac{2\pi }{3}+\frac{k3\pi }{2}\left(k\in \mathbb{Z}\right).$$

$$x=\frac{\pi }{3}+k\pi \left(k\in \mathbb{Z}\right).$$

$$x=\frac{\pi }{2}+\frac{k3\pi }{2}\left(k\in \mathbb{Z}\right).$$

$$x=\frac{1}{3}+\frac{5\pi }{18}+k\frac{\pi }{3}\left(k\in \mathbb{Z}\right).$$

$$x=\frac{1}{3}+\frac{\pi }{18}+k\frac{\pi }{3}\left(k\in \mathbb{Z}\right).$$

$$x=\frac{5\pi }{18}+k\frac{\pi }{3}\left(k\in \mathbb{Z}\right).$$

$$x=\frac{1}{3}-\frac{\pi }{6}+k\pi \left(k\in \mathbb{Z}\right).$$

$u_n=1-n$.

$u_n=3+2n$.

$u_n=2-n$.

$u_n=9-n^2$.

$$a_4=10$$.

$$a_4=210$$.

$$a_4=30$$.

$$a_4=360$$.

$$\frac{1}{20}$$.

$$\frac{1}{30}$$.

$$\frac{1}{25}$$.

$$\frac{1}{51}$$.

$$0,\frac{1}{2},\frac{3}{2},-\frac{1}{2}$$.

$$1,\frac{1}{2},\frac{3}{2},\frac{1}{2}$$.

$$1,\frac{1}{2},\frac{3}{2},\frac{3}{2}$$.

$$0,\frac{1}{2},-\frac{1}{2},\frac{1}{2}$$.

Dãy $$\left(u_n\right)$$ bị chặn trên và không bị chặn dưới.

Dãy $$\left(u_n\right)$$ bị chặn dưới và không bị chặn trên.

Dãy $$\left(u_n\right)$$ bị chặn.

Dãy $$\left(u_n\right)$$ không bị chặn.

-6.

-18.

12.

$18$.

$u_{17}=242$.

$u_{17}=235$.

$u_{17}=11$.

$u_{17}=4$.

$u_1=1,d=-3.$

$u_1=1,d=3.$

$u_1=-1,d=-3.$

$u_1=-1,d=3.$

$S=16875$.

$S=63375$.

$S=63562,5$.

$S=16687,5$.

$u_n=3.(-2{)}^n$.

$u_n=3.(-2{)}^{n-1}$.

$u_n=3.(2{)}^{n-1}$.

$u_n=3.(2{)}^n$.

$S_5=121$ hoặc $S_5=\frac{181}{16}.$

$S_5=121$ hoặc $S_5=\frac{35}{16}.$

$S_5=114$ hoặc $S_5=\frac{185}{16}.$

$S_5=141$ hoặc $S_5=\frac{183}{16}.$

$u_n=3^{n-1}$.

$u_n=9^{n-1}$.

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

$u_n=9^{n+1}$.

9

14

5

12

14

15

24

12

12

4

5

9

51,81kg

59,81kg

60,81kg

41,81kg

60%

40%

30%

63%

$14.$

$24.$

$26.$

$28.$

13,1

10,3

18,1

23,0

154,18 cm

153,81 cm

153,18 cm

155,18 cm