Đề thi giữa kì 1 Toán 11 THPT Chuyên Trần Phú (Hải Phòng) năm 2023-2024 (có đáp án)

$$\frac{a_{i+1}-a_i}{2}.$$

$$\frac{a_i+a_{i+1}}{2}.$$

$$a_i.$$

$$\frac{a_i+a_{i-1}}{2}.$$

$$M_e=5,9.$$

$$M_e=6.$$

$$M_e=6,3.$$

$$M_e=6,25.$$

$$-32^{0}55^{\prime }$$

$$-29^{0}30^{\prime }$$

$$33^{0}45^{\prime }$$

$$-33^{0}45^{\prime }$$

$$\frac{-13}{27}.$$

$$\frac{-11}{27}.$$

$$\frac{11}{27}.$$

$$0.$$

$-8$

$-9$

$-20$

$0$

$$u_n=\frac{n-1}{n}.$$

$$u_n=\frac{n}{n+1}.$$

$$u_n=\frac{n+1}{n}.$$

$$u_n=\frac{n^2-n}{n+1}.$$

$-3.$

$-5.$

$5.$

$3.$

$A=2\sin x.$

$A=-2\mathrm{cotx}.$

$A=0$

$A=-2s\mathrm{inx}$

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

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

$[\begin{array}{l}x=k2\pi \\ x=\pm \frac{\pi }{3}+k2\pi \end{array}.$

$x=\pm \frac{\pi }{6}+k2\pi .$

2.

0.

3.

1.

$[\begin{array}{l}m<-1\\ m>1\end{array}.$

$m>1.$

$-1\le m\le 1.$

$m<-1.$

$$T=\frac{2\pi }{3}.$$

$$T=\pi .$$

$$T=2\pi .$$

$$T=\frac{\pi }{3}.$$

$$Q_1=\frac{116}{5};Q_3=\frac{80}{21}.$$

$$Q_1=\frac{1360}{37};Q_3=\frac{800}{21}.$$

$$Q_1=\frac{136}{5};Q_3=\frac{800}{12}.$$

$$Q_1=\frac{136}{5};Q_3=\frac{800}{21}.$$

$$\frac{56}{65}.$$

$$\frac{-33}{65}.$$

$$0.$$

$$\frac{63}{65}.$$

$\frac{3\sqrt{10}}{8}.$

$\frac{5\sqrt{6}}{16}.$

$\frac{3\sqrt{10}}{16}.$

$\frac{5\sqrt{6}}{8}.$

$$-\frac{3}{7}\le m\le -\frac{2}{7}.$$

$$-\frac{1}{2}\le m\le -\frac{2}{3}.$$

$$-\frac{1}{2}\le m\le -\frac{2}{7}.$$

$$-\frac{4}{7}\le m\le -\frac{2}{7}.$$

$$[\begin{array}{l}x=\frac{5\pi }{4}+k2\pi \\ x=\frac{-5\pi }{4}+k2\pi \end{array}.$$

$$[\begin{array}{l}\mathrm{x}=\frac{\pi }{4}+k2\pi \\ x=\frac{-\pi }{4}+k2\pi \end{array}.$$

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

$$[\begin{array}{l}x=\frac{\pi }{4}+k2\pi \\ x=\frac{3\pi }{4}+k2\pi \end{array}.$$

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

$$u_n=3n+1.$$

$$u_n=3^n.$$

$$u_n=5n^2+n.$$

5.

6.

4.

3.

$$k_1$$ và $$k_2.$$

$$k_1+1,...,k_2.$$

$$k_1,k_1+1,..,k_2.$$

$$k_1,...,k_2+1.$$

$$\frac{5\pi }{2}cm.$$

$$\frac{5\pi }{3}cm.$$

$$\frac{2\pi }{5}cm.$$

$$\frac{\pi }{3}cm.$$

Trung vị.

Tứ phân vị thứ nhất.

Tứ phân vị thứ ba.

Số trung bình.

$$\overline{x}=7,5.$$

$$\overline{x}=9,4.$$

$$\overline{x}=8,9.$$

$$\overline{x}=9.$$

$$20.$$

$$16,5.$$

$$18,22.$$

$$17,2.$$

$$2548,5.$$

$$2550,5.$$

$$5096,5.$$

$$1274,5.$$

$$2985.$$

$$300.$$

$$150.$$

$$597.$$

$$x\ne k2\pi .$$

$$\{\begin{array}{l}x\ne \frac{\pi }{2}+k\pi \\ x\ne \frac{\pi }{3}+k\pi \end{array}.$$

$$x=\frac{\pi }{3}+k2\pi .$$

$$\{\begin{array}{l}x\ne \frac{\pi }{2}+k\pi \\ x\ne k2\pi \end{array}.$$

$$\frac{35}{6}.$$

$$\frac{36}{5}.$$

$$\frac{35}{8}.$$

$$\frac{5}{6}.$$

$\frac{1}{\sqrt{2}}.$

$\frac{3}{4}.$

$\frac{3}{8}.$

$-\frac{3}{4}.$

$\cot \alpha >0.$

$\cos \alpha >0.$

$$\tan \alpha >0.$$

$\sin \alpha >0.$

$$y=\frac{{\sin }^{2}x}{1+\cos x}.$$

$$y=\frac{\cos x}{x+x^2}.$$

$$y=\frac{\sin x}{1-\sin x}.$$

$$y=\frac{\tan x}{1+{\sin }^{2}x}.$$

$$279240000$$ (đồng).

$$300500000$$ (đồng).

$$250392000$$ (đồng).

$$293200000$$ (đồng).

$$121.$$

$$120.$$

$$122.$$

$$200.$$

$$u_1=\frac{6}{5};q=3.$$

$$u_1=\frac{6}{5};q=5.$$

$$u_1=\frac{6}{5};q=-2.$$

$$u_1=\frac{6}{5};q=2.$$

$$a>2.$$

$$a<1.$$

$$a<-5.$$

$$a<0.$$