Đề thi giữa kì 1 Toán 11 năm 2023-2024 THPT (TP.HCM) có đáp án

$$\mathrm{11.500.000}$$.

$$\mathrm{11.030.000}$$.

$$\mathrm{11.213.000}$$.

$$\mathrm{10.847.000}$$.

$\cos \left(x-y\right)$.

$\cos \left(x+y\right)$.

$\sin \left(x+y\right)$.

$\sin \left(x-y\right)$.

$$25,1m$$

$$25,5m$$

$$25m$$

$$26m$$

$x=\pm \frac{2\pi }{3}+k2\pi ,k\in Z$

$x=\pm \frac{\pi }{6}+k\pi ,k\in Z$

$x=\pm \frac{\pi }{3}+k2\pi ,k\in Z$

$x=\pm \frac{5\pi }{6}+k2\pi ,k\in Z$

$$u_1=16$$.

$$u_1=-\frac{1}{16}$$.

$$u_1=\frac{1}{16}$$.

$$u_1=-16$$.

$$u_2=u_1+n$$.

$$u_2=7$$.

$$u_2=5+n$$.

$$u_2=6$$.

$$[\begin{array}{l}x=-\frac{\pi }{6}+k\frac{2\pi }{3}\\ x=\frac{\pi }{2}+k2\pi \end{array};k\in \mathbb{Z}$$.

$$[\begin{array}{l}x=-\frac{\pi }{6}+k\frac{2\pi }{3}\\ x=-\frac{\pi }{2}+k2\pi \end{array};k\in \mathbb{Z}$$.

$$[\begin{array}{l}x=\frac{\pi }{6}+k\frac{2\pi }{3}\\ x=\frac{\pi }{2}+k2\pi \end{array};k\in \mathbb{Z}$$.

$$[\begin{array}{l}x=\frac{\pi }{6}+k2\pi \\ x=\frac{\pi }{2}+k2\pi \end{array};k\in \mathbb{Z}$$

$$[\begin{array}{l}x=\alpha +k2\pi \\ x=\pi -\alpha +k2\pi \end{array};k\in \mathbb{Z}$$.

$$[\begin{array}{l}x=\alpha +k\pi \\ x=-\alpha +k\pi \end{array};k\in \mathbb{Z}$$.

$$[\begin{array}{l}x=\alpha +k2\pi \\ x=-\alpha +k2\pi \end{array};k\in \mathbb{Z}$$.

$$[\begin{array}{l}x=\alpha +k\pi \\ x=\pi -\alpha +k\pi \end{array};k\in \mathbb{Z}$$.

$$\sin \alpha -\sin \beta =2\cos \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$$

$$\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$$.

$$\sin \alpha +\sin \beta =2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$$

$$\cos \alpha +\cos \beta =2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}$$

$$14$$.

$13$.

$15$.

$16$.

$$y=\cot x$$ là hàm số lẻ.

$$y=\tan x$$là hàm số lẻ

$$y=\cos x$$ là hàm số lẻ.

$$y=\sin x$$ là hàm số lẻ.

$$\{\begin{array}{l}{u}_1=1\\ u_{n+1}=u_n+(-1{)}^{2n};n\in N^{\ast }\end{array}$$.

$$\{\begin{array}{l}{u}_1=1\\ u_{n+1}=\frac{1}{2}{u}_n+n^2;n\in N^{\ast }\end{array}$$.

$$\{\begin{array}{l}{u}_{20}=8u_{17}\\ u_1+u_5=272\end{array}$$.

$u_n=\frac{\sqrt{3n^2-2n+1}}{2n+1};n\in N^{\ast }$.

$$x=\frac{\pi }{6}+k\pi ,k\in Z$$

$$x=\frac{\pi }{3}+k\pi ,k\in Z$$

$$x=\frac{\pi }{6}+k2\pi ,k\in Z$$

$$x=\frac{\pi }{3}+k2\pi ,k\in Z$$.

$$d=-5$$

$$d=-7$$

$$d=-\frac{2}{5}$$

$$d=7$$.

$$x\ne k2\pi ;k\in Z$$.

$x\ne k\pi ;k\in Z$.

$x\ne \frac{\pi }{2}+k\pi ;k\in Z$.

$x\ne \frac{k\pi }{2};k\in Z$.

$3,5,7,9,11$ .

$2,4,6,...2n,...$.

$1,3,5,7,9,...2n+1,...$

$1,2,3,4,5,6,7,8,...,n,...$

$6$.

$9$.

$10$.

$8$.

Số hạng tổng quát: $$u_n=u_1+(n-1)d$$ với $$n\in N^{\ast }$$.

Số hạng tổng quát: $$u_{n+1}=u_1+(n-1)d$$ với $$n\in N^{\ast }$$.

Tổng $$n$$ số hạng đầu: $$S_n=\frac{n(u_1+u_n)}{2}$$.

Hệ thức truy hồi: $$u_n=u_{n-1}+d$$ với $$n\ge 2$$.

$$\sin \alpha =\frac{b}{a}$$.

$$\sin \alpha =\frac{a}{b}$$.

$\sin \alpha =b$.

$\sin \alpha =a$.

$\cos \left(–{60}^0\right)=-\cos {60}^0$

$\sin \left(–{30}^0\right)=-\sin {30}^0$

$\sin \left(–{30}^0\right)=\cos {60}^0$

$\cos \left(–{60}^0\right)=\sin {60}^0$

$\frac{4\sqrt{6}}{5}$

$\frac{2\sqrt{6}}{5}$

$\frac{2\sqrt{6}}{25}$

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

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

$D=\mathbb{R}$.

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

$D=\mathbb{R}\backslash \left\{\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}\right\}.$

$$B=tana$$

$$B=-tana$$

$$B=-\cot 2a$$

$$B=-tan2a$$

$$1,1,1,1,1$$.

$$-8,-6,-4,-2,0$$.

$$\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2},\frac{9}{2}$$.

$$3,1,-1,-2,-4$$.

$$\sqrt{3}$$.

$$1$$.

$$-\sqrt{3}$$.

$$\frac{\pi }{6}$$.

$\cos x=\frac{3}{\sqrt{10}}$.

$\cos x=\frac{1}{\sqrt{10}}$.

$\cos x=\pm \frac{3}{\sqrt{10}}$.

$\cos x=\pm \frac{1}{\sqrt{10}}$.

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

$$\frac{\sqrt{3}}{2}.$$

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

$$-\frac{\sqrt{3}}{2}.$$

$-180^{\circ }$ .

$360^{\circ }$ .

$180^{\circ }$.

$0^{\circ }$.

$100$.

$64$.

$102$.

$81$.

$$\cos \alpha <0.$$

$$\cot \alpha <0.$$

$$\tan \alpha <0.$$

$$\sin \alpha <0.$$

$\frac{\sqrt{3}}{2}$ .

$\frac{1}{2}$.

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

$-\frac{\sqrt{3}}{2}$.

$q=\frac{1}{8}$.

$q=-14$.

$q=14$.

$q=8$.

Tồn tại một số $$m$$ sao cho $$u_n\ge m$$ với mọi $$n\in N^{\ast }$$

Tồn tại một số $$M$$ sao cho $$u_n\le M$$ với mọi $$n\in N^{\ast }$$.

Đó là dãy số tăng hoặc dãy số giảm.

Tồn tại các số $$m$$,$$M$$ sao cho $$m\le u_n\le M$$ với mọi $$n\in N^{\ast }$$.

$$u_n=\frac{n}{2}.\left[-6+\frac{1}{2}\left(n-1\right)\right]$$

$$u_n=-3+\frac{1}{2}\left(n+1\right)$$.

$$u_n=\frac{1}{2}n-\frac{7}{2}$$.

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

$2,4,6,8,...$

$1,3,6,9,12,...$

$1,3,5,7,...$.

$2,4,8,16,...$