Đề thi giữa kì 1 Toán 11 THPT Trần Hưng Đạo (Gia Lai) năm 2023-2024 (có đáp án)

$84.$

$30.$

$22$

$19$.

$y=\sin x.$

$y=\cot x.$

$y=\cos x.$

$y=\sin x+\cos x.$

$2;4;6;8;10;...$

$2;3;5;8;13;...$

$2;6;18;54;...$

$3;9;12;20;...$

$y=\tan x.$

$y=\sin x.$

$y=\cos x.$

$y=\cot x.$

$1;5;7;8;\mathrm{..}$

$$2;3;6;7;10;...$$

$$6;8;10;12;14;...$$

$$5;8;4;12;...$$

$M_e=a_p+\frac{\frac{3n}{4}-\left(m_1+\dots +m_{p-1}\right)}{m_p}\cdot \left(a_{p+1}-a_p\right)$

$M_e=a_p+\frac{\frac{n}{2}-\left(m_1+\dots +m_{p-1}\right)}{m_p}\cdot \left(a_{p+1}+a_p\right)$.

$M_e=a_p+\frac{\frac{n}{2}-\left(m_1+\dots +m_{p-1}\right)}{m_p}\cdot \left(a_{p+1}-a_p\right)$.

$M_e=a_p+\frac{\frac{n}{4}-\left(m_1+\dots +m_{p-1}\right)}{m_p}\cdot \left(a_{p+1}-a_p\right)$.

$$u_n=u_1.q^n(n\ge 2)$$ .

$$u_n=u_1.q^{n+1}(n\ge 2)$$.

$$u_n=q^n(n\ge 2)$$.

$$u_n=u_1.q^{n-1}(n\ge 2)$$.

$\sin \left(a-b\right)=\cos a\cos b-\sin a\sin b$

$\sin \left(a-b\right)=\sin a\cos b-\sin b\cos a.$

$\sin \left(a-b\right)=\sin a\sin b-\cos a\cos b.$

$\sin \left(a-b\right)=\sin a\cos b+\sin b\cos a.$

$45$.

$4$.

$39$.

$42$.

$\cos x=\frac{5}{3}$.

$\cos x=-\frac{\sqrt{5}}{2}$.

$\sin x=-\frac{6}{5}$

$$\cos x=-\frac{\sqrt{2}}{3}$$.

${\sin }^2\alpha +{\cos }^2\alpha =1$.

${\sin }^2\alpha -{\cos }^2\alpha =1$.

$1+{\tan }^2\alpha =-\frac{1}{{\cos }^2\alpha }$.

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

$21$.

$42$.

$30$.

$15.$.

Dãy số $$\left(u_n\right)$$ được gọi là bị chặn nếu nó vừa bị chặn trên vừa bị chặn dưới, tức là tồn tại các số $m$, $M$ sao cho $m\le M\le u_n,\forall n\in {\mathbb{N}}^{\ast }$

Dãy số $$\left(u_n\right)$$ được gọi là bị chặn trên nếu tồn tại một số $M$ sao cho $u_n\le M,\forall n\in {\mathbb{N}}^{\ast }$

Dãy số $$\left(u_n\right)$$ được gọi là bị chặn trên nếu tồn tại một số $M$ sao cho $u_n>M,\forall n\in {\mathbb{N}}^{\ast }$

Dãy số $$\left(u_n\right)$$ được gọi là bị chặn dưới nếu tồn tại một số $m$ sao cho $m\le u_n,\forall n\in {\mathbb{N}}^{\ast }$

$x=-\alpha +k2\pi ,k\in \mathbb{Z}$.

$x=\alpha +k\pi ,k\in \mathbb{Z}$.

$x=-\alpha +k\pi ,k\in \mathbb{Z}.$

$x=\alpha +k2\pi ,k\in \mathbb{Z}.$

$\bar{x}=\frac{m_1{x}_1+\dots +m_k{x}_k}{n}$ .

$\bar{x}=\frac{n}{m_1{x}_1+\dots +m_k{x}_k}$.

$\bar{x}=\frac{m_1{x}_1+\dots +m_k{x}_k}{n-1}$

$\bar{x}=\frac{\left(m_1{x}_1\right)\dots \left(m_k{x}_k\right)}{n}$.

$\sin 2a=2\sin a.\cos a.$

$\cos 2a=1-2{\sin }^{2}a.$

$\sin 2a=\sin a.\cos a.$

$\cos 2a={\cos }^{2}a-{\sin }^{2}a$

$\tan \alpha <0$.

$\sin \alpha <0$.

$\cos \alpha <0$.

$\cot \alpha <0$.

$\tan (\pi -\alpha )=\tan \alpha .$

$\cot (\pi -\alpha )=\cot \alpha .$

$\sin (\pi -\alpha )=\sin \alpha .$

$\cos (\pi -\alpha )=\cos \alpha .$

$\cos x=\cos \alpha \iff [\begin{array}{l}x=\alpha +k2\pi \\ x=\pi -\alpha +k2\pi \end{array},\left(k\in \mathbb{Z}\right)$.

$\cos x=\cos \alpha \iff [\begin{array}{l}x=\alpha +k\pi \\ x=-\alpha +k\pi \end{array},\left(k\in \mathbb{Z}\right)$.

$\cos x=\cos \alpha \iff [\begin{array}{l}x=\alpha +k\pi \\ x=-\alpha +k\pi \end{array},\left(k\in \mathbb{Z}\right)$.

$\cos x=\cos \alpha \iff [\begin{array}{l}x=\alpha +k2\pi \\ x=-\alpha +k2\pi \end{array},\left(k\in \mathbb{Z}\right)$.

$u_n=u_{n-1}-d,n\ge 2$.

$u_n=u_{n-1}+d,n\ge 2$.

$u_n=\frac{u_{n-1}}{d},n\ge 2$.

$u_n=u_{n-1}.d,n\ge 2$.

$2$.

$-3$.

$4$.

$-4$.

$[\begin{array}{l}x=-\frac{5\pi }{12}+k2\pi \\ x=-\frac{11\pi }{12}+k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

$[\begin{array}{l}x=-\frac{5\pi }{12}+k2\pi \\ x=\frac{11\pi }{12}+k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

$[\begin{array}{l}x=-\frac{5\pi }{12}+k2\pi \\ x=\frac{7\pi }{12}+k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

$[\begin{array}{l}x=\frac{5\pi }{12}+k2\pi \\ x=\frac{7\pi }{12}+k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

$q=4.$

$q=5$.

$q=-4.$

$q=-5$.

$\cos \alpha =-\frac{2\sqrt{2}}{3}.$

$\cos \alpha =\frac{2\sqrt{2}}{3}.$

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

$\cos \alpha =\pm \frac{2\sqrt{2}}{3}.$

$u_6=21$.

$u_6=16$.

$u_6=-15$.

$u_6=26$.

$25;20;15;10;5;\mathrm{...}$

$2;4;6;8;10;...$

$1;6;2;7;3;8;...$

$2;-2;2;-2;-2;2;...$

$15,5v\grave{a}35,5.$

$10,5v\grave{a}30,5.$

$11v\grave{a}4.$

$20,5v\grave{a}40,5.$

$u_4=-502.$

$u_4=-1250.$

$u_4=80.$

$u_4=-250.$

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

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

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

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

$\frac{36}{11}\pi$ .

$55\pi$.

$\frac{11}{36}\pi$.

$\frac{18}{5\pi }$.

$$S_{19}=1995.$$

$$S_{19}=1083.$$

$$S_{19}=969.$$

$$S_{19}=-399.$$

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

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

$P=\frac{1}{2}\cdot$

$P=-\frac{1}{2}\cdot$

$18,1$.

$17,8.$

$18,5.$

$17,3$

$56,5cm.$

$113\sqrt{2}cm.$

$113cm.$

$226cm.$

$$S_{10}=14762.$$

$$S_{10}=7381.$$

$$S_{10}=-\frac{4921}{2}.$$

$$S_{10}=\frac{29524}{3}.$$