Đề thi giữa kì 1 Toán 11 THPT Tam Đảo 2 (Vĩnh Phúc) năm 2023-2024 (có đáp án)

$\frac{4\pi }{7}$.

$\frac{7\pi }{2}$.

$\frac{2\pi }{7}$.

$\frac{7\pi }{4}$.

$\tan \left(\frac{3\pi }{2}-\alpha \right)\le 0.$

$\tan \left(\frac{3\pi }{2}-\alpha \right)\ge 0.$

$\tan \left(\frac{3\pi }{2}-\alpha \right)>0.$

$\tan \left(\frac{3\pi }{2}-\alpha \right)<0.$

$$-3;-1;1;3;5;...$$.

$$-3;0;1;3;5;...$$.

$$1;2^2;3^2;...$$.

$$1;-1;1;-1;1;.....$$.

Hàm số $$y=\cot x$$ là hàm số lẻ.

Hàm số $$y=\sin x$$ là hàm số chẵn.

Hàm số $$y=\cos x$$ là hàm số chẵn.

Hàm số $$y=\tan x$$ là hàm số lẻ.

$$u_n=\sin n$$.

$$u_n=n+1$$.

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

$$u_n=-n+3$$.

$\left[10;14\right).$

$\left[18;22\right).$

$\left[14;18\right).$

$\left[26;30\right).$

$$x=k2\pi$$.

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

$$x=\frac{\pi }{2}+k\pi$$.

$$x=\pi +k2\pi$$.

$$u_n=u_1+\left(n+1\right)d$$.

$$u_n=u_1-nd$$.

$$u_n=u_1+\left(n-1\right)d$$.

$$u_n=u_1+nd$$.

$$S=\left\{{12}^0+k{360}^0,k\in \mathbb{Z}\right\}$$.

$$S=\left\{{12}^0+k{360}^0;{178}^0+k{360}^0,k\in \mathbb{Z}\right\}$$.

$$S=\left\{\pm {12}^0+k{180}^0,k\in \mathbb{Z}\right\}$$.

$$S=\left\{\pm {12}^0+k{360}^0,k\in \mathbb{Z}\right\}$$.

$\max y=11;\min y=-21$.

$\max y=8;\min y=-8$.

$\max y=-4;\min y=-6$.

$\max y=3;\min y=-13$.

$$-1<m<1$$.

$$-1\le m\le 1$$.

$$m\ge 1$$.

$$m\le -1$$.

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

$$u_n=\frac{{(-1)}^n}{n}$$.

$$u_n=2n+1$$.

$$u_n=-n+1$$.

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

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

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

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

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

$$0$$.

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

$$1$$.

$$M=\sin x$$.

$M=\cos x$.

$M=\sin \left(x+2y\right)$.

$M=\cos \left(x+2y\right)$.

$$u_n=4n-1$$.

$$u_n=4n+3$$.

$$u_n=4n+1$$.

$$u_n=4n$$.

$$y=\cos x$$.

$$y=\cot x$$.

$$y=\tan x$$.

$$y=\sin x$$.

$$\frac{\pi }{6};\frac{\pi }{3};\frac{\pi }{2}.$$

$$\frac{\pi }{5};\frac{2\pi }{5};\frac{4\pi }{5}.$$

$$\frac{\pi }{6};\frac{2\pi }{6};\frac{4\pi }{6}.$$

$$\frac{\pi }{7};\frac{2\pi }{7};\frac{4\pi }{7}.$$

$8.$

$11.$

$6.$

$7.$

$\pi$.

$1$.

$2$.

$3$.

$$u_3=4$$.

$$u_3=8$$.

$$u_3=-3$$.

$$u_3=10$$.

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

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

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

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

$\frac{2}{3}$.

$-\frac{1}{3}$.

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

$\frac{1}{3}$.

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

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

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

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

$$d=-3$$.

$$d=7$$.

$$d=-7$$.

$$d=4$$.

$\frac{1}{2}$.

$9$.

$2$.

$3$.

$0^{\circ }$.

$-30^{\circ }$.

$-60^{\circ }$.

$30^{\circ }$.

$1.$

$2.$

$3.$

$4.$

$D=\mathbb{R}\backslash \left\{\frac{\pi }{4}+k\frac{\pi }{2}\mid k\in \mathbb{Z}\right\}$.

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

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

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

$$u_n=8n-11$$.

$$u_n=5-8n$$.

$$u_n=8n-5$$.

$$u_n=11-8n$$.

$$u_3=2$$.

$$u_3=3$$.

$$u_3=9$$.

$$u_3=8$$.

$22.$

$4.$

$3$

$24.$

$\left[7;9\right).$

$\left[9;11\right).$

$\left[11;13\right)$

$\left[13;15\right).$

$2.$

$6.$

$4.$

$8.$

$18.$

$20.$

$4.$

$7.$