Đề thi giữa kì 1 Toán 11 THPT Ngô Mây (Bình Định) năm 2023-2024 (có đáp án)

$[\begin{array}{l}x=\frac{\pi }{6}+k2\pi \\ x=\frac{\pi }{2}+k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

$[\begin{array}{l}x=\frac{\pi }{6}+\frac{k\pi }{3}\\ x=\frac{\pi }{3}+k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

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

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

1.

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

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

-1.

$\cos 2a={\cos }^{2}a–{\sin }^{2}a.$

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

$\cos 2a=1–2{\sin }^{2}a.$

$\cos 2a=2{\cos }^{2}a–1.$

$l=\frac{3\pi }{8}$.

$l=\frac{5\pi }{8}$.

$l=\frac{7\pi }{8}$.

$l=\frac{\pi }{8}$.

$1$.

$2$.

$3$.

$\pi$.

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

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

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

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

$$y=\cos x$$.

$$y=\sin x$$.

$$y=\cot x$$.

$$y=\tan x$$.

$$u_n=-n+5$$.

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

$$u_n=3n-2$$.

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

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

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

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

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

$x=\pm \frac{2\pi }{3}+k2\pi$.

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

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

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

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

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

$x=\alpha +k4\pi \left(k\in \mathbb{Z}\right)$.

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

Hàm số tuần hoàn với chu kì $$\pi$$.

Hàm số tuần hoàn với chu kì $$\pi$$.

Hàm số tuần hoàn với chu kì $$2\pi$$.

Hàm số $$y=\cos x$$ tuần hoàn với chu kì $$\pi$$.

$$u_n=5n+1$$.

$$u_n=5n-1$$.

$$u_n=5n$$.

$$u_n=5n-3$$.

IV.

I.

III.

II.

$$u_3=3$$.

$$u_3=8$$.

$$u_3=9$$.

$$u_3=2$$.

$\cos \alpha =\frac{8}{13}$.

$\cos \alpha =-\frac{8}{13}$.

$\cos \alpha =-\frac{12}{13}$.

$\cos \alpha =\frac{12}{13}$.

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

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

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

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

$$\cos 2\alpha =\frac{1}{2}$$.

$$\cos 2\alpha =-\frac{7}{8}$$.

$$\cos 2\alpha =\frac{7}{8}$$.

$$\cos 2\alpha =\frac{9}{8}$$.

$\left[-3;3\right]$.

$\left[-7;-1\right]$.

$\left[-4;4\right]$.

$\left[-1;1\right]$.

$\frac{1}{2}$.

$9$.

$2$.

$3$.

$16.$

$32.$

$4.$

$8.$

$$3;3;3;3;\mathrm{3....}$$.

$$0;5;10;15;20;25$$.

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

$$1;2;3;4;5;6$$.

Số hạng thứ $16$.

Số hạng thứ $13$.

Số hạng thứ $15$.

Số hạng thứ $14$.

$6,0$.

$6,4$.

$6,2$.

$6,6$.

$16$.

$19$.

$29$.

$26$.

$S_n=\frac{u_1\left(1-q^n\right)}{1-q}$.

$S_n=\frac{u_1\left(1-q^n\right)}{q-1}$.

$S_n=\frac{\left(1-q^n\right)}{1-q}$.

$S_n=\frac{u_1{\left(1+q\right)}^n}{1-q}$.

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

$$u_n=u_1-nd$$.

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

$$u_n=u_1+nd$$.

11.

12.

10.

9.

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

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

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

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

$$x=-\frac{\pi }{2}+k2\pi ,k\in \mathbb{Z}$$.

$$x=\frac{3\pi }{2}+k\pi ,k\in \mathbb{Z}$$.

$$x=-\frac{\pi }{2}+k\pi ,k\in \mathbb{Z}$$ .

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

$\left[20;40\right)$.

$\left[60;80\right)$.

$\left[40;60\right)$.

$675$.

$725$.

$765$.

$715$.

25

12

77

37

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

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

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

$u_n=3^n$.

$\mathrm{16.250.000}$đồng.

$\mathrm{17.250.000}$ đồng.

$\mathrm{15.250.000}$ đồng.

$\mathrm{14.000.000}$ đồng.