Đề thi giữa kì 1 Toán 11 năm 2023-2024 THPT Nguyễn Trãi (Hà Nội) có đáp án

$$\frac{35\pi }{18}$$.

$$\frac{29\pi }{18}$$.

$$\frac{25\pi }{12}$$.

$$\frac{25\pi }{9}$$.

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

$l=\frac{40}{\pi }$ cm.

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

$l=\frac{5.180}{8}$cm

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

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

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

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

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

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

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

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

Hàm số $y=\sin x$ tuần hoàn với chu kì $2\pi .$

Hàm số $y=\cos x$ tuần hoàn với chu kì $2\pi .$

Hàm số $$y=\tan x$$ tuần hoàn với chu kì $2\pi .$

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

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

$$(-1;1)$$.

$$\left[-2023;2023\right]$$.

$$\left[-\frac{1}{2};\frac{1}{2}\right]$$.

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

$x=\pm \frac{\pi }{4}+k2\pi ,k\in \mathbb{Z}.$

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

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

$5$.

$6$.

$7$.

$4$.

$1;3;5;7;9$.

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

$1;5;3;7;9$.

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

$3$.

$5$.

$1$.

$7$.

$2$.

$4$.

$6$.

$3$.

$u_n=4n+2$.

$u_n=4n-2$.

$u_n=2n-4$.

$u_n=2n+4$.

$1;-2;4;8;-16.$

$2;22;222;2222;22222.$

$3;6;12;24;48.$

$x;2x;3x;4x;5x.$ với $x\ne 0$.

$q=5$.

$q=8$.

$q=-5$.

$q=12$.

$-62$.

$62$.

$22$.

$-22$.

$156,5$.

$157$.

$157,5$.

$158$.

$$20$$.

$$26$$.

$$4$$.

$$42$$.

$$[40;60)$$.

$$[20;40)$$.

$$[60;80)$$.

$$[80;100)$$.

$$\left[156;158\right)$$.

$$\left[154;156\right)$$.

$$\left[152;154\right)$$.

$$\left[158;160\right)$$.

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

$\left[5;7\right)$.

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

$\left[11;13\right)$.

$-\frac{3}{5}.$

$\frac{3}{5}.$

$\pm \frac{3}{5}.$

$\frac{9}{25}.$

$\sin a+\sin \frac{\pi }{3}$.

$\frac{1}{2}\sin a+\frac{\sqrt{3}}{2}\cos a$.

$\frac{1}{2}\sin a-\frac{\sqrt{3}}{2}\cos a$.

$\frac{\sqrt{3}}{2}\sin a+\frac{1}{2}\cos a$.

$3.$

$2.$

$1.$

$4.$

$\left(u_n\right)$ với $$u_n=-n^2,n\in {\mathbb{N}}^{\ast }$$.

$\left(u_n\right)$ với $u_n=n,n\in {\mathbb{N}}^{\ast }$.

$\left(u_n\right)$ với $u_n=\tan n,n\in {\mathbb{N}}^{\ast }$.

$\left(u_n\right)$ với $u_n=\sin n,n\in {\mathbb{N}}^{\ast }$.

$u_{10}=-31$.

$u_{10}=-23$.

$u_{10}=-20$.

$u_{10}=15$.

$u_1=5$, $$q=4$$.

$u_1=5$, $$q=6$$.

$u_1=4$, $$q=5$$.

$u_1=6$, $$q=5$$.

$6,5$.

$7,5$.

$7,25$.

$8$.

$$M_e=\frac{392}{3}$$.

$$M_e=\frac{394}{3}$$.

$$M_e=\frac{391}{3}$$.

$$M_e=\frac{395}{3}$$.

$56$.

$50$.

$60$.

$55$.

$5$ giây.

$10$giây.

$2,5$ giây.

$20$giây.

$[\begin{array}{l}x=\frac{\pi }{4}+k\frac{2\pi }{3}\\ x=k2\pi \end{array}\left(k\in \mathbb{Z}\right)$.

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

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

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

12 năm.

9 năm.

10 năm.

11 năm.

$n=8$.

$n=9$.

$n=6$.

$n=7$.

$$81$$.

$$77$$.

$$71.$$

$$74$$.