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

1

$$(u_n)$$

$$\mathrm{d}=-60,u_1=820$$

$$u_5={\mathrm{u}}_1+4d=820+4(-60)=580$$

$$\cos \left(x+\frac{\pi }{4}\right).$$

$$t=\sin x-\cos x=\sqrt{2}\sin \left(x-\frac{\pi }{4}\right)$$

$$-\sqrt{2}\le t\le \sqrt{2}.$$

$$t^2={\left(\sin x-\cos x\right)}^2={\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x\Rightarrow \sin x\cos x=\frac{1-t^2}{2}.$$

$$6t+\frac{1-t^2}{2}+6=0\iff [\begin{array}{l}t=-1\\ t=13\left(loa\ddot{i}i\right)\end{array}$$

$$\Rightarrow \sqrt{2}\sin \left(x-\frac{\pi }{4}\right)=-1\iff \sin \left(x-\frac{\pi }{4}\right)=-\frac{1}{\sqrt{2}}\iff \sin \left(\frac{\pi }{4}-x\right)=\frac{1}{\sqrt{2}}$$

$$\Rightarrow \cos \left[\frac{\pi }{2}-\left(\frac{\pi }{4}-x\right)\right]=\frac{1}{\sqrt{2}}\iff \cos \left(x+\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}.$$

$$S=1+2.3+3.3^2+...+11.3^{10}=a+\frac{{21.3}^b}{4}.$$

$$3S=3+2.3^2+3.3^3+...+11.3^{11}$$

$-2S=S-3S=1+3+3^2+...+3^{10}-10.3^{11}=\frac{1-3^{11}}{1-3}-11.3^{11}=-\frac{1}{2}-\frac{{21.3}^{11}}{2}\Rightarrow S=\frac{1}{4}+\frac{21}{4}.3^{11}.$

$\pi$

$\frac{70}{\pi }.$

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

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

$x=-\frac{\pi }{12}+k\pi ,k\in Z.$

$x=\frac{7\pi }{12}+k\pi ,k\in Z.$

$88^0$.

$80^0$.

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

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

$\left(-1;1\right).$

$\mathbb{R}$.

$\left\{-1;1\right\}.$

$[-1;1].$

$y=\cos 2x$

$y=\sin x.$

$y=\cos x.$

$y=x\cot x.$

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

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

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

$x=\pm \frac{\pi }{12}+k\pi ,k\in Z.$

$$d=\frac{1}{3}.$$

$$d=2.$$

$$d=3.$$

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

$y=\sin x.$

$y=\cot x.$

$y=\tan x.$

$y=\cos x.$

$-\frac{1}{2};0;1;\frac{1}{2};1.$

$-\frac{1}{2};0;\frac{1}{2};0;\frac{1}{2}.$

$\frac{1}{2};1;\frac{3}{2};2;\frac{5}{2}.$

$-\frac{1}{2};0;\frac{1}{2};1;\frac{3}{2}.$

$$\{\begin{array}{l}{u}_1=5\\ q=3\end{array}.$$

$$\{\begin{array}{l}{u}_1=6\\ q=2\end{array}.$$

$$\{\begin{array}{l}{u}_1=6\\ q=3\end{array}.$$

$$\{\begin{array}{l}{u}_1=5\\ q=2\end{array}.$$

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

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

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

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

$-\frac{1}{2};-\frac{2}{3};-\frac{3}{4};-\frac{4}{5};-\frac{5}{6}.$

$-\frac{2}{3};-\frac{3}{4};-\frac{4}{5};-\frac{5}{6};-\frac{6}{7}.$

$\frac{1}{2};\frac{2}{3};\frac{3}{4};\frac{4}{5};\frac{5}{6}.$

$\frac{2}{3};\frac{3}{4};\frac{4}{5};\frac{5}{6};\frac{6}{7}.$

$u_6=96.$

$u_6=32.$

$u_6=104.$

$u_6=48.$

Thứ $$20.$$

Thứ $$15.$$

Thứ $$36.$$

Thứ $$35.$$

8.

7.

6.

5.

$$\cos \alpha <0.$$

$$\tan \alpha <0.$$

$$\cot \alpha <0.$$

$$\sin \alpha >0.$$

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

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

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

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

Cung có độ dài bằng đường kính.

Cung tương ứng với góc ở tâm $60^0$.

Cung có độ dài bằng nửa đường kính.

Cung có độ dài bằng 1.

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

$$x=\frac{3\pi }{4}+k2\pi ,k\in Z.$$

$x=\pi +k2\pi ,k\in Z.$

$x=k2\pi ,k\in Z.$

$y=\sin 2x+\cos x.$

$y=\sin x-\tan x.$

$y=\cot x.$

$y=\frac{1+2\cos x}{\sin x}.$

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

$[\begin{array}{l}x=\frac{\pi }{3}+k\pi \\ x=k\pi \end{array},k\in Z.$

$[\begin{array}{c}x=\frac{\pi }{4}+k\pi \\ x=\frac{\pi }{2}+k\pi \end{array},k\in Z.$

$[\begin{array}{l}x=\frac{\pi }{6}+k\pi \\ x=k\pi \end{array},k\in Z.$

8; 13; 18.

6; 12; 18.

7; 12; 17,

6; 10; 14.

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

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

$x=\pi +k2\pi ,k\in Z.$

$x=k2\pi ,k\in Z.$

$15;5;1;\frac{1}{5};...$

$5;6;7;8;...$

$\sqrt{2};2;4;4\sqrt{2};....$

$128;-64;32;-16;8;...$

$u_1=-2.$

$u_2=4.$

$u_3=-6.$

$u_4=-8.$

$$u_n=\frac{1}{2^n}.$$

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

$$u_n=n^2.$$

$$u_n=\sqrt{n+2}.$$

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

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

$x=\frac{\pi }{6}+\frac{k\pi }{2},k\in Z.$

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

$\mathbb{R}$.

$\left(1;+\infty \right).$

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

$\left(-\infty ;-1\right).$

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

D = R\{0}.

D = R.

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

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

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

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

$u_n=-3+\frac{1}{4}\left(n-1\right).$

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

1.

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

0.

$x=64.$

$x=68.$

$x=32.$

$x=14.$