Bộ 15 đề thi giữa kì 2 Toán 12 có đáp án năm 2022-2023 (Đề 11)

${2019}^x\ln 2019.$

${2019}^{x+1}.$

${2019}^x.$

$\frac{{2019}^x}{\ln 2019}.$

$\frac{1}{{\left(2-3x\right)}^2}+C$

$-\frac{3}{{\left(2-3x\right)}^2}+C$

$-\frac{1}{3}\ln \left|3x-2\right|+C.$

$\frac{1}{3}\ln \left|2-3x\right|+C$

F(x) = $\frac{x^3}{3}-\frac{3x^2}{2}-\ln \left|x\right|+C$.

F(x) = $\frac{x^3}{3}-\frac{3x^2}{2}+\ln x+C.$

F(x) = $\frac{x^3}{3}-\frac{3x^2}{2}+\ln \left|x\right|+C.$

F(x) = $\frac{x^3}{3}+\frac{3x^2}{2}+\ln x+C.$

$F\left(x\right)=\frac{3x\sqrt[3]{x}}{4}+C.$

$F\left(x\right)=\frac{3\sqrt[3]{x^2}}{4}+C.$

$F\left(x\right)=\frac{4x}{3\sqrt[3]{x}}+C.$

$F\left(x\right)=\frac{4x}{3\sqrt[3]{x^2}}+C.$

$F\left(x\right)=x^4-x^3+2x+3.$

$F\left(x\right)=x^4-x^3-2x-3.$

$F\left(x\right)=x^4-x^3+2x-3.$

$F\left(x\right)=x^4-x^3+2x.$

$\frac{2}{3}x+2\cos x-\frac{1}{4}\sin 2x+C.$

$\frac{3}{2}x-2\cos x+\frac{1}{4}\sin 2x+C.$

$\frac{2}{3}x-2\cos 2x-\frac{1}{4}\sin 2x+C.$

$\frac{3}{2}x-2\cos x-\frac{1}{4}\sin 2x+C.$

$m=-\frac{4}{3}$

$m=\frac{4}{3}$

$m=-\frac{3}{4}$

$m=\frac{3}{4}$

$M=3\sqrt{11}$; $m=\sqrt{3}$

$M=4\sqrt{11}$; $m=\sqrt{3}$.

$M=20$; $m=2$.

$M=20$; $m=\sqrt{2}$.

2

1

$\frac{1}{2}.$

$\frac{1}{4}.$

$I>J$.

$I=J$.

$I<J$.

Không so sánh được.

Bước 1.

Bước 2.

Bước 3.

Bước 4.

29.

5.

19.

9.

$\underset{-2}{\overset{2}{\int }}f\left(x\right)dx$

$\underset{0}{\overset{-2}{\int }}f\left(x\right)dx+\underset{0}{\overset{2}{\int }}f\left(x\right)dx$

$\underset{2}{\overset{0}{\int }}f\left(x\right)dx+\underset{-2}{\overset{0}{\int }}f\left(x\right)dx.$

$\underset{-2}{\overset{1}{\int }}f\left(x\right)dx+\underset{1}{\overset{2}{\int }}f\left(x\right)dx$

$V=\underset{a}{\overset{b}{\int }}{\left[f_1\left(x\right)-f_2\left(x\right)\right]}^{2}dx.$

$V=\pi \underset{a}{\overset{b}{\int }}\left[{f_1}^2\left(x\right)-{f_2}^2\left(x\right)\right]dx.$

$V=\pi \underset{a}{\overset{b}{\int }}{\left[f_1\left(x\right)-f_2\left(x\right)\right]}^{2}dx.$

$V=\pi \underset{a}{\overset{b}{\int }}\left[f_1\left(x\right)-f_2\left(x\right)\right]dx.$

${\int }_0^4(2x-x^2)dx.$

${\int }_0^2(x^2-2x)dx.$

${\int }_0^2(2x-x^2)dx.$

${\int }_0^4(x^2-2x)dx.$

$S=\left|\underset{a}{\overset{b}{\int }}\left[f\left(x\right)-g\left(x\right)\right]dx\right|$

$S=\underset{a}{\overset{b}{\int }}\left[g\left(x\right)-f\left(x\right)\right]dx$

$S=\underset{a}{\overset{b}{\int }}f\left(x\right)dx-\underset{a}{\overset{b}{\int }}g\left(x\right)dx$

$S=\underset{a}{\overset{b}{\int }}\left|f\left(x\right)-g\left(x\right)\right|dx$

9

$\sqrt[3]{3}.$

3

$3\sqrt[3]{3}.$

$M(0,0,4)$

$N(0,9,0)$

$P(3,0,0)$

$Q(3,9,4)$

$\overrightarrow{b}\left(2;4;6\right)$

$\overrightarrow{c}\left(-2;-4;3\right)$

$\overrightarrow{d}\left(-1;-2;-3\right)$

$\overrightarrow{e}\left(-1;0;3\right).$

$D(2,3,4)$

$D(3,4,2)$

$D(-2,-3,4)$

$D(-2,-3,-4)$

$AB=\sqrt{5}$

$AB=\sqrt{3}$

$AB=\sqrt{10}$

$AB=2\sqrt{3}$

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ không đồng phẳng $\iff \left[\overrightarrow{a},\overrightarrow{b}\right].\overrightarrow{c}\ne 0$

$\overrightarrow{a}$ cùng phương $\overrightarrow{b}\iff \left[\overrightarrow{a},\overrightarrow{b}\right]=\overrightarrow{0}.$

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ đồng phẳng $\iff \left[\overrightarrow{a},\overrightarrow{b}\right].\overrightarrow{c}=0.$

$\left|\left[\overrightarrow{a},\overrightarrow{b}\right]\right|=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|.\cos \left(\widehat{\overrightarrow{a},\overrightarrow{b}}\right).$

10

20

30

40

$D(0;0;0),D(-6;0;0)$

$D(0;0;0),D(6;0;0)$

$D(0;0;2),D(6;0;0)$

$D(0;0;1),D(6;0;0)$

$\overrightarrow{n_4}\left(1;-2;2\right).$

$\overrightarrow{n_2}\left(1;2;2\right).$

$\stackrel{\rightharpoonup }{n_3}\left(1;8;2\right).$

$\overrightarrow{n_1}\left(1;2;0\right).$

$\text{[}\overrightarrow{a},\overrightarrow{b}\text{]}=(-4;-4;-6)$

$\text{[}\overrightarrow{a},\overrightarrow{b}\text{]}=(4;-4;-6)$

$\text{[}\overrightarrow{a},\overrightarrow{b}\text{]}=(-4;4;-6)$

$\text{[}\overrightarrow{a},\overrightarrow{b}\text{]}=(-4;-4;6)$

$x^2+y^2+z^2-3x-3y+3z+6=0$

$x^2+y^2+z^2+3x-3y-3z+6=0$

$x^2+y^2+z^2-3x-3y-3z+6=0$

$x^2+y^2+z^2-3x-3y+3z-6=0$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=16.$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=8.$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=9.$

${\left(x-1\right)}^2+{\left(y+2\right)}^2+{\left(z-3\right)}^2=10.$

${\left(x-1\right)}^2+{\left(y+3\right)}^2+{\left(z-2\right)}^2=14.$

${\left(x+2\right)}^2+{\left(y-6\right)}^2+{\left(z+4\right)}^2=56.$

${\left(x+1\right)}^2+{\left(y-3\right)}^2+{\left(z+2\right)}^2=14.$

${\left(x-2\right)}^2+{\left(y+6\right)}^2+{\left(z-4\right)}^2=56.$

$I\left(1;-3;4\right);r=25.$

$I\left(1;-3;4\right);r=5.$

$I\left(-1;3;-4\right);r=5.$

$I\left(1;-3;4\right);r=-5.$

$-x+2y-2=0$

$-x+2y-1=0$

$-x-2y-2=0$

$-x+2y-3=0$

$2x-y+3z=0$.

$2x-y+3z-4=0$.

$2x-y+3z+4=0$.

$x-2y-4=0$.

$x+4y-z-3=0.$

$2x+4y-z-9=0.$

$2x+8y-2z-1=0.$

$x+4y-z-7=0.$

$12x-14y-5z-25=0.$

$12x-14y-5z+3=0$.

$12x+14y-5z-81=0.$

$12x+14y-5z+3=0$.

$(2;-3;7)$

$(-2;-3;7)$

$(2;3;7)$

$(2;-3;-7)$

$\overrightarrow{u_1}=\left(1;0;-1\right).$

$\overrightarrow{u_1}=\left(0;1;2\right).$

$\overrightarrow{u_1}=\left(0;0;2\right).$

$\overrightarrow{u_1}=\left(0;2;-2\right).$

$120°$.

$30°$.

$60°$.

$150°$.

$\frac{x-2}{2}=\frac{y}{-3}=\frac{z+3}{5}$.

$\frac{x+2}{2}=\frac{y}{-3}=\frac{z-3}{5}$.

$\frac{x+2}{1}=\frac{y}{-3}=\frac{z-3}{5}$.

$\frac{x-2}{2}=\frac{y}{3}=\frac{z+3}{5}$

$\left\{\begin{array}{l}x=1+t\\ y=-1+5t\\ z=3-4t\end{array}\right..$

$\frac{x-1}{-1}=\frac{y+1}{-6}=\frac{z-3}{3}.$

$\left\{\begin{array}{l}x=4+3t\\ y=3-2t\\ z=-1+3t\end{array}\right..$

$\frac{x}{-1}=\frac{y}{-5}=\frac{z-3}{4}.$