Đề kiểm tra Giữa kì 2 Toán 12 có đáp án (Mới nhất) (Đề 12)

(1; −3; 2)

(2; −3; 1)

(1; 2; −3)

(2; 1; −3)

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

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

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

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

$\int f\left(x\right)\mathrm{dx}=-\sqrt{3-2x}+C$

$\int f\left(x\right)\mathrm{dx}=\sqrt{3-2x}+C$

$\int f\left(x\right)\mathrm{dx}=-\frac{1}{2}\sqrt{3-2x}+C$

$\int f\left(x\right)\mathrm{dx}=\frac{1}{2}\sqrt{3-2x}+C$

4x⁴ – 9x + C

$\frac{1}{2}$x⁴ – 9x + C

$\frac{1}{4}$x⁴ + C

4x³ – 9x + C

$\int {\mathrm{xe}}^x\mathrm{dx}={\mathrm{xe}}^x-e^x+C$

$\int {\mathrm{xe}}^x\mathrm{dx}=\frac{x^2}{2}{e}^x+e^x+C$

$\int {\mathrm{xe}}^x\mathrm{dx}=e^x+{\mathrm{xe}}^x+C$

$\int {\mathrm{xe}}^x\mathrm{dx}=\frac{x^2}{2}{e}^x+C$

$\frac{1}{5}$ln( 5x + 4) + C

$\frac{1}{\ln 5}\ln \left|5x+4\right|$+ C

ln|5x + 4| + C

$\frac{1}{5}\ln \left|5x+4\right|$+C

I = ln2 + 2

I = ln2 + 1

I = ln2 – 1

I = ln2 + 3

2

3

4

1

ln5

4ln5

2ln5

$\frac{1}{2}$ ln5

$\overrightarrow{\mathrm{AB}}$ = (0; 2; 7)

$\overrightarrow{\mathrm{AB}}$ = (0; −2; −7)

$\overrightarrow{\mathrm{AB}}$ = $\left(0;1;\frac{7}{2}\right)$

$\overrightarrow{\mathrm{AB}}$ = (4; 12; 7)

-6

4

6

-4

S = 7

S = $\frac{8}{3}$

S = $\frac{7}{3}$

S = 8

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

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

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

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

$\int \mathrm{kf}\left(x\right)\mathrm{dx}=k\int f\left(x\right)\mathrm{dx},(k\ne 0)$

$\int \frac{f\left(x\right)}{g\left(x\right)}\mathrm{dx}=\frac{\int f\left(x\right)\mathrm{dx}}{\int g\left(x\right)\mathrm{dx}}$

$\int f\text{'}\left(x\right)\mathrm{dx}=f\left(x\right)+C$

$\int \left[f\left(x\right)-g\left(x\right)\right]\mathrm{dx}=\int f\left(x\right)\mathrm{dx}-\int g\left(x\right)\mathrm{dx}$

M = 14.

M = 106.

M = 8.

M = 28.

Q(1; −2; 2).

N(1; −1; −1).

P(2; −1; −1).

M(1; 1; −1).

x² + (y – 2)² + (z – 1)² = 20.

x² + (y – 2)² + (z – 1)² = 5.

x² + (y – 2)² + (z −1)² = $\sqrt{5}$ .

x² + (y – 2)² + (z – 1)² = $\sqrt{20}$ .

I (1; 2; −3); R = 4.

I (−1; −2; 3); R = 2.

I (1; 2; −3); R = 2.

I (−1; −2; 3); R = 4.

I(1; −2; 1)

I(2;0; −2)

I(4; 0; −4)

I(1; 0; −2)

I = $\frac{1}{2}a+1$

I = 2a + 1

I = 2a

I = $\frac{1}{2}a$

4x – 5z + 4 =0

4x – 5y + 4 = 0

4x – 5z – 4 = 0

4x – 5y – 4 = 0

I = $\underset{1}{\overset{e}{\int }}(3t+1)\mathrm{dx}$

I = $\underset{0}{\overset{1}{\int }}\frac{3t+1}{e^t}\mathrm{dt}$

I = $\underset{0}{\overset{1}{\int }}(3t+1)\mathrm{dt}$

I = $\underset{1}{\overset{e}{\int }}\frac{3t+1}{t}\mathrm{dt}$

I = −4

I = 4

I = 0

I = 3

−cosx + x² + C

– cosx + 2x² + C

cosx + x² + C

2x² + cosx + C

y + z = 0.

y – z = 0.

x = 0.

x = y + z.

3

6

4

7

$\overrightarrow{n}$ = (2; 1; −1).

$\overrightarrow{n}$ = (1; 2; 0).

$\overrightarrow{n}$ = (−2; −1; 1).

$\overrightarrow{n}$ = (2; 1; 0).

Trùng nhau

Vuông góc

Cắt nhưng không vuông

Song song

5 – e

5 + e

e + 3

3 – e

A₁ (0; 2; 3)

A₁ (1; 0; 3)

A₁ (1; 2; 0)

A₁ (1; 0; 0)

S = $\underset{-2}{\overset{3}{\int }}{\mathrm{xe}}^x\mathrm{dx}$

S = $\underset{-2}{\overset{3}{\int }}\left|{\mathrm{xe}}^x\right|\mathrm{dx}$

S = $\left|\underset{-2}{\overset{3}{\int }}{\mathrm{xe}}^x\mathrm{dx}\right|$

S = $\pi \underset{-2}{\overset{3}{\int }}{\mathrm{xe}}^x\mathrm{dx}$

$\frac{1}{2}$cos²x + C

sinx + C

$-\frac{1}{2}$cos²x +C

–sinx + C

4

8

10

6

$\ln \frac{7}{3}$

$\frac{1}{2}\log \frac{7}{3}$

$\frac{1}{2}\ln \frac{3}{7}$

$\frac{1}{2}\ln \frac{7}{3}$

$\int \left[f_1\left(x\right)+f_2\left(x\right)\right]\mathrm{dx}=\int f_1\left(x\right)\mathrm{dx}+\int f_2\left(x\right)\mathrm{dx}$

Nếu F(x) và G(x) đều là nguyên hàm của hàm số f(x) thì F(x) = G(x).

$\int \mathrm{kf}\left(x\right)\mathrm{dx}=k\int f\left(x\right)\mathrm{dx}$ (k là hằng số và k ≠ 0).

Nếu $\int f\left(x\right)\mathrm{dx}=F\left(x\right)+C$ thì $\int f\left(u\right)\mathrm{du}=F\left(u\right)+C$.