Đề kiểm tra Giữa kì 2 Toán 12 có đáp án (Mới nhất) (Đề 10)

S = $\underset{1}{\overset{3}{\int }}\left|x^2+4x+3\right|\mathrm{dx}$

S = $\underset{1}{\overset{3}{\int }}\left|-x^2-4x+3\right|\mathrm{dx}$

S = $\underset{1}{\overset{3}{\int }}\left(x^2-4x+3\right)\mathrm{dx}$

S = $\underset{1}{\overset{3}{\int }}\left(\left|x^2+3\right|-\left|4x\right|\right)\mathrm{dx}$

(−1; −2; 3)

(1; 2; 3)

(3; 5; 1)

(3; 4; 1)

$\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=0$

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

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

$\frac{x}{1}+\frac{y}{2}+\frac{z}{3}=6$

M(1; 1; 1)

Q(2; −1; 0)

P(−1; 2; 1)

N(1; 2; 3)

(4; −2; 10)

(2; 6; 4)

(2; −1; 5)

(1; 3; 2)

S = $\frac{1}{4}$

S = 2

S = 1

S = $\frac{1}{2}$

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

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$ = −F(b) – F(a)

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$ = F(a) – F(b)

$\underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$ = F(b) – F(a)

1

4

2

3

S = 4

S = 7

S = 1

S = 2

$\int \mathrm{lnxdx}=\frac{1}{x}+C$

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

$\int \mathrm{xdx}=\frac{x^2}{2}+C$

$\int \frac{1}{x}\mathrm{dx}=\ln \left|x\right|+C$

OA = 4

OA = 9

OA = 5

OA = 3

1

−1

$\frac{3}{2}$

$\frac{2}{3}$

m < 6

m ≥ 6

m > 6

m ≤ 6

0

$\frac{3}{2}$

1

$\frac{2}{3}$

$S=\underset{a}{\overset{c}{\int }}f\left(x\right)\mathrm{dx}+\underset{c}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$

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

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

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

3

6

5

4

9

1

6

5

(3; −1; 0)

(0; −1; 0)

(0; 0; 1)

(3; 0; 0)

$\int \sin \text{x}\mathrm{dx}$ = cosx + C

$\int \sin \text{x}\mathrm{dx}$ = sinx + C

$\int \mathrm{cosdx}$ = sinx + C

$\int \mathrm{cosdx}$ = −sinx + C

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

F(x) = 2^{x – 1} + C

F(x) = −2^x + C

F(x) = 2^x + C

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

$\int k.f\left(x\right)\mathrm{dx}=k\int f\left(x\right)\mathrm{dx}$

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

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

6x + 7y – 2z + 17 = 0

6x – 7y + 2z + 17 = 0

6x + 7y + 2z – 17 = 0

6x + 7y + 2z + 17 = 0

I(−1; −2; −3) và R = 2

I(1; 2; 3) và R = 4

I(−1; −2; −3) và R = 16

I(1; 2; 3) và R = 2

(x – 1)² + (y – 2)² + (z – 3)² = 4

(x – 1)² + (y – 2)² + (z – 3)² = 2

(x + 1)² + (y + 2)² + (z + 3)² = 2

(x + 1)² + (y + 2)² + (z + 3)² = 4

V = $\pi \left(\underset{-1}{\overset{1}{\int }}{f}^2\left(x\right)\mathrm{dx}-\underset{1}{\overset{4}{\int }}{f}^2\left(x\right)\mathrm{dx}\right)$

V = $\underset{-1}{\overset{4}{\int }}{\mathrm{\pi f}}^2\left(x\right)\mathrm{dx}$

V = $\pi \underset{-1}{\overset{4}{\int }}{f}^2\left(x\right)\mathrm{dx}$

V = $\pi \left(\underset{-1}{\overset{1}{\int }}{f}^2\left(x\right)\mathrm{dx}+\underset{1}{\overset{4}{\int }}{f}^2\left(x\right)\mathrm{dx}\right)$

F(x) = sin2x + C

F(x) = $\frac{\sin 2x}{2}+C$

F(x) = −cos2x + C

F(x) = $-\frac{\cos 2x}{2}+C$

$\underset{a}{\overset{b}{\int }}f\text{'}\left(x\right)\mathrm{dx}$ = f(a) – f(b).

$\underset{a}{\overset{b}{\int }}f\text{'}\left(x\right)\mathrm{dx}$ = f(a)− f(b).

$\underset{a}{\overset{b}{\int }}f\text{'}\left(x\right)\mathrm{dx}$ = f(b) – f(a).

$\underset{0}{\overset{1}{\int }}\frac{\mathrm{tdt}}{2}$

$\underset{3}{\overset{4}{\int }}\frac{\mathrm{tdt}}{2}$

$\underset{3}{\overset{4}{\int }}\mathrm{tdt}$

$-\underset{0}{\overset{1}{\int }}\mathrm{tdt}$

2x – y + 3z – 4 = 0

2x – y + 3z + 4 = 0

2x – y + 3z + 9 = 0

x – 2y – 4 = 0

$\frac{5}{3}$

0

$\frac{4}{3}$

1

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

F(x) = 2x + C

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

F(x) = x² + C

(−1; −2; 3)

(3; −2; 1)

(−2; 3; 1)

(1; −2; 3)

2x – y + z + 3 = 0

2x –y + z – 3 = 0

2x + y + z + 3 = 0

2x – y – z – 3 = 0

F(x) = −e^x + C

F(x) = e^x + C

F(x) = $\frac{e^x}{x}$ + C

F(x) = $\frac{e^x}{2}$ + C

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

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

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

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

$\underset{-1}{\overset{2}{\int }}(2x-2)\mathrm{dx}$

$\underset{-1}{\overset{2}{\int }}(-2x+2)\mathrm{dx}$

$\underset{-1}{\overset{2}{\int }}(-2x^2+2x+4)\mathrm{dx}$

$\underset{-1}{\overset{2}{\int }}(2x^2-2x-4)\mathrm{dx}$

$\overrightarrow{n}$ = (2; −3; −4)

$\overrightarrow{n}$ = (−2; 3; 1)

$\overrightarrow{n}$ = (2; 3; −4)

$\overrightarrow{n}$ = (2; −3; 4)

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

V = $\pi \underset{a}{\overset{b}{\int }}f\left(x\right)\mathrm{dx}$

V = $\pi \underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|\mathrm{dx}$

V = $\pi \underset{a}{\overset{b}{\int }}{f}^2\left(x\right)\mathrm{dx}$

I = $\underset{0}{\overset{2}{\int }}2\mathrm{xdx}={4x^2|}_0^2$

I = $\underset{0}{\overset{2}{\int }}2\mathrm{xdx}={x^2|}_2^0$

I = $\underset{0}{\overset{2}{\int }}2\mathrm{xdx}={2|}_0^2$

I = $\underset{0}{\overset{2}{\int }}2x\text{dx=}{x^2|}_0^2$

$\left(0;\frac{3}{16}\right)$

$\left(\frac{3}{16};\frac{7}{32}\right)$

$\left(\frac{1}{4};\frac{9}{32}\right)$

$\left(\frac{7}{32};\frac{1}{4}\right)$

I = $\frac{2}{3}$

I = $\frac{1}{3}$

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

I = $-\frac{2}{3}$

V » 251,33 cm³

V » 502,65 cm³

V » 100,53 cm³

V » 320 cm³

I = 10π

I = $\frac{5}{2}$ π

5π

5

S = 4π(7 – m)

S = π(7 – m)

S = 16π(m – 7)

S = 4π(m – 7)

10

65

41

5

-2

2

1

3

x² + (y + 2)² + (z + 1)² = 3

x² + (y + 2)² + (z + 1)² = 1

x² + (y + 2)² + (z – 1)² = 3

x² + (y + 2)² + (z + 1)² = 2

4x – 2y + z − 10 = 0

3x + 2y + z + 10 = 0

2x – 2y + z + 9 = 0

x – 2y + z − 10 = 0

I = 8

I = 2

I = 16

I = 4

$-\frac{2022}{2021}$

$-\frac{2022}{2023}$

$-\frac{2019}{2020}$

$-\frac{2021}{2022}$