Đề kiểm tra Giữa kì 2 Toán 12 có đáp án (Mới nhất) (Đề 8)

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

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

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

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

$\frac{3}{2}$

$\frac{2}{3}$

6

2

-5

5

-10

10

I(−1; −2; 2), R = 16

I(1; 2; −2), R = 16

I(1; 2; −2), R = 4

I(−1; −2; 2), R = 4

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

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

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

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

$2\sqrt{3}$

$2\sqrt{2}$

$2\sqrt{6}$

$4\sqrt{2}$

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

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

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

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

$\int f\left(x\right)\mathrm{dx}=\frac{x^2}{2}-2\mathrm{lnx}+C$

$\int f\left(x\right)\mathrm{dx}=\frac{2}{x^2}+C$

$\int f\left(x\right)\mathrm{dx}=x^2-2\ln \left|x\right|+C$

$\int f\left(x\right)\mathrm{dx}=\frac{x^2}{2}-2\ln \left|x\right|+C$

V = $\frac{14}{3}$

V = $\frac{206\pi }{15}$

V = $\frac{256}{15}$

V = $\frac{14\pi }{3}$

sinx + C

$\frac{1}{2}$cos²x + C

–sinx + C

–cosx + C

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

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

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

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

20,83 m

20,8 m

$\frac{125}{6}$m

20,8333333 m

3

2

1

5

-13

13

8

1

$\int f\left(x\right)\mathrm{dx}$ = 2e^{2x – 1} + C

$\int f\left(x\right)\mathrm{dx}$ = e^{2x – 1} + C

$\int f\left(x\right)\mathrm{dx}$ = e^2x + C

$\int f\left(x\right)\mathrm{dx}$ = e^{2x – 1} + C

4x⁴ + C

x⁴ + C

3x² + C

$\frac{1}{4}$ x⁴ + C

$\frac{1}{3}$

1

$\frac{5}{3}$

-1

x + 2y – 3z + 5 = 0

2x + 3y – z – 1 = 0

2x + 3y – z = 0

2x + 3y – z + 1 = 0

F(x) = (x + 1)e^x + 3

F(x) = (x + 4)e^x – 2

F(x) = (x – 1)e^x + 3

F(x) = −e^x + 3

8

5

4

7

8

4

2

1

x – 2y – 7 = 0

x – 2y – 3 z – 1 = 0

x – 2y + 3z – 1 = 0

3x – 2y – 2z – 1 = 0

M = (1; −2; −1)

M = (1; 3; 1)

M = (1; 1; 3)

M = (1; −1; −3)

(3; 3; 2)

(0; −9; 7)

(4; 1; 5)

(−1; −7; 4)

(x – 2)² + (y – 1)² + (z + 1)² = $\sqrt{6}$

(x – 4)² + (y – 2)² + (z + 2)² = 24

(x – 2)² + (y – 1)² + (z – 1)² = 6

(x + 2)² + (y + 1)² + (z – 1)² = 6

20

21

18

19

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

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

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

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

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

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

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

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

A = (2; 3; −1)

A = (2; −1; 3)

A = (2; 1; 3)

A = (−1; 2; 3)

$\overrightarrow{u}$ = (3; 4; −5)

$\overrightarrow{u}$ = (4; −5; 3)

$\overrightarrow{u}$ = (4; 3; 5)

$\overrightarrow{u}$ = (4; 3; −5)