Đề kiểm tra Giữa kì 2 Toán 12 có đáp án (Mới nhất) (Đề 7)

$\ln \left|x\right|-2x^3+C$

$-\ln \left|x\right|-2x^3+C$

$-\frac{1}{x^2}-12x+C$

$\ln \left|x\right|-6x^3+C$

$d_1=\frac{10\sqrt{21}}{21}\mathrm{và}{d}_2=2$

$d_1=\frac{10\sqrt{21}}{21}\mathrm{và}{d}_2=3$

$d_1=\frac{10}{\sqrt{20}}\mathrm{và}{d}_2=2$

$d_1=\frac{10}{\sqrt{20}}\mathrm{và}{d}_2=1$

$I=-\frac{\pi }{8}+\frac{1}{4}$

$I=-\frac{\pi }{8}-\frac{1}{4}$

$I=\frac{\pi }{8}+\frac{1}{4}$

$I=\frac{\pi }{8}-\frac{1}{4}$

6x – 3y – 2z = 0

−6x – 3y – 2z = 0

6x + 3y – 2z= 0

−6x + 3y – 2z = 0

1

5

4

8

f(x) = x⁴ + x³ + x² + Cx + C'

f(x) = 12x² – 6x + 2

f(x) = x⁴ – x³ + x² + Cx

12x² – 6x + 2 + C

$\overrightarrow{b}$ = (−2; 5; 0), $\overrightarrow{c}$ = (−1; −3; 0).

$\overrightarrow{b}$ = (−2; 5; 0), $\overrightarrow{c}$ = (0; −3; 0).

$\overrightarrow{b}$ = (0; −2; 5), $\overrightarrow{c}$ = (−1; −3; 0).

$\overrightarrow{b}$ = (1; −2; 5), $\overrightarrow{c}$ = (−1; −3; 1).

$F\left(x\right)=2\ln \left|x\right|-\frac{3}{x}-4$

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

$F\left(x\right)=2x+\frac{3}{x}-4$

$F\left(x\right)=2x-\frac{3}{x}+2$

F(x) = x³ + x² – 3x

F(x) = x³ + x² – 3x + 1

F(x) = x³ + x² – 3x + 2

F(x) = x³ + x² – 3x – 1

$x^2\left(1+\frac{6x^3}{5}\right)+C$

$2x\left(x+\frac{3}{4}{x}^4\right)+C$

$x^2\left(x+\frac{3}{4}{x}^3\right)+C$

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

(S): (x + 2)² + (y + 1)² + (z + 2)² = 9

(S): (x – 2)² + (y – 1)² + (z – 2)² = 9

(S): (x + 2)² + (y + 1)² + (z + 2)² = 3

(S): (x – 2)² + (y – 1)² + (z – 1)² = 3

I = $x\frac{s\text{in}2x}{2}+\frac{1}{2}\int s\text{in2xdx}$

I = $-x\frac{\sin 2x}{2}-\frac{1}{2}\int \sin 2xdx$

I = $-x\frac{\sin 2x}{2}+\frac{1}{2}\int \sin 2xdx$

I = $x\frac{\sin 2x}{2}-\frac{1}{2}\int \sin 2xdx$

OA = $\sqrt{5}$

OA = 5

OA = 3

OA = 9

F(9) = −6

F(9) = −12

F(9) = 6

F(9) = 12

6

-14

-6

14

64

12

7

$\frac{4}{3}$

(x – 2)² + (y + 2)² + z² = 3

(x + 2)² + (y – 2)² + z² = 9

(x – 2)² + (y + 2)² + z² = 9

(x – 2)² + (y + 2)² + z² = 3

3

1

4

2

I = 3

I = −5

I = 5

I = −3

x + 3ln(x – 1) + C

x – 3ln(x – 1) + C

x − $\frac{3}{{(x-1)}^2}$ + C

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

m = -30

$m=\frac{5}{2}$

m = 4

$m=-\frac{5}{2}$

$-\frac{1}{3}\cos (1-3x)+C$

3cos(1 −3x) + C

−3cos(1 – 3x) + C

$\frac{1}{3}$cos(1 – 3x) + C

I = 4

I = 7

I = 6

I = 5

(−3; 2; 2)

(−3; 2; −2)

(−3; 0; −2)

(−3; 0; 2)

$\int \frac{1}{x^2}\mathrm{dx}=\frac{1}{2x}+C$

$\int \frac{1}{x^2}\mathrm{dx}={\mathrm{lnx}}^2+C.$

$\int \frac{1}{x^2}\mathrm{dx}=\frac{1}{x}+C$

$\int \frac{1}{x^2}\mathrm{dx}=-\frac{1}{x}+C$

3

2

$\frac{3}{4}$

$\frac{3}{2}$

$\frac{3^x}{\ln 3}-\cos 8x+C$

$\frac{3^x}{\ln 3}-\frac{1}{8}\cos 8x+C$

$\frac{3^x}{\ln 3}+\frac{1}{8}\cos 8x+C$

3^x ln3$-\frac{1}{8}\cos 8x+C$

$I=\frac{1}{2}\ln 3$

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

$I=\frac{1}{4}\ln 3$

$I=\frac{1}{5}\ln 3$

m = 3

m = 0

m = 1

m = 2

$\overrightarrow{n}$ = (5; −6; 7)

$\overrightarrow{n}$ = (−5; 6; 7)

$\overrightarrow{n}$ = (−5; 6; −7)

k = 28

k = 0

k = 30

k = 2

2

1

3

4

1

-12

12

7

$I=-\frac{{\left(\sqrt{7-3x^2}\right)}^3}{3}+C$

$I=\frac{{\left(\sqrt{7-3x^2}\right)}^3}{3}+C$

$I=-3{\left(\sqrt{7-3x^2}\right)}^3+C$

$I=-{\left(\sqrt{7-3x^2}\right)}^3+C$

I = $2\sqrt{2}$

I = $2-\frac{1}{\sqrt{2}}$

I = $2-\sqrt{2}$

I = $1-\frac{1}{\sqrt{2}}$