Đề kiểm tra Giữa kì 2 Toán 12 có đáp án (Mới nhất) (Đề 9)

$\int f\left(x\right)d\text{x=}\sqrt{2\text{x}+1}\text{+C}$

$\int f\left(x\right)d\text{x=}\frac{1}{2}\sqrt{2\text{x}+1}\text{+C}$

$\int f\left(x\right)d\text{x=2}\sqrt{2\text{x}+1}\text{+C}$

$\int f\left(x\right)d\text{x=}\frac{1}{\left(2\text{x}+1\right)\sqrt{2\text{x}+1}}\text{+C}$

13

4

$\frac{13}{3}$

$\frac{4}{3}$

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

$\ln |2x+3|+C.$

$\frac{1}{2}\ln (2x+3)+C.$

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

f (x) = 3x² + e^x.

f (x) = $\frac{x^4}{12}$+ e^x.

f (x) = x² + e^x.

f (x) = $\frac{x^4}{3}$+ e^x.

I (3; −2; 4), R = 5.

I (3; −2; 4), R = 25.

I (−3; 2; −4), R = 5.

I (−3; 2; −4), R = 25.

2

6

4

8

$\frac{32\pi }{15}$

$\frac{4\pi }{3}$

$\frac{496\pi }{15}$

$\frac{16\pi }{15}$

(x + 1)² + y² + (z − 2)² = 3.

(x − 1)² + y² + (z + 2)² = 9.

(x − 1)² + y² + (z + 2)² = 3.

(x + 1)² + y² + (z − 2)² = 9.

P = 4.

P = −4.

P = 10.

P = 7.

I = ${\left(2x+1\right)e^x|}_0^1-2\underset{0}{\overset{1}{\int }}{e}^{x}d\text{x}$

I = ${\left(2x+1\right)e^x|}_0^1+2\underset{0}{\overset{1}{\int }}{e}^{2x}d\text{x}$

I = ${\left(2x+1\right)e^x|}_0^1-\underset{0}{\overset{1}{\int }}{e}^{x}d\text{x}$

I = ${\left(2x+1\right)e^x|}_0^1+\underset{0}{\overset{1}{\int }}{e}^{2x}d\text{x}$

(2; −1; −3).

(−1; 2; −3).

(2; −3; −1).

(−3; 2; −1).

y + 2z − 5 = 0.

2x − y − 1 = 0.

2x − y + 1 = 0.

− y + 2z − 3 = 0.

$\frac{\sqrt{3}\pi +3}{6}$

$\frac{\sqrt{3}\pi -3}{3}$

$\frac{\sqrt{3}\pi -3}{6}$

$\frac{\sqrt{3}\pi }{6}$

$\frac{2}{15}$

$\frac{16}{225}$

$\log \frac{5}{3}$

$\ln \frac{5}{3}$

$\underset{a}{\overset{b}{\int }}{f}^2\left(x\right)d\text{x}$

$\pi \underset{a}{\overset{b}{\int }}f\left(x\right)d\text{x}$

$\underset{a}{\overset{b}{\int }}\left|f\left(x\right)\right|d\text{x}$

$\underset{a}{\overset{b}{\int }}f\left(x\right)d\text{x}$

F (3) = ln2 − 1.

F (3) = $\frac{1}{2}$.

F (3) = ln2 + 1.

F (3) = $\frac{7}{4}$.

P (0; −1; 0).

M (3; 0; 0).

N (0; −1; 1).

Q (0; 0; 1).

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

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

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

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

2x + y − z + 1 = 0.

2x + y − z − 1 = 0.

x + y + 2z + 1 = 0.

x + y + 2z − 1 = 0.

V = $\pi \underset{0}{\overset{1}{\int }}\sqrt{2\text{x}+1}d\text{x}$

V = $\underset{0}{\overset{1}{\int }}\sqrt{2x+1}\mathrm{dx}$

V = $\pi \underset{0}{\overset{1}{\int }}\left(2\text{x}+1\right)d\text{x}$

V = $\underset{0}{\overset{1}{\int }}\left(2\text{x}+1\right)d\text{x}$

G (6; 3; 3).

G (2; 1; 1).

G (3; 1; 1).

G (1; 2; 1).

$\frac{e^3-2}{9}$

$\frac{e^3+2}{9}$

$\frac{2e^3-1}{9}$

$\frac{2e^3+1}{9}$

$\int 2f\left(x\right)d\text{x}$ = 2$\int f\left(x\right)d\text{x}$.

$\int \left[f\left(x\right)+g\left(x\right)\right]d\text{x}$ = $\int f\left(x\right)d\text{x}$ + $\int g\left(x\right)d\text{x}$.

$\int f\left(x\right).g\left(x\right)d\text{x}$ = $\int f\left(x\right)d\text{x}$.$\int g\left(x\right)d\text{x}$.

$\int \left[f\left(x\right)-g\left(x\right)\right]d\text{x}$ = $\int f\left(x\right)d\text{x}$ − $\int g\left(x\right)d\text{x}$.

I = 4

I = 3

I = 2

I = 1

S = $\frac{49}{4}$

S = $\frac{21}{4}$

S = $\frac{31}{4}$

S = $\frac{39}{4}$