Đề kiểm tra Giữa học kì 2 Toán 12 có đáp án (Mới nhất) (Đề 1)

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

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

$x^3-3x^2+\ln x+C$

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

$F\left(x\right)=\frac{3x^2}{2}-\frac{3}{2x}-\frac{1}{2}$

$F\left(x\right)=\frac{3x^2}{4}-\frac{3}{2x}-\frac{7}{4}$

$F\left(x\right)=\frac{3x^2}{2}+\frac{3}{4x}-\frac{7}{4}$

$F\left(x\right)=\frac{3x^2}{4}+\frac{3}{2x}+\frac{7}{4}$

$-\frac{1}{5}\cos 5x-\cos x+C$

$\frac{1}{5}\cos 5x+\cos x+C$

$5\cos 5x+\cos x+C$

Kết quả khác

$F\left(x\right)=\frac{1}{2}{e}^{x^2}+2$

$F\left(x\right)=\frac{1}{2}\left(e^{x^2}+5\right)$

$F\left(x\right)=\frac{1}{2}{e}^{x^2}+C$

$F\left(x\right)=-\frac{1}{2}\left(2-e^{x^2}\right)$

$I=\ln x.\ln \left(\ln x\right)+C.$

$I=\ln x.\ln \left(\ln x\right)+\ln x+C.$

$I=\ln x.\ln \left(\ln x\right)-\ln x+C.$

$I=\ln \left(\ln x\right)+\ln x+C.$

2 .

4 .

6 .

8

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

$F\text{'}\left(x\right)=f\left(x\right)$ với mọi $x\in (a;b)$

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

Hàm số Gcho bởi $G\left(x\right)=F\left(x\right)+5$ cũng thỏa mãn $\underset{a}{\overset{b}{\int }}f\left(x\right)dx=G\left(b\right)-G\left(a\right)$

I = 1.

I = 2.

I = 3

I = -1.

$K=\ln 2$

$K=2\ln 2$

$K=\ln \frac{8}{3}$

$K=\frac{1}{2}\ln \frac{8}{3}$

b = 0 hoặc b = 2.

b = 0 hoặc b = 4.

b= 1 hoặc b= 2.

b = 1 hoặc b = 4.

-6 .

6 .

-3 .

3.

$\frac{\pi }{4}-\frac{\sqrt{2}}{2}$

$\frac{\pi }{4}+\frac{\sqrt{2}}{2}$

$-\frac{\pi }{4}-\frac{\sqrt{2}}{2}$

$-\frac{\pi }{4}+\frac{\sqrt{2}}{2}$

6

5

4

$\frac{1}{64}$

L = p

L = -p

L = -2

L = 0

$a=\pi ,b=0$

$a=\pi ,b=2$

$a=2\pi ,b=2$

$a=2\pi ,b=3$

$\frac{1}{2}\left(1-\ln 2\right)$

$\frac{1}{2}\left(1+\ln 2\right)$

$\frac{1}{2}\left(\ln 2-1\right)$

$\frac{1}{4}\left(1+\ln 2\right)$

$\frac{1}{6}\left(dvdt\right)$

$\frac{5}{2}\left(dvdt\right)$

$\frac{6}{5}\left(dvdt\right)$

$\frac{1}{2}\left(dvdt\right)$

$\underset{-3}{\overset{0}{\int }}f\left(x\right)\text{d}x+\underset{4}{\overset{0}{\int }}f\left(x\right)\text{d}x$

$\underset{-3}{\overset{1}{\int }}f\left(x\right)\text{d}x+\underset{1}{\overset{4}{\int }}f\left(x\right)\text{d}x$

$\underset{0}{\overset{-3}{\int }}f\left(x\right)\text{d}x+\underset{0}{\overset{4}{\int }}f\left(x\right)\text{d}x$

$\underset{-3}{\overset{4}{\int }}f\left(x\right)\text{d}x$

$\pi \underset{a}{\overset{b}{\int }}{\left[f\left(x\right)-g\left(x\right)\right]}^2\text{d}x$

$\pi \underset{a}{\overset{b}{\int }}\left[f^2\left(x\right)-g^2\left(x\right)\right]\text{d}x$

${\left\{\pi \underset{a}{\overset{b}{\int }}\left[f\left(x\right)-g\left(x\right)\right]\text{d}x\right\}}^2$

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

$-\pi$

$\frac{\pi }{3}$

$\frac{\pi }{6}$

$\pi$

P = -2

P = 3

P = 0

P = 4

S = 4

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

S = 0

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

$\left(-1;0;3\right)$

$\left(0;-1;3\right)$

$\left(-1;3;0\right)$

$\left(-1;3\right)$

m = 2 và n = –1.

m = –2 và n = 1.

m = 1 và n = –2.

m = –1 và n = 2.

m = 0 V m = –2.

m = –1 V m = 2.

m = 0 V m = –1.

m = 2 V m = 0.

$Q\left(-2;-3;-4\right).$

$Q\left(2;3;-4\right).$

$Q\left(-2;-3;4\right).$

$Q\left(4;4;2\right).$

(-1;2;0)

(-1;0;0)

$\left(0;0;3\right)$

(0;2;0)

x + y = 1

x + y = 17

x + y = $\frac{11}{5}$

x + y = $-\frac{11}{5}$

${135}^0$

${45}^0$

${60}^0$

${120}^0$

${45}^0$

${60}^0$

${90}^0$

${135}^0$

$M\left(-2;6;-4\right)$

$M\left(2;-6;4\right)$

$M\left(-2;-6;4\right)$

$M\left(5;5;0\right)$

$\frac{\sqrt{21}}{6}$

$\frac{\sqrt{42}}{3}$

$\frac{\sqrt{14}}{6}$

$\frac{\sqrt{14}}{3}$

8

4

6

10

V = $\frac{1}{6}$

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

V = 2

V = 1

$\frac{1}{\sqrt{21}}$

$\frac{\sqrt{21}}{7}$

$\frac{2}{\sqrt{13}}$

$\frac{\sqrt{21}}{3}$