Đề kiểm tra Giữa kì 2 Toán 12 có đáp án (Mới nhất) (Đề 3)

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

$\ln \left|x\right|-\frac{1}{x}+C$

$\ln \left|x\right|+\frac{1}{x}+C$

$\ln \left|x\right|-\frac{1}{x}+C$

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

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

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

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

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

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

$-\cos (3x-1)+C$

Kết quả khác

$F\left(x\right)=e^{\sin x}+4$

$F\left(x\right)=e^{\sin x}+C$

$F\left(x\right)=e^{\cos x}+4$

$F\left(x\right)=e^{\mathrm{cosx}}+C$

$F\left(x\right)=\left(x-1\right)e^x$

$F\left(x\right)=\left(x-2\right)e^x$

$F\left(x\right)=\left(x+1\right)e^x+1$

$F\left(x\right)=\left(x-2\right)e^x+3$

5.

1.

-1

-5

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

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

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

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

$\frac{11}{12}$

$-\frac{145}{12}$

$-\frac{11}{12}$

$\frac{145}{12}$

$I=1$

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

$I=\ln 2$

$I=-\ln 2$

$\frac{4-\sqrt{2}}{3}$

$\frac{8-2\sqrt{2}}{3}$

$\frac{4+\sqrt{2}}{3}$

$\frac{8+2\sqrt{2}}{3}$

$\frac{\pi }{6}+1$

$\frac{\pi }{2}$

$\frac{\pi }{4}$

Đáp án khác

$\left\{5\right\}$

$\left\{5;-1\right\}$

$\left\{4\right\}$

$\left\{4;-1\right\}$

${\pi }^2-4$

${\pi }^2+4$

$2{\pi }^2-3$

$\underset{0}{\overset{\frac{\pi }{6}}{\int }}t\text{d}t$

$\underset{0}{\overset{\frac{\pi }{6}}{\int }}\text{d}t$

$\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{1}{t}\text{d}t$

$\underset{0}{\overset{\frac{\pi }{3}}{\int }}\text{d}t$

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

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

$K=3\ln 2$

$K=2\ln 2-\frac{1}{2}$

-2

$\frac{3\pi }{16}$

$\ln 2-\frac{3}{4}$

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

4

$\frac{9}{2}$

5

$\frac{7}{2}$

$\frac{1}{4}\left(dvdt\right)$

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

$\frac{3}{2}\left(dvdt\right)$

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

$\frac{8\pi }{15}$

2$\pi$

$\frac{46\pi }{15}$

$\frac{5\pi }{2}$

$8\pi$

$\frac{16\pi }{3}$

$\frac{8\pi }{3}$

$\frac{8\pi }{15}$

$\left(11;22\right)$

$\left(0;9\right)$

$\left(7;21\right)$

$\left(2017;2020\right)$

$\frac{e-1}{2}$

$\frac{e^2}{4}$

e-2

$\frac{e}{2}$

0

-3

3

4

$\sqrt{22}$

$2\sqrt{22}$

$\frac{\sqrt{22}}{2}$

$\frac{\sqrt{11}}{2}$

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

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

V =2

V =1

$\left(6;0;-6\right)$

$\left(6;-6;0\right)$

$\left(-6;0;6\right)$

$\left(-6;6;0\right)$

$\frac{3\sqrt{14}}{7}$

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

$\frac{4\sqrt{14}}{7}$

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

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

$D\left(0;2;-1\right)$

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

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

$x^2+{(y-3)}^2+{(z-1)}^2=9$

$x^2+{(y+3)}^2+{(z-1)}^2=9$

$x^2+{(y-3)}^2+{(z+1)}^2=3$

$x^2+{(y-3)}^2+{(z+1)}^2=9$

45^o

60^o.

30^o

120^o