Bộ 15 đề thi giữa kì 2 Toán 12 có đáp án năm 2022-2023 (Đề 1)

$F\left(x\right)=\frac{x^4}{4}$.

$F\left(x\right)=\frac{x^4}{4}+C$.

$F\left(x\right)=x^3+C$.

$3x^2+C$.

Cho hàm số f(x) xác định trên K và F(x) là một nguyên hàm của f(x) trên K. Khi đó $F^{\text{'}}\left(x\right)=f\left(x\right)$, $\forall x\in K$.

$\int f\text{'}\left(x\right)\text{\hspace{0.17em}d}x=f\left(x\right)+C$.

$\int kf\left(x\right)\text{\hspace{0.17em}\hspace{0.17em}d}x=k\int f\left(x\right)\text{\hspace{0.17em}d}x$ với là hằng số khác 0.

Nếu F(x) và G(x) đều là nguyên hàm của hàm số f(x) thì $F\left(x\right)=G\left(x\right).$

$\int \cos x\text{\hspace{0.17em}\hspace{0.17em}d}x=\sin x$.

$\int \cos x\text{\hspace{0.17em}\hspace{0.17em}d}x=\sin x+C$.

$\int \frac{1}{x}\text{\hspace{0.17em}d}x=\ln x+C$.

$\int x^2\text{\hspace{0.17em}d}x=2x+C$.

$\frac{8}{3}$.

$\frac{-8}{3}$.

2.

-5.

1.

2.

3.

0.

$f\left(x\right)=x-2\cos x+2$.

$f\left(x\right)=x-2\cos x-1$.

$f\left(x\right)=x+2\cos x+2$.

$f\left(x\right)=x+2\cos x-1$.

$F\left(x\right)=\frac{{\left(2x+1\right)}^9}{18}+C$.

$F\left(x\right)=\frac{{\left(2x+1\right)}^{11}}{11}+C$.

$F\left(x\right)=\frac{{\left(2x+1\right)}^{11}}{22}+C$.

$F\left(x\right)=\frac{{\left(2x+1\right)}^9}{9}+C$.

21.

-14.

10.

-24.

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

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

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

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

$I=\underset{0}{\overset{2}{\int }}2x\text{d}x=2\left|\begin{array}{l}2\\ 0\end{array}\right.$.

$I=\underset{0}{\overset{2}{\int }}2x\text{d}x=4x^2\left|\begin{array}{l}2\\ 0\end{array}\right.$.

$I=\underset{0}{\overset{2}{\int }}2x\text{d}x=x^2\left|\begin{array}{l}0\\ 2\end{array}\right.$.

$I=\underset{0}{\overset{2}{\int }}2x\text{d}x=x^2\left|\begin{array}{l}2\\ 0\end{array}\right.$.

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

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

$\underset{a}{\overset{b}{\int }}\left[f\left(x\right).g\left(x\right)\right]\text{d}x=\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x.\underset{a}{\overset{b}{\int }}g\left(x\right)\text{d}x$.

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

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

$\underset{0}{\overset{2}{\int }}f\left(x\right)\text{d}x=\underset{0}{\overset{1}{\int }}f\left(x\right)\text{d}x-\underset{1}{\overset{2}{\int }}f\left(x\right)\text{d}x$.

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

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

$\underset{a}{\overset{a}{\int }}f\left(x\right)\text{\hspace{0.17em}d}x=0$.

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

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

$\underset{a}{\overset{b}{\int }}\left[f\left(x\right).g\left(x\right)\right]\text{d}x=\underset{a}{\overset{b}{\int }}f\left(x\right)\text{d}x.\underset{a}{\overset{b}{\int }}g\left(x\right)\text{d}x$.

-1.

-3.

4.

-5.

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

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

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

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

$\left(2;3;0\right)$.

$\left(2;0;3\right)$.

$\left(0;2;3\right)$.

$\left(2;3\right)$.

$I\left(1;2;3\right)$, $R=5$.

$I\left(1;-2;3\right)$, $R=-5$.

$I\left(1;2;-3\right)$, $R=-5$.

$I\left(1;2;3\right)$, $R=-5$.

$\overrightarrow{n}=\left(3;-2;0\right)$.

$\overrightarrow{n}=\left(3;0;2\right)$.

$\overrightarrow{n}=\left(3;0;-2\right)$.

$\overrightarrow{n}=\left(3;2;0\right)$.

$\overrightarrow{n}=\left(1;-2;0\right)$.

$\overrightarrow{n}=\left(2;1;2\right)$.

$\overrightarrow{n}=\left(0;1;2\right)$.

$\overrightarrow{n}=\left(2;-1;2\right)$.

m = 1.

m = -1.

m = 3.

m = 2.

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

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

$F\left(x\right)=3e^{3x}-6e^{2x}+3e^x$.

$F\left(x\right)=3e^{3x}-6e^{2x}+3e^x-2$.

9.

10.

11.

12.

$2x^3+2x^2+x-3$.

$2x^3+2x^2-3x-3$.

$2x^3-2x^2+x-3$.

$12x+4$.

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

$\frac{x^3}{3}+x+C$.

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

$x+C$.

${\ln }^{3}x+\ln x+C$.

${\ln }^{3}x+C$.

${\ln }^{3}x+x+C$.

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

$\ln \frac{2}{3}$.

ln 6.

$\ln \frac{4}{3}$.

ln 3.

$I=-3$.

$I=-5$.

$I=-2$.

$I=-6$

-5

-7.

-9.

-10.

1.

2.

7.

9.

3.

6.

5.

4.

1.

-7.

-1.

7.

$m=-5$.

$m=5$.

$m=-1$.

$m=1$.

$-1<m<7$.

$-7<m<1$

$\left[\begin{array}{l}m<-1\\ m>7\end{array}\right.$.

$\left[\begin{array}{l}m<-7\\ m>1\end{array}\right.$.

-8.

9.

8.

4.

2

$\frac{1}{2}$

-2

3