Bộ 15 đề thi giữa kì 2 Toán 12 có đáp án năm 2022-2023 (Đề 4)

$3x^3+x+C$.

$x^3+x+C$.

$x^3+C$.

$\int \left(3x^2+1\right)\text{\hspace{0.17em}d}x=3\frac{x^3}{3}+x+C=x^3+x+C.$

$2\sin x-\cos x+C$.

$-2\sin x-\cos x+C$.

$2\sin x+\cos x+C$.

$-2\sin x+\cos x+C$.

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

$\frac{{\left(x^2+1\right)}^5}{4}+C$.

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

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

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

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

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

$-\frac{1}{3}\sin \left(3x-\frac{1}{3}\right)+C$

$\frac{x^2}{2}+\frac{5^x}{\ln 5}+C$.

$\frac{x^2}{2}+5^x.\ln 5+C$.

$1+\frac{5^x}{\ln 5}+C$.

$x^2+\frac{5^x}{\ln 5}+C$

$\frac{2}{9}{\left(1+3\ln x\right)}^2\left[{\left(1+3\ln x\right)}^2-1\right]+C$.

$\left(1+3\ln x\right)\sqrt{1+3\ln x}\left(\frac{1+3\ln x}{5}-\frac{1}{3}\right)+C$.

$\frac{2}{9}\left(1+3\ln x\right)\sqrt{1+3\ln x}\left(\frac{1+3\ln x}{5}-\frac{1}{3}\right)+C$.

$\frac{2}{3}\left(1+3\ln x\right)\sqrt{1+3\ln x}\left(\frac{1+3\ln x}{5}-\frac{1}{3}\right)+C$.

$I=\frac{1}{12}$.

$I=-\frac{1}{12}$.

$I=\frac{37}{320}$.

$I=\frac{7}{640}$.

0.

$\pi +1$.

$\pi +2$.

1.

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

$I=2$.

$I=\frac{2}{3}$.

$\frac{-3}{\text{e}}$.

${\text{e}}^2$

$3{\text{e}}^2$.

$\frac{3}{\text{e}}$.

12.

4.

-12.

8.

$-2\ln 2$.

$-4\ln 2$.

$\ln 2$.

$4\ln 2$

$b-a=1$.

$b-a=-1$.

$b-a=7$.

$b-a=-7$.

$I=4\ln 2-3$.

$I=2\ln 2-3$.

$I=2\ln 2+3$.

$I=3\ln 2-4$.

$T=-4$.

$T=21$.

$T=9$.

$T=-12$.

$I=\frac{3}{5}$.

$I=\frac{1}{2}$.

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

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

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

$\underset{\frac{1}{3}}{\overset{2}{\int }}f\left(x\right)\text{d}x$.

$\underset{-1}{\overset{\frac{1}{3}}{\int }}f\left(x\right)\text{d}x-\underset{\frac{1}{3}}{\overset{2}{\int }}f\left(x\right)\text{d}x$.

$-\underset{-1}{\overset{\frac{1}{3}}{\int }}f\left(x\right)\text{d}x+\underset{\frac{1}{3}}{\overset{2}{\int }}f\left(x\right)\text{d}x$.

$\frac{11}{20}$.

$\frac{1}{20}$.

$\frac{19}{20}$.

$\frac{117}{20}$.

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

$S=\frac{11}{2}.$

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

$S=\frac{9}{4}.$

$2869834$ đồng.

$1434917$ đồng.

$2119834$ đồng.

$684917$ đồng.

16000.

21750.

12750.

11750.

6.

2.

4.

8.

6.

7.

8.

9.

$\underset{1}{\overset{4}{\int }}\sqrt{x}\text{\hspace{0.17em}\hspace{0.05em}d}x$.

$\pi \underset{1}{\overset{4}{\int }}x\text{\hspace{0.17em}d}x$.

$\pi \underset{1}{\overset{4}{\int }}\sqrt{x}\text{d}x$.

$\pi \underset{1}{\overset{4}{\int }}{x}^2\text{\hspace{0.17em}d}x$.

$A=13$.

$A=19$.

$A=21$.

$A=29$.

$\left(1;0;4\right)$.

$\left(1;6;1\right)$.

$\left(1;-4;6\right)$.

$\left(1;-10;9\right)$.

$\sqrt{30}$.

$\sqrt{10}$.

$\sqrt{22}$.

2.

$I\left(-1;2;1\right)$ và $R=2$.

$I\left(1;-2;-1\right)$ và $R=2$.

$I\left(-1;2;1\right)$ và $R=4$.

$I\left(1;-2;-1\right)$ và $R=4$.

${\left(x-2\right)}^2+{\left(y+1\right)}^2+{(z-2)}^2=\sqrt{24}$.

${\left(x-2\right)}^2+{\left(y+1\right)}^2+{(z-2)}^2=24$.

${\left(x+2\right)}^2+{\left(y-1\right)}^2+z^2=24$.

${\left(x-2\right)}^2+{\left(y-1\right)}^2+{(z-2)}^2=24$.

$x^2+y^2+{(z-2)}^2=3\text{.}$

$x^2+y^2+{(z+2)}^2=3\text{.}$

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

$x^2+y^2+{(z+2)}^2=9.$

$V=\frac{\pi a^3\sqrt{6}}{8}$.

$V=\frac{\pi a^3\sqrt{6}}{4}$.

$V=\frac{\pi a^3\sqrt{3}}{8}$.

$V=\frac{\pi a^2\sqrt{6}}{8}$.

$\frac{\sqrt{6}}{2}$.

$\sqrt{3}$.

$\sqrt{2}$.

$\frac{\sqrt{2}}{2}$.

$x+y-3z+6=0$.

$x-y-3z+5=0$.

$x-y-3z+1=0$.

$2x+y+2z+10=0$.

$8x-y+5z+23=0$.

$4x+y-5z+25=0$.