Đề thi Học kì 2 Toán 11 có đáp án (Đề 4)

A. $2\stackrel{\to }{OI}=\frac{1}{2}\left(\stackrel{\to }{u}+\stackrel{\to }{v}+\stackrel{\to }{x}+\stackrel{\to }{y}\right)$

B. $2\stackrel{\to }{OI}=-\frac{1}{2}\left(\stackrel{\to }{u}+\stackrel{\to }{v}+\stackrel{\to }{x}+\stackrel{\to }{y}\right)$

C. $2\stackrel{\to }{OI}=\frac{1}{4}\left(\stackrel{\to }{u}+\stackrel{\to }{v}+\stackrel{\to }{x}+\stackrel{\to }{y}\right)$

D. $2\stackrel{\to }{OI}=-\frac{1}{4}\left(\stackrel{\to }{u}+\stackrel{\to }{v}+\stackrel{\to }{x}+\stackrel{\to }{y}\right)$

A. 3

B. 18

C. -1

D. 1

A. $\frac{-x}{\left(x^2+2\right)\sqrt{x^2+2}}$

B. $\frac{x+8}{\left(x^2+2\right)\sqrt{x^2+2}}$

C. $\frac{-x+8}{\left(x^2+3\right)\sqrt{x^2+2}}$

D. $\frac{-x+8}{\left(x^2+2\right)\sqrt{x^2+2}}$

A. $\frac{1126}{33}$

B. $\frac{1263}{34}$

C. $\frac{1311}{35}$

D. $\frac{1411}{37}$

A. $\stackrel{\to }{IK}=\frac{1}{2}\stackrel{\to }{AC}=\frac{1}{2}\stackrel{\to }{A\text{'}C\text{'}}$

B. Bốn điểm I,K,C,A đồng phẳng

C. $\stackrel{\to }{BD}+2\stackrel{\to }{IK}=2\stackrel{\to }{BC}$

D.  Ba vecto $\stackrel{\to }{BD},\stackrel{\to }{IK},\stackrel{\to }{B\text{'}C\text{'}}$ không đồng phẳng

A. $\cos \phi =\frac{1}{4}$

B. $\phi =60°$

C. $\phi =30°$

D. $\cos \phi =\frac{3}{4}$

A. SA ⊥ BC

B. AH ⊥ BC

C. AH ⊥ AC

D. AH ⊥ SC

A. 24 $m/s^2$

B. 17 $m/s^2$

C. 14 $m/s^2$

D. 12 $m/s^2$

A. $f\text{'}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$

B. $f\text{'}\left(x_0\right)=\underset{∆x\to 0}{\lim }\frac{f\left(x_0+∆x\right)-f\left(x_0\right)}{∆x}$

C. $f\text{'}\left(x_0\right)=\underset{h\to 0}{\lim }\frac{f\left(x_0+h\right)-f\left(x_0\right)}{h}$

D. $f\text{'}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x+x_0\right)-f\left(x_0\right)}{x-x_0}$

A. -4

B. 4

C. 2

D. 1

A. 1

B. 3

C. 4

D. 2

A. $\left(cotx\right)\text{'}=-\frac{1}{si{n}^{2}x}$

B. (sinx)' = -cosx

C. (cosx)' = sinx

D. $\left(tamx\right)\text{'}=-\frac{1}{co{s}^{2}x}$

A. y = 8x-6; y = -8x-6

B. y = 8x-6; y = -8x+6

C. y = 8x-8; y = -8x+8

D. y = 40x-57

A. $y\text{'}=\frac{1}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}$

B. $y\text{'}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}$

C. $y\text{'}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\left(1-\frac{1}{x^2}\right)$

D. $y\text{'}=\frac{1+{\tan }^2\left(x+\frac{1}{x}\right)}{2\sqrt{2+\tan \left(x+\frac{1}{x}\right)}}\left(1+\frac{1}{x^2}\right)$

A. $dy=\frac{3}{\sqrt{3x+2}}dx$

B. $dy=\frac{1}{2\sqrt{3x+2}}dx$

C. $dy=\frac{1}{\sqrt{3x+2}}dx$

D. $dy=\frac{3}{2\sqrt{3x+2}}dx$

A. $lim\frac{n^2+n}{\sqrt{n^2+10n}}$

B. $lim\frac{2n+700}{\sqrt{n^3+1}}$

C. $lim\left(10-3n\right)$

D. $lim\frac{1-n}{\sqrt{n^2+10n}}$

A. $y\text{'}=\frac{x^2+6x+7}{{\left(x+2\right)}^2}$

B. $y\text{'}=\frac{x^2+8x+7}{{\left(x+2\right)}^2}$

C. $y\text{'}=\frac{x^2+4x+7}{{\left(x+2\right)}^2}$

D. $y\text{'}=\frac{x^2+6x+5}{{\left(x+2\right)}^2}$

A. $SO\perp \left(ABCD\right)$

B. $CD\perp \left(SBD\right)$

C. $AB\perp \left(SAC\right)$

D. $CD\perp AC$

A. 7

B. 8

C. 11

D. 4

A. -3

B. 3

C. 20

D. 15

A. Góc giữa AC và (BCD) là góc ACB.

B. Góc giữa AD và (ABC) là góc ADB.

C. Góc giữa AC và (ABD) là góc CAB.

D. Góc giữa CD và (ABD) là góc CBD.

A. $+\infty$

B. $-\infty$

C. $-\frac{1}{3}$

D. 1

A. 30°

B. 45°

C. 60°

D. 75°

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

B. $\frac{2}{5}$

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

D. $\frac{1}{2}$

A. $\left(\frac{1}{\sqrt{3}};+\infty \right)$

B. $\left(-\infty ;\frac{1}{\sqrt{3}}\right)$

C. $[\frac{2}{\sqrt{3}};+\infty )$

D. $[\frac{1}{\sqrt{3}};+\infty )$