Đề thi giữa HK2 môn Toán 12 năm 2021 - Trường THPT...

A. \(-\frac{1}{6} \sqrt{\left(5-4 x^{2}\right)^{3}}+C\)

B. \(-\frac{3}{8} \sqrt{\left(5-4 x^{2}\right)}+C\)

C. \(\frac{1}{6} \sqrt{\left(5-4 x^{2}\right)^{3}}+C\)

D. \(-\frac{1}{12} \sqrt{\left(5-4 x^{2}\right)^{3}}+C\)

A. \(F\left( x \right) = 2x - \frac{3}{x} + 2\)

B. \(F\left( x \right) = 2\ln \left| x \right| + \frac{3}{x} + 2\)

C. \(F\left( x \right) = 2x + \frac{3}{x} - 4\)

D. \(F\left( x \right) = 2\ln \left| x \right| - \frac{3}{x} + 4\)

A. \(-\frac{1}{4 \sin ^{4} x}\)

B. \(\frac{1}{4 \sin ^{4} x}\)

C. \(\frac{4}{\sin ^{4} x}\)

D. \(\frac{-4}{\sin ^{4} x}\)

A. \(F\left( x \right) = \frac{{2{x^3}}}{3} - \frac{3}{x} + C\)

B. \(F\left( x \right) = \frac{{{x^3}}}{3} - \frac{3}{x} + C\)

C. \(F\left( x \right) =  - 3{x^3} - \frac{3}{x} + C\)

D. \(F\left( x \right) = \frac{{2{x^3}}}{3} + \frac{3}{x} + C\)

A. \(f(x)=x^{3}-\sqrt{x}-\frac{1}{x}-x\)

B. \(f(x)=x^{3}-2 \sqrt{x}-\frac{1}{x}-x\)

C. \(f(x)=x^{3}-2 \sqrt{x}+\frac{1}{x}\)

D. \(f(x)=x^{3}-\frac{1}{2} \sqrt{x}-\frac{1}{x}-x\)

A. \(2 \alpha\)

B. \( \alpha\)

C. \(4 \alpha\)

D. \(\frac{\alpha}{2}\)

A. F(2)-F(1)

B. -F(1)

C. F(2)

D. F(1)-F(2)

A. \(\int_{0}^{2}\left(x^{2}+x-3\right) d x\)

B. \(3 \int_{0}^{3 \pi} \sin x d x\)

C. \(\int_{0}^{\ln \sqrt{10}} e^{2 x} d x\)

D. \(\int_{0}^{\pi} \cos (3 x+\pi) d x\)

A. Nếu \(m \leq f(x) \leq M \forall x \in[a ; b] \text { thì } m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(a-b)\)

B. Nếu \(\begin{array}{l} f(x) \geq m \forall x \in[a ; b] \end{array}\) thì \( \int_{a}^{b} f(x) d x \geq m(b-a)\)

C. Nếu \(f(x) \leq M \forall x \in[a ; b] \) thì \(\int_{a}^{b} f(x) d x \leq M(b-a)\)

D. Nếu \(f(x) \geq m \forall x \in[a ; b]\) thì \(\int_{a}^{b} f(x) d x \geq m(a-b)\)

A. \( S = \left| {\mathop \smallint \nolimits_{ - 1}^1 \left( {3x - {x^3}} \right)dx} \right|\)

B. \( S = \mathop \smallint \nolimits_{ - 1}^0 \left( {3x - {x^3}} \right)dx + \mathop \smallint \nolimits_0^1 \left( {{x^3} - 3x} \right)dx\)

C. \( S = \mathop \smallint \nolimits_{ - 1}^1 \left( {3x - {x^3}} \right)dx\)

D. \( S = \mathop \smallint \nolimits_{ - 1}^0 \left( {{x^3} - 3x} \right)dx + \mathop \smallint \nolimits_0^1 \left( {3x - {x^3}} \right)dx\)

A. \( S = \mathop \smallint \limits_0^\pi \cos x{\mkern 1mu} {\rm{d}}x.\)

B. \( S = \mathop \smallint \limits_0^\pi \cos^2 x{\mkern 1mu} {\rm{d}}x.\)

C. \( S = \mathop \smallint \limits_0^\pi \left| {\cos x} \right|{\mkern 1mu} {\rm{d}}x.\)

D. \( S =\pi \mathop \smallint \limits_0^\pi \left| {\cos x} \right|{\mkern 1mu} {\rm{d}}x.\)

A. \( S = \mathop \smallint \limits_0^e \left| {{e^x} + x} \right|dx\)

B. \( S = \mathop \smallint \limits_0^e \left| {{e^x} - x} \right|dx\)

C. \( S = \mathop \smallint \limits_e^0 \left| {{e^x} - x} \right|dx\)

D. \( S = \mathop \smallint \limits_e^0 \left| {{e^x} + x} \right|dx\)

A. \( S = \mathop \smallint \limits_a^b \left[ {f\left( x \right) - g\left( x \right)} \right]{\mkern 1mu} {\rm{d}}x.\)

B. \( S = \mathop \smallint \limits_a^b \left[ {g\left( x \right) - f\left( x \right)} \right]{\mkern 1mu} {\rm{d}}x.\)

C. \( S = \mathop \smallint \limits_a^b \left[ {g\left( x \right) - f\left( x \right)} \right]{\mkern 1mu} {\rm{d}}x.\)

D. \( S = \mathop \smallint \limits_a^b \left| {f\left( x \right) - g\left( x \right)} \right|{\mkern 1mu} {\rm{d}}x.\)

A. M(2;0;1)

B. M(2;1;0)

C. M(0;2;1)

D. M(1;2;0)

A. M(1;1;−3)

B. M(1;−1;−3)

C. M(1;−3;1)

D. M(−1;−3;1)

A. (a;b;c)

B. (a;c;b)

C. (c;b;a)

D. (c;a;b)

A. \(\overrightarrow {OM} = x.\vec i + y.\vec j + z.\vec k\)

B. \(\overrightarrow {OM} = z.\vec i + y.\vec j + x.\vec k\)

C. \(\overrightarrow {OM} = z.\vec i + x.\vec j + y.\vec k\)

D. \(\overrightarrow {OM} = z.\vec i + y.\vec j + x \vec k\)

A. H(0 ; 7 ;-13)

B. H(5 ; 7 ; 0)

C. H(0 ;-7 ; 13)

D. H(5 ; 0 ;-13)

A. Q(0 ; 0 ; 5)

B. M(3 ; 0 ; 0)

C. N(0 ;-4 ; 5)

D. P(3 ; 0 ; 5)

A. F(0 ; 2 ; 0)

B. E(1 ; 0 ; 3)

C. K(0 ; 2 ; 3)

D. H(1 ; 2 ; 0)

A. H(5 ;-6 ; 7)

B. H(2 ; 0 ; 4)

C. H(3 ;-2 ; 5)

D. H(-1 ; 6 ; 1)

A. M(-1 ; 2 ; 0)

B. P(0 ; 2 ; 1)

C. N(-1 ; 0 ; 1)

D. Q(0 ; 2 ; 0)

A. \(\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+1}{-2}\)

B. \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{1}\)

C. \(\frac{x-1}{-2}=\frac{y-2}{-6}=\frac{z+1}{-4}\)

D. \(\frac{x+1}{1}=\frac{y+2}{3}=\frac{z-1}{2}\)

A. \(\frac{x+2}{1}=\frac{y+2}{2}=\frac{z+1}{3}\)

B. \(\frac{x-2}{1}=\frac{y-2}{2}=\frac{z-1}{3}\)

C. \(\frac{x+1}{2}=\frac{y+2}{2}=\frac{z+3}{1}\)

D. \(\frac{x-1}{2}=\frac{y-2}{2}=\frac{z-3}{1}\)

A. \(\left\{\begin{array}{l}x=3 \\ y=-1+t \\ z=t\end{array}\right.\)

B. \(\left\{\begin{array}{l}x=3 \\ y=-1 \\ z=t\end{array}\right.\)

C. \(\left\{\begin{array}{l} x=3+t \\ y=-1 \\ z=0 \end{array}\right.\)

D. \(\begin{aligned} &\left\{\begin{array}{l} x=3 \\ y=-1+t \\ z=0 \end{array}\right.\\ \end{aligned}\)

A. \(\frac{x-1}{1}=\frac{y-2}{-2}=\frac{z}{1}\)

B. \(\frac{x-1}{1}=\frac{y+2}{2}=\frac{z}{2}\)

C. \(\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z}{1}\)

D. \(\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z}{1}\)

A. \(\frac{x-1}{-3}=\frac{y}{-4}=\frac{z+1}{8}\)

B. \(\frac{x+1}{-3}=\frac{y-3}{-4}=\frac{z}{8}\)

C. \(\frac{x-1}{-3}=\frac{y-3}{-4}=\frac{z}{8}\)

D. \(\frac{x-1}{-3}=\frac{y}{-4}=\frac{z-1}{8}\)

A. \(d: \frac{x-1}{2}=\frac{y-4}{2}=\frac{z+7}{1}\)

B. \(d: \frac{x-1}{4}=y+4=\frac{z+7}{2}\)

C. \(d: \frac{x-1}{1}=\frac{y-4}{2}=-\frac{z+7}{2}\)

D. \(d: \frac{x-1}{1}=\frac{y-4}{2}=\frac{z+7}{2}\)

A. \(\Delta:\left\{\begin{array}{l}x=1+3 t \\ y=-3+2 t \\ z=-2+2 t\end{array}\right.\)

B. \(\Delta:\left\{\begin{array}{l}x=1+4 t \\ y=-3-t \\ z=-2\end{array}\right.\)

C. \(\Delta:\left\{\begin{array}{l}x=3+4 t \\ y=2-t \\ z=2\end{array}\right.\)

D. \(\Delta:\left\{\begin{array}{l}x=3-t \\ y=2+3 t \\ z=2+2 t\end{array}\right.\)

A. \(\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+4}{-1}\)

B. \(\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-4}{-1}\)

C. \(\frac{x-2}{2}=\frac{y+3}{-1}=\frac{z-5}{1}\)

D. \(\frac{x-2}{1}=\frac{y-1}{-3}=\frac{z+1}{4}\)

A. \(\left\{\begin{array}{l}x=0 \\ y=1 \\ z=t\end{array}\right.\)

B. \(\left\{\begin{array}{l}x=0 \\ y=t \\ z=0\end{array}\right.\)

C. \(\left\{\begin{array}{l}x=t \\ y=0 \\ z=0\end{array}\right.\)

D. \(\left\{\begin{array}{l}x=0 \\ y=0 \\ z=t\end{array}\right.\)

A. \(\frac{x-1}{2}=\frac{y+3}{-4}=\frac{z+2}{1}\)

B. \(\frac{x+1}{-2}=\frac{y-3}{-2}=\frac{z-2}{-4}\)

C. \(\frac{x+1}{2}=\frac{y-3}{-4}=\frac{z-2}{1}\)

D. \(\frac{x-2}{-1}=\frac{y+4}{3}=\frac{z-1}{2}\)

A. \([\vec a, \vec b]=(-3;-3;-6)\)

B. \([\vec a, \vec b]=(3;3;-6)\)

C. \([\vec a, \vec b]=(1;1;-2)\)

D. \([\vec a, \vec b]=(-1;-1;2)\)

A. \(\frac{1}{2}\)

B. \(\frac{1}{6}\)

C. \(\frac{1}{4}\)

D. \(\frac{1}{3}\)

A. \(\sqrt{45}\over7\)

B. \(270\over7\)

C. \(45\over7\)

D. \(90\over7\)