Đề ôn tập Chương 3 Giải tích lớp 12 năm 2021 Trường THPT Nguyễn Hữu Thọ

A. \(a = \frac{1}{5}\)

B. \(a = \frac{2}{5}\)

C. \(a = \frac{3}{5}\)

D. \(a = \frac{4}{5}\)

A. \(I =  - \frac{{15a}}{{16}} + \ln 2\)

B. \(I = \frac{{15a}}{{16}} - \ln 2\)

C. \(I = \frac{{15a}}{{16}} + \ln 2\)

D. \(I =  - \frac{{15a}}{{16}} - \ln 2\)

A. a = 5

B. a = 6

C. a = 7

D. a = 8

A. \(I = 2\ln 2 - \frac{5}{4}\)

B. \(I = 2\ln 2 + \frac{3}{4}\)

C. \(I = 2\ln 2 + \frac{5}{4}\)

D. \(I = 2\ln 2 - \frac{3}{4}\)

A. \(I = \frac{{{a^2}\ln a}}{2} + \frac{{1 - {a^2}}}{4}\)

B. \(I = \frac{{{a^2}\ln a}}{2} - \frac{{1 - {a^2}}}{4}\)

C. \(I = \frac{{{a^2}\ln \left| a \right|}}{2} + \frac{{1 - {a^2}}}{4}\)

D. \(I = \frac{{{a^2}\ln \left| a \right|}}{2} - \frac{{1 - {a^2}}}{4}\)

A. \(I = \frac{{\pi + 6 - 3\sqrt[{}]{3}}}{{6a}}\)

B. \(I = \frac{{\pi + 3 - 3\sqrt[{}]{3}}}{{6a}}\)

C. \(I = \frac{{\pi + 6 + 3\sqrt[{}]{3}}}{{6a}}\)

D. \(I = \frac{{\pi + 3 + 3\sqrt[{}]{3}}}{{6a}}\)

A. \(I = \ln 2 - \frac{1}{2}\)

B. \(I = 2\ln 2 - \frac{1}{2}\)

C. \(I = 2\ln 2 - 1\)

D. \(I = \ln 2 - 1\)

A. \(I = \frac{{{e^2} + 1}}{4}\)

B. \(I = \frac{{{e^2} + 3}}{4}\)

C. \(I = \frac{{{e^2} + 5}}{4}\)

D. \(I = \frac{{{e^2} + 7}}{4}\)

A. \(I = \frac{{4\sqrt 2  + 3}}{3}\)

B. \(I = \frac{{4\sqrt 2  + 1}}{3}\)

C. \(I = \frac{{4\sqrt 2  + 5}}{3}\)

D. \(I = \frac{{4\sqrt 2  - 3}}{3}\)

A. \(I = \frac{{5{\pi ^4}}}{{324}} + \frac{{2{\pi ^2}}}{9} + \frac{\pi }{4} - \frac{{\sqrt 3 }}{2}\)

B. \(I = \frac{{5{\pi ^4}}}{{324}} - \frac{{2{\pi ^2}}}{9} + \frac{\pi }{4} - \frac{{\sqrt 3 }}{2}\)

C. \(I = \frac{{5{\pi ^4}}}{{324}} + \frac{{2{\pi ^2}}}{9} - \frac{\pi }{4} - \frac{{\sqrt 3 }}{2}\)

D. \(I = \frac{{5{\pi ^4}}}{{324}} + \frac{{2{\pi ^2}}}{9} + \frac{\pi }{4} + \frac{{\sqrt 3 }}{2}\)

A. \(I = \ln \left| {\frac{{{e^{\frac{\pi }{3}}}\left( {{e^{\frac{\pi }{3}}} + 2} \right)}}{{{e^{\frac{{2\pi }}{3}}} - 2}}} \right|\)

B. \(I = \ln \left| {\frac{{{e^{\frac{\pi }{3}}}\left( {{e^{\frac{\pi }{3}}} - 2} \right)}}{{{e^{\frac{{2\pi }}{3}}} - 2}}} \right|\)

C. \(I = \ln \left| {\frac{{{e^{\frac{\pi }{3}}}\left( {{e^{\frac{\pi }{3}}} + 2} \right)}}{{{e^{\frac{{2\pi }}{3}}} + 2}}} \right|\)

D. \(I = \ln \left| {\frac{{{e^{\frac{\pi }{3}}}\left( {{e^{\frac{\pi }{3}}} - 2} \right)}}{{{e^{\frac{{2\pi }}{3}}} + 2}}} \right|\)

A. I = -2e

B. I = -e

C. I = e

D. I = 2e

A. \(I = \sqrt 2  - 1 + \ln \left( {\sqrt 2  - 1} \right)\)

B. \(I = \sqrt 2  - 1 - \ln \left( {\sqrt 2  - 1} \right)\)

C. \(I =  - \sqrt 2  + 1 + \ln \left( {\sqrt 2  - 1} \right)\)

D. \(I =  - \sqrt 2  + 1 - \ln \left( {\sqrt 2  - 1} \right)\)

A. \(I = \frac{\pi }{4}\tan \frac{\pi }{8} - 2\ln \left( {\cos \frac{\pi }{8}} \right)\)

B. \(I = \frac{\pi }{4}\tan \frac{\pi }{8} + 2\ln \left( {\cos \frac{\pi }{8}} \right)\)

C. \(I = \frac{\pi }{4}\tan \frac{\pi }{4} - 2\ln \left( {\cos \frac{\pi }{8}} \right)\)

D. \(I = \frac{\pi }{4}\tan \frac{\pi }{4} + 2\ln \left( {\cos \frac{\pi }{8}} \right)\)

A. \(I = \frac{1}{2}\left( { - \pi + \frac{{2\pi \sqrt 3 }}{3} + 4\ln \sqrt 2 + \ln 2} \right)\)

B. \(I = \frac{1}{2}\left( { - \pi + \frac{{2\pi \sqrt 3 }}{3} + 2\ln \sqrt 2 - \ln 2} \right)\)

C. \(I = \frac{1}{2}\left( { - \pi + \frac{{2\pi \sqrt 3 }}{3} + 4\ln \sqrt 2 - \ln 2} \right)\)

D. \(I = \frac{1}{2}\left( { - \pi + \frac{{2\pi \sqrt 3 }}{3} + 2\ln \sqrt 2 + \ln 2} \right)\)

A. \(a =  - \frac{\pi }{2}\)

B. \(a =  - \frac{\pi }{4}\)

C. \(a = \frac{\pi }{3}\)

D. \(a = \frac{\pi }{6}\)

A. \(I = \frac{\pi }{{12}} + \ln \left( {\sqrt 3  + 1} \right)\)

B. \(I = \frac{\pi }{{12}} + \ln \frac{{\sqrt 3  + 1}}{4}\)

C. \(I = \frac{\pi }{{12}} - \frac{{\ln \left( {\frac{{\sqrt 3  + 1}}{2}} \right)}}{2}\)

D. \(I = \frac{\pi }{{12}} + \ln \frac{{\sqrt 3  + 1}}{2}\)

A. \(I = \frac{1}{{2\sqrt 2 }}\left( {\ln \frac{{\sqrt 2  - 2}}{{\sqrt 2  + 2}} + \ln \frac{{\sqrt 2  - 1}}{{\sqrt 2  + 1}}} \right)\)

B. \(I = \frac{1}{{2\sqrt 2 }}\left( {\ln \frac{{\sqrt 2  - 2}}{{\sqrt 2  + 2}} - \ln \frac{{\sqrt 2  + 1}}{{\sqrt 2  - 1}}} \right)\)

C. \(I = \frac{1}{{2\sqrt 2 }}\left( {\ln \frac{{\sqrt 2  - 2}}{{\sqrt 2  + 2}} - \ln \frac{{\sqrt 2  - 1}}{{\sqrt 2  + 1}}} \right)\)

D. \(I = \frac{1}{{2\sqrt 2 }}\left( {\ln \frac{{\sqrt 2  + 2}}{{\sqrt 2  - 2}} - \ln \frac{{\sqrt 2  - 1}}{{\sqrt 2  + 1}}} \right)\)

A. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} - 1} \right) - \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)

B. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} + 1} \right) - \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)

C. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} - 1} \right) + \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)

D. \(I = \ln \left( {\frac{{{\pi ^2}}}{4} + 1} \right) + \ln \left( {\frac{{{\pi ^2}}}{{16}} + \frac{{\sqrt 2 }}{2}} \right)\)

A. a = 1

B. a = 2

C. a = 3

D. a = 4

A. \({I_2} = \frac{{17}}{3}\)

B. \({I_2} = \frac{{19}}{3}\)

C. \({I_2} = \frac{{16}}{3}\)

D. \({I_2} = \frac{{13}}{3}\)

A. -2

B. -4

C. 2

D. 4

A. -1

B. -2

C. -3

D. -4

A. -1

B. 1

C. -2

D. 2

A. -1

B. -2

C. 1

D. 2

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

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

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

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

A. \(I = \int\limits_a^b {\left( {{x^2} + 1} \right)dx}  = \int\limits_a^b {{x^2}dx + \int\limits_a^b {dx} } \)

B. \(I = \left. {\left( {{x^3} + x} \right)} \right|_a^b\)

C. \(I = \frac{1}{3}{b^3} + b - \frac{1}{3}{a^3} - a\)

D. Chỉ có A và C đúng.

A. 1

B. 2

C. 3

D. 4

A. 1

B. 2

C. 3

D. 4

A. 4

B. 3

C. 2

D. 1

A. \(\frac{9}{2}\)

B. \(\frac{7}{2}\)

C. \(\frac{5}{2}\)

D. \(\frac{3}{2}\)

A. \(\frac{2}{3}\)

B. \(\frac{4}{3}\)

C. \(\frac{8}{3}\)

D. 2

A. \(\frac{{27}}{2}\)

B. \(\frac{{21}}{2}\)

C. \(\frac{{23}}{2}\)

D. \(\frac{{25}}{2}\)

A. 2

B. 4

C. 6

D. 8

A. \(\frac{{11}}{3}\)

B. \(\frac{{13}}{3}\)

C. \(\frac{{14}}{3}\)

D. \(\frac{{17}}{3}\)

A. \(\frac{{21}}{4}\)

B. \(\frac{{23}}{4}\)

C. \(\frac{{25}}{4}\)

D. \(\frac{{27}}{4}\)