C\u00e2u 15:<\/b>\u00a0T\u00ecm c\u00e1c \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n

<\/p>\n

A. y = 0 \u00a0\u00a0\u00a0C. y = 0 v\u00e0 y = 3\/2<\/p>\n

B. y = 3 \u00a0\u00a0\u00a0D. y = 0 v\u00e0 y = 3<\/p>\n

C\u00e2u 16:<\/b>\u00a0M\u1ed9t qu\u1ea7n th\u1ec3 vi khu\u1ea9n l\u00fac \u0111\u1ea7u c\u00f3 200 c\u00e1 th\u1ec3 v\u00e0 c\u1ee9 sau m\u1ed9t ng\u00e0y th\u00ec s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 t\u0103ng l\u00ean g\u1ea5p ba l\u1ea7n. T\u00ecm c\u00f4ng th\u1ee9c bi\u1ec3u th\u1ecb s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 (k\u00ed hi\u1ec7u N) c\u1ee7a qu\u1ea7n th\u1ec3 n\u00e0y sau t ng\u00e0y k\u1ec3 t\u1eeb l\u00fac ban \u0111\u1ea7u.<\/p>\n

A. N(t) = 200.t^{3<\/sup>\u00a0\u00a0\u00a0\u00a0C. N(t) = 200.e3t<\/sup><\/p>\n}

B. N(t) = 200.3^{t<\/sup>\u00a0\u00a0\u00a0\u00a0D. N(t) = 200.et\/3<\/sup><\/p>\n}

C\u00e2u 17:<\/b>\u00a0S\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 c\u1ee7a m\u1ed9t lo\u00e0i sinh v\u1eadt b\u1ecb suy gi\u1ea3m trong 10 n\u0103m theo c\u00e1ch : s\u1ed1 l\u01b0\u1ee3ng n\u0103m sau b\u1eb1ng 95% s\u1ed1 l\u01b0\u1ee3ng n\u0103m tr\u01b0\u1edbc \u0111\u00f3. T\u1ea1i th\u1eddi \u0111i\u1ec3m ch\u1ecdn l\u00e0m m\u1ed1c th\u1eddi gian lo\u00e0i n\u00e0y c\u00f3 5000 c\u00e1 th\u1ec3. C\u00f4ng th\u1ee9c n\u00e0o sau \u0111\u00e2y di\u1ec5n t\u1ea3 s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 (k\u00ed hi\u1ec7u N) c\u1ee7a lo\u00e0i theo th\u1eddi gian t (t\u00ednh b\u1eb1ng n\u0103m, 0 \u2264 t \u2264 10 ) ?<\/p>\n

A. N = 5000.(1 + 0,95)^{t<\/sup>\u00a0\u00a0\u00a0\u00a0C. N = 5000.e-0,95t<\/sup><\/p>\n}

B. N = 5000.(0,95)^{t<\/sup>\u00a0\u00a0\u00a0\u00a0D. N = 5000.e-0,05t<\/sup><\/p>\n}

C\u00e2u 18:<\/b>\u00a0M\u1ed9t ng\u01b0\u1eddi g\u1eedi ti\u1ebft ki\u1ec7m v\u00e0o ng\u00e2n h\u00e0ng s\u1ed1 ti\u1ec1n 50 tri\u1ec7u \u0111\u1ed3ng v\u1edbi h\u00ecnh th\u1ee9c l\u00e3i k\u00e9p v\u00e0 l\u00e3i su\u1ea5t 6,8% m\u1ed9t n\u0103m. H\u1ecfi sau 3 n\u0103m trong t\u00e0i kho\u1ea3n ti\u1ebft ki\u1ec7m c\u1ee7a ng\u01b0\u1eddi \u0111\u00f3 c\u00f3 bao nhi\u00eau ti\u1ec1n (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ngh\u00ecn) ?<\/p>\n

A. 60200000 \u0111\u1ed3ng \u00a0\u00a0\u00a0C. 61280000 \u0111\u1ed3ng<\/p>\n

B. 60909000 \u0111\u1ed3ng \u00a0\u00a0\u00a0D. 61315000 \u0111\u1ed3ng<\/p>\n

C\u00e2u 19:<\/b>\u00a0Cho hai s\u1ed1 th\u1ef1c a v\u00e0 b, v\u1edbi 0 < a < 1 < b. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 \u0111\u00fang?<\/p>\n

A. log_{b<\/sub>a + loga<\/sub>b < 0 \u00a0\u00a0\u00a0C. logb<\/sub>a + loga<\/sub>b = 0<\/p>\n}

B. 0 < log_{b<\/sub>a + loga<\/sub>b < 2 \u00a0\u00a0\u00a0D. logb<\/sub>a + loga<\/sub>b \u2265 2<\/p>\n}

C\u00e2u 20:<\/b>\u00a0T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t, gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 y = x^{2<\/sup>\u00a0– 2x + ln(2x + 1) tr\u00ean [0; 1]<\/p>\n}

<\/p>\n

C\u00e2u 21:<\/b>\u00a0D\u00e2n s\u1ed1 Vi\u1ec7t Nam n\u0103m 2015 l\u00e0 91,71 tri\u1ec7u ng\u01b0\u1eddi v\u00e0 t\u1ec9 l\u1ec7 t\u0103ng d\u00e2n s\u1ed1 l\u00e0 1,08%. H\u1ecfi n\u1ebfu v\u1eabn gi\u1eef nguy\u00ean t\u1ec9 l\u1ec7 t\u0103ng d\u00e2n s\u1ed1 h\u00e0ng n\u0103m n\u00e0y th\u00ec n\u0103m 2020 d\u00e2n s\u1ed1 Vi\u1ec7t Nam s\u1ebd l\u00e0 bao nhi\u00eau (l\u00e0m tr\u00f2n k\u1ebft qu\u1ea3 \u0111\u1ebfn h\u00e0ng ch\u1ee5c ngh\u00ecn) ?<\/p>\n

A. 96,66 tri\u1ec7u ng\u01b0\u1eddi\u00a0\u00a0\u00a0C. 96,80 tri\u1ec7u ng\u01b0\u1eddi<\/p>\n

B. 96,77 tri\u1ec7u ng\u01b0\u1eddi\u00a0\u00a0\u00a0D. 97,85 tri\u1ec7u ng\u01b0\u1eddi<\/p>\n

C\u00e2u 22:<\/b>\u00a0Gi\u1ea3 s\u1eed s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 trong m\u1ed9t m\u1ebb c\u1ea5y vi khu\u1ea9n thay \u0111\u1ed5i theo th\u1eddi gian t theo c\u00f4ng th\u1ee9c<\/p>\n

<\/p>\n

T\u00ecm s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 vi khu\u1ea9n l\u1edbn nh\u1ea5t (k\u00ed hi\u1ec7u M) v\u00e0 nh\u1ecf nh\u1ea5t (k\u00ed hi\u1ec7u m) c\u1ee7a m\u1ebb c\u1ea5y n\u00e0y trong kho\u1ea3ng th\u1eddi gian 0 \u2264 t \u2264 100<\/p>\n

A. M = 161788, m = 128369 \u00a0\u00a0\u00a0C. M = 225000, m = 125000<\/p>\n

B. M = 161788, m = 125000\u00a0\u00a0\u00a0D. M = 225000, m = 128369<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n

15-C<\/td>\n

16-C<\/td>\n

17-B<\/td>\n

18-B<\/td>\n

19-A<\/td>\n

20-C<\/td>\n

21-C<\/td>\n

22-B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n C\u00e2u 15:<\/b><\/p>\n <\/p>\n T\u1eeb \u0111\u00f3 suy ra h\u00e0m s\u1ed1 c\u00f3 hai ti\u1ec7m c\u1eadn ngang l\u00e0 y = 3\/2 v\u00e0 y = 0<\/p>\n C\u00e2u 16:<\/b><\/p>\n Theo gi\u1ea3 thi\u1ebft, s\u1ed1 l\u01b0\u1ee3ng vi khu\u1ea9n sau 1, 2, 3,\u2026 ng\u00e0y l\u00e0 200.3 ; 200 .3.3 ; 200.3.3.3 ;\u2026 T\u1eeb \u0111\u00f3 ta th\u1ea5y c\u00f4ng th\u1ee9c \u0111\u00fang l\u00e0 N(t) = 200.3t<\/sup><\/p>\n C\u00e2u 18:<\/b><\/p>\n S\u1ed1 ti\u1ec1n trong t\u00e0i kh\u1ecfan ng\u01b0\u1eddi \u0111\u00f3 sau n n\u0103m n\u1ebfu ng\u01b0\u1eddi \u0111\u00f3 kh\u00f4ng r\u00fat ti\u1ec1n v\u00e0 l\u00e3i su\u1ea5t kh\u00f4ng thay \u0111\u00f4\u1ec9 \u0111\u01b0\u1ee3c t\u00ednh theo c\u00f4ng th\u1ee9c : P(t) = 50000000(1 + 0,068)t<\/sup>\u00a0(\u0111\u1ed3ng)<\/p>\n S\u1ed1 ti\u1ec1n c\u1ea7n t\u00ednh : P(3) = 50000000(1 + 0,068)3<\/sup>\u00a0\u2248 60909000(\u0111\u1ed3ng)<\/p>\n C\u00e2u 19:<\/b><\/p>\n Do 0 < a < 1 n\u00ean h\u00e0m s\u1ed1 y = loga<\/sub>x ngh\u1ecbch bi\u1ebfn, c\u00f2n h\u00e0m s\u1ed1 y = logb<\/sub>x \u0111\u1ed3ng bi\u1ebfn tr\u00ean (0; +\u221e). Ta c\u00f3 loga<\/sub>b < loga<\/sub>1 = 0 v\u00e0 logb<\/sub>a < logb<\/sub>1 = 0.<\/p>\n Do \u0111\u00f3 loga<\/sub>b + logb<\/sub>a < 0<\/p>\n C\u00e2u 20:<\/b><\/p>\n <\/p>\n C\u00e2u 21:<\/b><\/p>\n D\u00e2n s\u1ed1 l\u00fac \u0111\u00f3: 91,71.e5.0,0108<\/sup>\u00a0\u2248 96,80 tri\u1ec7u ng\u01b0\u1eddi<\/p>\n C\u00e2u 22:<\/b><\/p>\n N'(t) = 250(20 – t)e-t\/20<\/sup>; N'(t) = 0 <=> t = 20<\/p>\n Ta c\u00f3: N(0) = 125000, N(20) \u2248 161788, N(100) \u2248 128369<\/p>\n T\u1eeb \u0111\u00f3 M = 161788 v\u00e0 m = 125000<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 15:\u00a0T\u00ecm c\u00e1c \u0111\u01b0\u1eddng ti\u1ec7m c\u1eadn ngang c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 A. y = 0 \u00a0\u00a0\u00a0C. y = 0 v\u00e0 y = 3\/2 B. y = 3 \u00a0\u00a0\u00a0D. y = 0 v\u00e0 y = 3 C\u00e2u 16:\u00a0M\u1ed9t qu\u1ea7n th\u1ec3 vi khu\u1ea9n l\u00fac \u0111\u1ea7u c\u00f3 200 c\u00e1 th\u1ec3 v\u00e0 c\u1ee9 sau m\u1ed9t ng\u00e0y th\u00ec s\u1ed1 […]<\/p>\n","protected":false},"author":3,"featured_media":27419,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_mi_skip_tracking":false,"tdm_status":"","tdm_grid_status":""},"categories":[157],"tags":[1447,1446],"yoast_head":"\n