{"id":392,"date":"2026-05-05T04:43:21","date_gmt":"2026-05-05T04:43:21","guid":{"rendered":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/06-de-on-tap-kiem-tra-giua-hoc-ki-2-mon-toan-12-nam-2024-2025\/"},"modified":"2026-05-05T07:16:04","modified_gmt":"2026-05-05T07:16:04","slug":"06-de-on-tap-kiem-tra-giua-hoc-ki-2-mon-toan-12-nam-2024-2025","status":"publish","type":"post","link":"https:\/\/79.buffdemo.com\/index.php\/2026\/05\/05\/06-de-on-tap-kiem-tra-giua-hoc-ki-2-mon-toan-12-nam-2024-2025\/","title":{"rendered":"06 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025"},"content":{"rendered":"

B\u1ed9 06 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025\u00a0<\/strong>l\u00e0 t\u00e0i li\u1ec7u gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh 12 chinh ph\u1ee5c 9+ trong k\u1ef3 thi gi\u1eefa k\u1ef3 m\u00f4n To\u00e1n 12. \u0110\u00e2y l\u00e0 b\u1ed9 t\u00e0i li\u1ec7u \u0111\u01b0\u1ee3c th\u1ea7y \u0110\u1eb7ng Vi\u1ec7t \u0110\u00f4ng (THPT Nho Quan A, Ninh B\u00ecnh) bi\u00ean so\u1ea1n, b\u00e1m s\u00e1t ch\u01b0\u01a1ng tr\u00ecnh h\u1ecdc. \u0110\u1eb7c bi\u1ec7t, t\u00e0i li\u1ec7u \u0111i k\u00e8m \u0111\u00e1p \u00e1n v\u00e0 l\u1eddi gi\u1ea3i chi ti\u1ebft, gi\u00fap h\u1ecdc sinh 12 d\u1ec5 d\u00e0ng \u00f4n t\u1eadp. C\u00f9ng \u00d4n Thi<\/strong><\/a> xem qua n\u1ed9i dung chi ti\u1ebft c\u1ee7a 06 \u0111\u1ec1 \u00f4n t\u1eadp sau \u0111\u00e2y.<\/p>\n

N\u1ed9i dung 06 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025<\/strong><\/h2>\n

B\u1ed9 t\u00e0i li\u1ec7u g\u1ed3m 06 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025<\/strong> c\u00f3 n\u1ed9i dung b\u00e1m s\u00e1t ki\u1ebfn th\u1ee9c \u0111\u00e3 h\u1ecdc v\u1edbi \u0111\u1ea7y \u0111\u1ee7 c\u00e1c d\u1ea1ng b\u00e0i t\u1eadp t\u1eeb nh\u1eadn bi\u1ebft \u2013 th\u00f4ng hi\u1ec3u \u2013 v\u1eadn d\u1ee5ng \u2013 v\u1eadn d\u1ee5ng cao. \u0110\u00e2y l\u00e0 ch\u00eca kh\u00f3a h\u1eefu \u00edch gi\u00fap h\u1ecdc sinh c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c, r\u00e8n luy\u1ec7n t\u01b0 duy logic v\u00e0 n\u00e2ng cao k\u1ef9 n\u0103ng gi\u1ea3i to\u00e1n m\u1ed9t c\u00e1ch hi\u1ec7u qu\u1ea3.
\n\u0110\u1ec1 \u00f4n t\u1eadp g\u1ed3m 12 c\u00e2u tr\u1eafc nghi\u1ec7m nhi\u1ec1u ph\u01b0\u01a1ng \u00e1n l\u1ef1a ch\u1ecdn + 04 c\u00e2u tr\u1eafc nghi\u1ec7m \u0111\u00fang sai + 06 c\u00e2u tr\u1eafc nghi\u1ec7m tr\u1ea3 l\u1eddi ng\u1eafn. H\u1ecdc sinh c\u00f3 90 ph\u00fat \u0111\u1ec3 l\u00e0m b\u00e0i H\u1ecdc sinh c\u00f3 90 ph\u00fat \u0111\u1ec3 ho\u00e0n th\u00e0nh b\u00e0i thi v\u00e0 c\u00f3 th\u1ec3 tra c\u1ee9u \u0111\u00e1p \u00e1n \u0111\u1ec3 t\u1ef1 \u0111\u00e1nh gi\u00e1 sau khi luy\u1ec7n t\u1eadp.<\/p>\n

Ph\u1ea7n 1 \u2013 C\u00e1c c\u00e2u h\u1ecfi d\u1ea1ng tr\u1eafc nghi\u1ec7m (12 c\u00e2u)<\/span><\/strong><\/p>\n

Th\u00ed sinh c\u1ea7n tr\u1ea3 l\u1eddi t\u1eeb c\u00e2u 1 \u0111\u1ebfn c\u00e2u 12 v\u00e0 ch\u1ecdn 1 ph\u01b0\u01a1ng \u00e1n ch\u00ednh x\u00e1c nh\u1ea5t, c\u0169ng xem qua c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y nh\u00e9:<\/p>\n

C\u00e2u 5.<\/strong> M\u1eb7t ph\u1eb3ng x + 2y – 3z = 0 kh\u00f4ng \u0111i qua \u0111i\u1ec3m n\u00e0o d\u01b0\u1edbi \u0111\u00e2y?<\/p>\n

A. M(1; 1; 1) .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. Q(2; -1; 0) .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 C. P(-1; 2; 1) .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. N(1; 2; 3) .<\/p>\n

C\u00e2u 6.<\/strong> Trong kh\u00f4ng gian v\u1edbi h\u1ec7 t\u1ecda \u0111\u1ed9 Oxyz , m\u1ed9t vecto ph\u00e1p tuy\u1ebfn c\u1ee7a m\u1eb7t ph\u1eb3ng (P) : 2x – y + 3z – 1 = 0 l\u00e0:<\/p>\n

A. n\u20d7 = ( 1\/2, -1, 1\/3 ) .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0B. n\u20d7 = (-4; 2; 6) .<\/p>\n

C. n\u20d7 = (-1; 1; -3) .\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 D. n\u20d7 = (2; -1; 3) .<\/p>\n

Ph\u1ea7n 2 \u2013 C\u00e2u h\u1ecfi \u0111\u00fang sai (4 c\u00e2u)<\/span><\/strong><\/p>\n

\u1ede ph\u1ea7n 2 n\u00e0y c\u00e1c b\u1ea1n h\u1ecdc sinh n\u00ean n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c \u0111\u1ec3 ch\u1ecdn \u0111\u01b0\u1ee3c ph\u01b0\u01a1ng \u00e1n n\u00e0o \u0111\u00fang \/ sai, ph\u1ea7n n\u00e0y r\u1ea5t d\u1ec5 b\u1ecb sai v\u00ec c\u00f3 nhi\u1ec1u \u0111\u00e1p \u00e1n nhi\u1ec5u, c\u00f9ng l\u00e0m quen v\u1edbi c\u00e1c c\u00e2u h\u1ecfi sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1.<\/strong> Cho c\u00e1c h\u00e0m s\u1ed1 f(x) = (2x – 3)\/(x) v\u00e0 g(x) = (3)\/(x\u00b2) x\u00e1c \u0111\u1ecbnh tr\u00ean t\u1eadp D = mathbbR setminus 0 .
\nC\u00e1c m\u1ec7nh \u0111\u1ec1 sau \u0111\u00fang hay sai?<\/p>\n

a) H\u00e0m s\u1ed1 f(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 g(x) tr\u00ean D .<\/p>\n

b) H\u00e0m s\u1ed1 F(x) = 2x – 3ln|x| + C l\u00e0 h\u1ecd c\u00e1c nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 f(x) .<\/p>\n

c) Cho F(1) = 5 , khi \u0111\u00f3 F(x) = 2x – 3ln|x| + 3 .<\/p>\n

d) G(x) l\u00e0 m\u1ed9t nguy\u00ean h\u00e0m c\u1ee7a h\u00e0m s\u1ed1 x f(x) th\u1ecfa m\u00e3n G(1) = 4 . Khi \u0111\u00f3 G(2) = 2 .<\/p>\n

C\u00e2u 2.<\/strong> Cho h\u00e0m s\u1ed1 f(x) li\u00ean t\u1ee5c tr\u00ean mathbbR . X\u00e9t t\u00ednh \u0111\u00fang sai c\u1ee7a c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau:<\/p>\n

a) N\u1ebfu int\u2081\u00b3 f(x)dx = 2 th\u00ec int\u2083\u00b9 f(x)dx = 3\/2 .<\/p>\n

b) N\u1ebfu int\u2082\u2077 f(x)dx = -3 v\u00e0 int\u2081\u2077 f(x)dx = 3 th\u00ec int\u2081\u00b2 f(x)dx = 0 .<\/p>\n

c) N\u1ebfu f(x) = 1\/x , F(1) = 0 th\u00ec F(2) = ln 2 .<\/p>\n

d) N\u1ebfu f(x) = x\u00b3 + 2x\u00b2 th\u00ec int\u208b\u2083\u00b9 |x\u00b3 + 2x\u00b2|dx = -9\/2 .<\/p>\n

Ph\u1ea7n 3 \u2013 C\u00e2u h\u1ecfi tr\u1ea3 l\u1eddi ng\u1eafn (6 c\u00e2u)<\/span><\/strong><\/p>\n

Tr\u00edch m\u1ed9t s\u1ed1 b\u00e0i t\u1eadp \u1edf ph\u1ea7n 3 trong \u0111\u1ec1 \u00f4n t\u1eadp sau \u0111\u00e2y:<\/p>\n

C\u00e2u 1.<\/strong> Cho h\u00e0m s\u1ed1 f(x) th\u1ecfa m\u00e3n f'(x) = 3 – 5cos x v\u00e0 f(0) = 5 . T\u00ednh f(\u03c0\/2) l\u00e0m tr\u00f2n \u0111\u1ebfn h\u00e0ng ph\u1ea7n tr\u0103m?<\/p>\n

C\u00e2u 2.<\/strong> T\u1ed1c \u0111\u1ed9 t\u0103ng tr\u01b0\u1edfng c\u1ee7a m\u1ed9t \u0111\u00e0n g\u1ea5u m\u00e8o t\u1ea1i th\u1eddi \u0111i\u1ec3m t th\u00e1ng k\u1ec3 t\u1eeb khi ng\u01b0\u1eddi ta th\u1ea3 100 c\u00e1 th\u1ec3 \u0111\u1ea7u ti\u00ean v\u00e0o m\u1ed9t khu r\u1eebng \u0111\u01b0\u1ee3c \u01b0\u1edbc l\u01b0\u1ee3ng b\u1edfi c\u00f4ng th\u1ee9c: P'(t) = 8t + 30 (con\/th\u00e1ng) v\u1edbi P(t) l\u00e0 s\u1ed1 l\u01b0\u1ee3ng c\u00e1 th\u1ec3 trong \u0111\u00e0n t\u1ea1i th\u1eddi \u0111i\u1ec3m t th\u00e1ng t\u01b0\u01a1ng \u1ee9ng. D\u1ef1a v\u00e0o t\u1ed1c \u0111\u1ed9 t\u0103ng tr\u01b0\u1edfng \u0111\u00e3 cho, h\u00e3y \u01b0\u1edbc t\u00ednh s\u1ed1 c\u00e1 th\u1ec3 c\u1ee7a \u0111\u00e0n g\u1ea5u m\u00e8o n\u00e0y t\u1ea1i th\u1eddi \u0111i\u1ec3m 3 th\u00e1ng k\u1ec3 t\u1eeb khi ch\u00fang \u0111\u01b0\u1ee3c th\u1ea3 v\u00e0o r\u1eebng.<\/p>\n

T\u1ea3i 06 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025<\/strong><\/h2>\n

T\u1eadp t\u00e0i li\u1ec7u 06 \u0111\u1ec1 \u00f4n t\u1eadp ki\u1ec3m tra gi\u1eefa h\u1ecdc k\u00ec 2 m\u00f4n To\u00e1n 12 n\u0103m 2024 \u2013 2025<\/strong> c\u00f3 k\u00e8m \u0111\u00e1p \u00e1n l\u00e0 tr\u1ee3 th\u1ee7 \u0111\u1eafc l\u1ef1c gi\u00fap c\u00e1c b\u1ea1n h\u1ecdc sinh chu\u1ea9n b\u1ecb cho m\u00ecnh m\u1ed9t l\u1ed9 tr\u00ecnh \u00f4n t\u1eadp, t\u1ef1 tin gi\u1ea3i \u0111\u1ec1 thi gi\u1eefa k\u00ec 2 to\u00e1n 12 <\/span><\/a>hi\u1ec7u qu\u1ea3 v\u00e0 b\u1ee9t ph\u00e1 \u0111i\u1ec3m s\u1ed1. T\u1ea3i \u0111\u1ea7y \u0111\u1ee7 b\u1ed9 t\u00e0i li\u1ec7u qua \u0111\u01b0\u1eddng link sau \u0111\u00e2y \u0111\u1ec3 \u00f4n t\u1eadp ngay nh\u00e9!<\/p>\n