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{"id":26656,"date":"2018-04-13T15:13:23","date_gmt":"2018-04-13T08:13:23","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26656"},"modified":"2018-04-13T15:13:23","modified_gmt":"2018-04-13T08:13:23","slug":"bai-tap-trac-nghiem-giai-tich-12-on-tap-cuoi-nam-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-giai-tich-12-on-tap-cuoi-nam-phan-2\/","title":{"rendered":"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp cu\u1ed1i n\u0103m (Ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 7:<\/b>\u00a0H\u00ea\u0323 s\u00f4\u0301 go\u0301c cu\u0309a ti\u00ea\u0301p tuy\u00ea\u0301n cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301<\/p>\n

\"\"<\/p>\n

ta\u0323i giao \u0111i\u00ea\u0309m cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 v\u01a1\u0301i tru\u0323c tung b\u0103\u0300ng<\/p>\n

A. \u20132 \u00a0\u00a0\u00a0B. 2 \u00a0\u00a0\u00a0C. 1 \u00a0\u00a0\u00a0 D. \u20131.<\/p>\n

C\u00e2u 8:<\/b>\u00a0Cho \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 2x . Go\u0323i x1<\/sub>, x2<\/sub>\u00a0la\u0300 hoa\u0300nh \u0111\u00f4\u0323 ca\u0301c \u0111i\u00ea\u0309m M, N tr\u00ean (C) ma\u0300 ta\u0323i \u0111o\u0301 ti\u00ea\u0301p tuy\u00ea\u0301n cu\u0309a (C) vu\u00f4ng go\u0301c v\u01a1\u0301i \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng y = -x + 2016. Khi \u0111o\u0301 x1<\/sub>\u00a0+ x2<\/sub>\u00a0b\u0103\u0300ng:<\/p>\n

\"\"<\/p>\n

C\u00e2u 9:<\/b>\u00a0Cho ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3x + 2 (C) . Ti\u0300m ph\u01b0\u01a1ng tri\u0300nh ti\u00ea\u0301p tuy\u00ea\u0301n cu\u0309a (C), bi\u00ea\u0301t ti\u00ea\u0301p tuy\u00ea\u0301n \u0111i qua \u0111i\u00ea\u0309m A(-1; 0).<\/p>\n

A. y = 0 \u00a0\u00a0\u00a0B. y = x + 1\u00a0\u00a0\u00a0C. y = x – 1\u00a0\u00a0\u00a0D. y = 2<\/p>\n

C\u00e2u 10:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ mx + 2m c\u0103\u0301t \u0111\u01b0\u01a1\u0300ng th\u0103\u0309ng y = -x + 2 ta\u0323i 3 \u0111i\u00ea\u0309m.<\/p>\n

\"\"<\/p>\n

C\u00e2u 11:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301<\/p>\n

\"\"<\/p>\n

co\u0301 \u0111\u01b0\u01a1\u0300ng ti\u00ea\u0323m c\u00e2\u0323n ngang<\/p>\n

A. m \u2260 0\u00a0\u00a0\u00a0B. m \u2260 \u00b11\u00a0\u00a0\u00a0 C. m \u2260 1 \u00a0\u00a0\u00a0D. Ca\u0309 A va\u0300 C.<\/p>\n

C\u00e2u 12:<\/b>\u00a0Ha\u0300m s\u00f4\u0301 y = (x – 1)ex<\/sup>\u00a0v\u01a1\u0301i x \u2208 [-1; 1] \u0111a\u0323t gia\u0301 tri\u0323 l\u01a1\u0301n nh\u00e2\u0301t ta\u0323i x b\u0103\u0300ng<\/p>\n

A. 1 \u00a0\u00a0\u00a0B. -1\u00a0\u00a0\u00a0C. 0\u00a0\u00a0\u00a0D. 1\/2<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
7-B<\/td>\n8-A<\/td>\n9-A<\/td>\n10-C<\/td>\n11-D<\/td>\n12-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 7:<\/b><\/p>\n

Giao \u0111i\u1ec3m v\u1edbi tr\u1ee5c tung B(0 ;-1). Ta c\u00f3<\/p>\n

\"\"<\/p>\n

C\u00e2u 8:<\/b><\/p>\n

Ta c\u00f3 y’ = 3x2<\/sup>\u00a0– 4x + 2 v\u00e0 y’ = 1 => x = 1, x = 1\/3<\/p>\n

C\u00e2u 12:<\/b><\/p>\n

V\u1ebd \u0111\u1ed3 th\u1ecb y’ = xex<\/sup>. y’ = 0 => x = 0<\/p>\n

y(0) = -1; y(-1) = -2\/e; y(1) = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 7:\u00a0H\u00ea\u0323 s\u00f4\u0301 go\u0301c cu\u0309a ti\u00ea\u0301p tuy\u00ea\u0301n cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 ta\u0323i giao \u0111i\u00ea\u0309m cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 v\u01a1\u0301i tru\u0323c tung b\u0103\u0300ng A. \u20132 \u00a0\u00a0\u00a0B. 2 \u00a0\u00a0\u00a0C. 1 \u00a0\u00a0\u00a0 D. \u20131. C\u00e2u 8:\u00a0Cho \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 y = x3\u00a0– 2×2\u00a0+ 2x . Go\u0323i x1, x2\u00a0la\u0300 hoa\u0300nh \u0111\u00f4\u0323 ca\u0301c \u0111i\u00ea\u0309m M, N tr\u00ean […]<\/p>\n","protected":false},"author":3,"featured_media":26657,"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":"\nB\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp cu\u1ed1i n\u0103m (Ph\u1ea7n 2)<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-giai-tich-12-on-tap-cuoi-nam-phan-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp cu\u1ed1i n\u0103m (Ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 7:\u00a0H\u00ea\u0323 s\u00f4\u0301 go\u0301c cu\u0309a ti\u00ea\u0301p tuy\u00ea\u0301n cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 ta\u0323i giao \u0111i\u00ea\u0309m cu\u0309a \u0111\u00f4\u0300 thi\u0323 ha\u0300m s\u00f4\u0301 v\u01a1\u0301i tru\u0323c tung b\u0103\u0300ng A. \u20132 \u00a0\u00a0\u00a0B. 2 \u00a0\u00a0\u00a0C. 1 \u00a0\u00a0\u00a0 D. \u20131. 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