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{"id":26663,"date":"2018-04-13T15:17:20","date_gmt":"2018-04-13T08:17:20","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=26663"},"modified":"2018-04-13T15:17:20","modified_gmt":"2018-04-13T08:17:20","slug":"bai-tap-trac-nghiem-giai-tich-12-on-tap-cuoi-nam-phan-1","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/bai-tap-trac-nghiem-giai-tich-12-on-tap-cuoi-nam-phan-1\/","title":{"rendered":"B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: \u00d4n t\u1eadp cu\u1ed1i n\u0103m (ph\u1ea7n 1)"},"content":{"rendered":"

C\u00e2u 1:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 y = x3<\/sup>\u00a0– 3x2<\/sup>\u00a0+mx – 1 co\u0301 hai \u0111i\u00ea\u0309m c\u01b0\u0323c tri\u0323 ta\u0323i x1<\/sub>, x2<\/sub>\u00a0tho\u0309a ma\u0303n<\/p>\n

\"\"<\/p>\n

C\u00e2u 2:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 ha\u0300m s\u00f4\u0301 y = (1\/3)x3<\/sup>\u00a0– x2<\/sup>\u00a0– mx + 1 lu\u00f4n \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean t\u01b0\u0300ng khoa\u0309ng xa\u0301c \u0111i\u0323nh cu\u0309a no\u0301<\/p>\n

A. m < – 1\u00a0\u00a0\u00a0B. m > -1 \u00a0\u00a0\u00a0C. m \u2264 -1 \u00a0\u00a0\u00a0D. m > -1<\/p>\n

C\u00e2u 3:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 ph\u01b0\u01a1ng tri\u0300nh |x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 9x + 2| = m co\u0301 6 nghi\u00ea\u0323m ph\u00e2n bi\u00ea\u0323t<\/p>\n

A. 0 < m < 3\u00a0\u00a0\u00a0B. m = 3 \u00a0\u00a0\u00a0 C. 3 < m < 29\u00a0\u00a0\u00a0D. m > -3<\/p>\n

C\u00e2u 4:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 ha\u0300m s\u00f4\u0301 y = -x3<\/sup>\u00a0+ (2m + 1)x2<\/sup>\u00a0– (m2<\/sup>\u00a0– 3m +2)x – 4 co\u0301 c\u01b0\u0323c \u0111a\u0323i, c\u01b0\u0323c ti\u00ea\u0309u n\u0103\u0300m v\u00ea\u0300 hai phi\u0301a so v\u01a1\u0301i tru\u0323c tung<\/p>\n

A. m \u2208 (1; 2) \u00a0\u00a0\u00a0B. m \u2208 [1; 2]<\/p>\n

C. m \u2208 (- \u221e; 1) \u222a (2; +\u221e)\u00a0\u00a0\u00a0D. m \u2208 (- \u221e; 1] \u222a [2; +\u221e)<\/p>\n

C\u00e2u 5:<\/b>\u00a0Ti\u0300m m \u0111\u00ea\u0309 ha\u0300m s\u00f4\u0301 y = x3<\/sup>\u00a0– 3mx2<\/sup>\u00a0+ 12x – 2 nghi\u0323ch bi\u00ea\u0301n tr\u00ean khoa\u0309ng (1; 4)<\/p>\n

A. m \u2265 5\/2\u00a0\u00a0\u00a0B. m \u2264 5\/2\u00a0\u00a0\u00a0C. m \u2264 2 \u00a0\u00a0\u00a0D. m > 2<\/p>\n

C\u00e2u 6:<\/b>\u00a0\u0110\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 co\u0301 ph\u01b0\u01a1ng tri\u0300nh la\u0300<\/p>\n

A. y = 1\u00a0\u00a0\u00a0B. y = 0\u00a0\u00a0\u00a0C. y = 1\/2 \u00a0\u00a0\u00a0D. y = \u00b11\/2<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
1-A<\/td>\n2-C<\/td>\n3-A<\/td>\n4-A<\/td>\n5-A<\/td>\n6-D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 1:<\/b><\/p>\n

y’ = 3x2<\/sup>\u00a0– 6x + m.<\/p>\n

H\u00e0m s\u1ed1 c\u00f3 c\u1ef1c tr\u1ecb khi y’ = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t :<\/p>\n

9 – 3m > <=> m < 3<\/p>\n

Khi \u0111\u00f3: x1<\/sub>\u00a0+ x2<\/sub>\u00a0= 2; x1<\/sub>x2<\/sub>\u00a0= m\/3<\/p>\n

\"\"<\/p>\n

C\u00e2u 2:<\/b><\/p>\n

y’ = x2<\/sup>\u00a0– 2x – m \u2265 0, \u2200x <=> 1 + m \u2264 0 <=> m \u2265 -1<\/p>\n

C\u00e2u 3:<\/b><\/p>\n

V\u1ebd \u0111\u1ed3 th\u1ecb y = |x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0– 9x + 2|. D\u1ef1a v\u00e0o \u0111\u1ed3 th\u1ecb ta c\u00f3 \u0111\u00e1p \u00e1n A.<\/p>\n

C\u00e2u 4:<\/b><\/p>\n

y = -3x2<\/sup>\u00a0+ 2(2m + 1)x – m2<\/sup>\u00a0+ 3m – 2 = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t.<\/p>\n

\"\"<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 1:\u00a0Ti\u0300m m \u0111\u00ea\u0309 y = x3\u00a0– 3×2\u00a0+mx – 1 co\u0301 hai \u0111i\u00ea\u0309m c\u01b0\u0323c tri\u0323 ta\u0323i x1, x2\u00a0tho\u0309a ma\u0303n C\u00e2u 2:\u00a0Ti\u0300m m \u0111\u00ea\u0309 ha\u0300m s\u00f4\u0301 y = (1\/3)x3\u00a0– x2\u00a0– mx + 1 lu\u00f4n \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean t\u01b0\u0300ng khoa\u0309ng xa\u0301c \u0111i\u0323nh cu\u0309a no\u0301 A. m < – 1\u00a0\u00a0\u00a0B. m > -1 \u00a0\u00a0\u00a0C. m \u2264 -1 \u00a0\u00a0\u00a0D. […]<\/p>\n","protected":false},"author":3,"featured_media":26664,"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 1)<\/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-1\/\" \/>\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 1)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 1:\u00a0Ti\u0300m m \u0111\u00ea\u0309 y = x3\u00a0– 3x2\u00a0+mx – 1 co\u0301 hai \u0111i\u00ea\u0309m c\u01b0\u0323c tri\u0323 ta\u0323i x1, x2\u00a0tho\u0309a ma\u0303n C\u00e2u 2:\u00a0Ti\u0300m m \u0111\u00ea\u0309 ha\u0300m s\u00f4\u0301 y = (1\/3)x3\u00a0– x2\u00a0– mx + 1 lu\u00f4n \u0111\u00f4\u0300ng bi\u00ea\u0301n tr\u00ean t\u01b0\u0300ng khoa\u0309ng xa\u0301c \u0111i\u0323nh cu\u0309a no\u0301 A. m < – 1\u00a0\u00a0\u00a0B. m > -1 \u00a0\u00a0\u00a0C. m \u2264 -1 \u00a0\u00a0\u00a0D. 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