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{"id":28108,"date":"2018-04-19T16:48:18","date_gmt":"2018-04-19T09:48:18","guid":{"rendered":"https:\/\/lop12.edu.vn\/?p=28108"},"modified":"2018-04-19T16:48:18","modified_gmt":"2018-04-19T09:48:18","slug":"chuong-1-bai-tap-trac-nghiem-giai-tich-12-su-dong-bien-nghich-bien-cua-ham-so-phan-2","status":"publish","type":"post","link":"https:\/\/lop12.edu.vn\/chuong-1-bai-tap-trac-nghiem-giai-tich-12-su-dong-bien-nghich-bien-cua-ham-so-phan-2\/","title":{"rendered":"Ch\u01b0\u01a1ng 1 – B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: S\u1ef1 \u0111\u1ed3ng bi\u1ebfn, ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 (Ph\u1ea7n 2)"},"content":{"rendered":"

C\u00e2u 6:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 3 . K\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y \u0111\u00fang?<\/p>\n

A. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) \u2229 (0; 1)<\/p>\n

B. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-1; 0) \u222a (1; +\u221e)<\/p>\n

C. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) \u222a (0; 1)<\/p>\n

D. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-1; 0) v\u00e0 (1; +\u221e)<\/p>\n

C\u00e2u 7:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = sin2x – 2x. H\u00e0m s\u1ed1 n\u00e0y<\/p>\n

A. Lu\u00f4n \u0111\u1ed3ng bi\u1ebfn tr\u00ean R \u00a0\u00a0\u00a0 B. Ch\u1ec9 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (0; +\u221e)<\/p>\n

C. Ch\u1ec9 ngh\u1ecbch bi\u1ebfn tr\u00ean (-\u221e; -1) \u00a0\u00a0\u00a0D. Lu\u00f4n ngh\u1ecbch bi\u1ebfn tr\u00ean R<\/p>\n

C\u00e2u 8:<\/b>\u00a0Trong c\u00e1c h\u00e0m s\u1ed1 sau, h\u00e0m s\u1ed1 n\u00e0o ch\u1ec9 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; 1) ?<\/p>\n

\"\"<\/p>\n

C\u00e2u 9:<\/b>\u00a0T\u00ecm m \u0111\u1ec3 h\u00e0m s\u1ed1<\/p>\n

\"\"<\/p>\n

lu\u00f4n ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng x\u00e1c \u0111\u1ecbnh.<\/p>\n

A.-2 < m \u2264 2 \u00a0\u00a0\u00a0 B. m < -2 ho\u1eb7c m > 2<\/p>\n

C. -2 < m < 2 \u00a0\u00a0\u00a0 D. m \u2260 \u00b12<\/p>\n

C\u00e2u 10:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = -x3<\/sup>\u00a0+ 3x2<\/sup>\u00a0+ 3mx – 1, t\u00ecm t\u1ea5t c\u1ea3 c\u00e1c gi\u00e1 tr\u1ecb c\u1ee7a tham s\u1ed1 m \u0111\u1ec3 h\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (0; +\u221e)<\/p>\n

A. m < 1\u00a0\u00a0\u00a0 B. m \u2265 1\u00a0\u00a0\u00a0 C. m \u2264 -1\u00a0\u00a0\u00a0 D. m \u2265 -1<\/p>\n

H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n\n\n\n
6-D<\/td>\n7-D<\/td>\n8-C<\/td>\n9-C<\/td>\n10-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

C\u00e2u 6:<\/b><\/p>\n

\"\"<\/p>\n

B\u1ea3ng x\u00e9t d\u1ea5u y\u2019:<\/p>\n

\"\"<\/p>\n

T\u1eeb \u0111\u00f3 ta c\u00f3: H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-1; 0) v\u00e0 (1; +\u221e) , ngh\u1ecbch bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; -1) v\u00e0 (0; 1) . Ch\u1ecdn \u0111\u00e1p \u00e1n D.<\/p>\n

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

T\u1eadp x\u00e1c \u0111\u1ecbnh D = R V\u1eady h\u00e0m s\u1ed1 lu\u00f4n ngh\u1ecbch bi\u1ebfn tr\u00ean R<\/p>\n

Ch\u1ecdn \u0111\u00e1p \u00e1n D.<\/p>\n

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

H\u00e0m s\u1ed1 y = x3<\/sup>\/3 – 2x2<\/sup>\u00a0+ 3x – 1 c\u00f3 y’= x2<\/sup>\u00a0– 4x + 3 . H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; 1) v\u00e0 (3; +\u221e) .<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean c\u00e1c kho\u1ea3ng (-\u221e; 1) v\u00e0 (1; +\u221e) .<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e;1) v\u00e0 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (1; +\u221e) .<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (1; +\u221e) .<\/p>\n

Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n

C\u00e2u 9:<\/b><\/p>\n

T\u1eadp x\u00e1c \u0111\u1ecbnh<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean t\u1eebng kho\u1ea3ng<\/p>\n

\"\"<\/p>\n

khi v\u00e0 ch\u1ec9 khi<\/p>\n

\"\"<\/p>\n

Suy ra -2 < m < 2. Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n

C\u00e2u 10:<\/b><\/p>\n

Ta c\u00f3 y’ = -3x2<\/sup>\u00a0+ 6x + 3m. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (0; +\u221e) n\u1ebfu y’ \u2264 0 tr\u00ean kho\u1ea3ng (o; +\u221e)<\/p>\n

C\u00e1ch 1: D\u00f9ng \u0111\u1ecbnh l\u00ed d\u1ea5u tam th\u1ee9c b\u1eadc hai.<\/p>\n

X\u00e9t ph\u01b0\u01a1ng tr\u00ecnh -3x2<\/sup>\u00a0+ 6x + 3m. Ta c\u00f3 \u0394’ = 9(1 + m)<\/p>\n

TH1: \u0394’ \u2264 0 => m \u2264 -1. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean R .<\/p>\n

TH2: \u0394’ > 0 => m > -1; y’ = 0 c\u00f3 hai nghi\u1ec7m ph\u00e2n bi\u1ec7t l\u00e0 x = 1 \u00b1\u221a(1+m) .<\/p>\n

\"\"<\/p>\n

H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean (0; +\u221e) <=> 1 + \u221a(1+m) \u2264 0, v\u00f4 l\u00ed.<\/p>\n

T\u1eeb TH1 v\u00e0 TH2, ta c\u00f3 m \u2264 -1<\/p>\n

C\u00e1ch 2: D\u00f9ng ph\u01b0\u01a1ng ph\u00e1p bi\u1ebfn thi\u00ean h\u00e0m s\u1ed1.<\/p>\n

Ta c\u00f3 y’ = -3x2<\/sup>\u00a0+ 6x + 3m \u2264 0, \u2200x > 0 <=> 3m \u2264 3x2<\/sup>\u00a0– 6x, \u2200x > 0<\/p>\n

T\u1eeb \u0111\u00f3 suy ra 3m \u2264 min(3x2<\/sup>\u00a0– 6x) v\u1edbi x > 0 Do \u0111\u00f3 m \u2264 -1.<\/p>\n

Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n","protected":false},"excerpt":{"rendered":"

C\u00e2u 6:\u00a0Cho h\u00e0m s\u1ed1 y = x4\u00a0– 2×2\u00a0+ 3 . K\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y \u0111\u00fang? A. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) \u2229 (0; 1) B. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-1; 0) \u222a (1; +\u221e) C. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) \u222a (0; 1) D. H\u00e0m s\u1ed1 […]<\/p>\n","protected":false},"author":3,"featured_media":28109,"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":"\nCh\u01b0\u01a1ng 1 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: S\u1ef1 \u0111\u1ed3ng bi\u1ebfn, ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 (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\/chuong-1-bai-tap-trac-nghiem-giai-tich-12-su-dong-bien-nghich-bien-cua-ham-so-phan-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Ch\u01b0\u01a1ng 1 - B\u00e0i t\u1eadp tr\u1eafc nghi\u1ec7m Gi\u1ea3i t\u00edch 12: S\u1ef1 \u0111\u1ed3ng bi\u1ebfn, ngh\u1ecbch bi\u1ebfn c\u1ee7a h\u00e0m s\u1ed1 (Ph\u1ea7n 2)\" \/>\n<meta property=\"og:description\" content=\"C\u00e2u 6:\u00a0Cho h\u00e0m s\u1ed1 y = x4\u00a0– 2x2\u00a0+ 3 . 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K\u1ebft lu\u1eadn n\u00e0o sau \u0111\u00e2y \u0111\u00fang? A. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) \u2229 (0; 1) B. H\u00e0m s\u1ed1 \u0111\u1ed3ng bi\u1ebfn tr\u00ean kho\u1ea3ng (-1; 0) \u222a (1; +\u221e) C. H\u00e0m s\u1ed1 ngh\u1ecbch bi\u1ebfn tr\u00ean kho\u1ea3ng (-\u221e; -1) \u222a (0; 1) D. 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