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 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":"\n\n\n
\n 6-D<\/td>\n 7-D<\/td>\n 8-C<\/td>\n 9-C<\/td>\n 10-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n