C\u00e2u 6:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = x3<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 3. \u0110i\u1ec3m M(0; 3) l\u00e0:<\/p>\n A. C\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 \u00a0\u00a0\u00a0 C. \u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n B. \u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 \u00a0\u00a0\u00a0 D. \u0110i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1<\/p>\n C\u00e2u 7:<\/b>\u00a0T\u00ecm \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 y = sin2<\/sup>x + \u221a3cosx + 1 v\u1edbi x \u2208 (0; \u03c0)<\/p>\n A. x = 0 \u00a0\u00a0\u00a0 B. x = \u03c0 \u00a0\u00a0\u00a0 C. \u03c0\/6 \u00a0\u00a0\u00a0D. \u03c0\/3<\/p>\n C\u00e2u 8:<\/b>\u00a0C\u00f3 bao nhi\u00eau m\u1ec7nh \u0111\u1ec1 \u0111\u00fang trong c\u00e1c ph\u00e1t bi\u1ec3u sau?<\/p>\n 1. H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 \u0111\u1ea1o h\u00e0m t\u1ea1i x = 0.<\/p>\n 2. H\u00e0m s\u1ed1 kh\u00f4ng lien t\u1ee5c t\u1ea1i x = 0.<\/p>\n 3. H\u00e0m s\u1ed1 kh\u00f4ng c\u00f3 c\u1ef1c tr\u1ecb t\u1ea1i x = 0.<\/p>\n 4. H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c tr\u1ecb t\u1ea1i x = 0.<\/p>\n A. 0 \u00a0\u00a0\u00a0 B. 1 \u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D. 3.<\/p>\n C\u00e2u 9:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = -3x4<\/sup>\u00a0– 2x3<\/sup>\u00a0+ 3<\/p>\n H\u00e0m s\u1ed1 c\u00f3<\/p>\n A. M\u1ed9t c\u1ef1c \u0111\u1ea1i v\u00e0 hai c\u1ef1c ti\u1ec3u<\/p>\n B. M\u1ed9t c\u1ef1c ti\u1ec3u v\u00e0 hai c\u1ef1c \u0111\u1ea1i<\/p>\n C. M\u1ed9t c\u1ef1c \u0111\u1ea1i v\u00e0 kh\u00f4ng c\u00f3 c\u1ef1c ti\u1ec3u<\/p>\n D. M\u1ed9t c\u1ef1c ti\u1ec3u v\u00e0 m\u1ed9t c\u1ef1c \u0111\u1ea1i.<\/p>\n C\u00e2u 10:<\/b>\u00a0\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 n\u00e0o sau \u0111\u00e2y c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb (a l\u00e0 tham s\u1ed1 nh\u1eadn gi\u00e1 tr\u1ecb \u00e2m)?<\/p>\n A. y = x4<\/sup>\u00a0– 2ax2<\/sup>\u00a0+ 3 \u00a0\u00a0\u00a0 B. y = ax4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 3<\/p>\n C. y = a2<\/sup>x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 3 \u00a0\u00a0\u00a0 D. y = x4<\/sup>\u00a0+ 2x2<\/sup>\u00a0+ 3a<\/p>\n C\u00e2u 11:<\/b>\u00a0T\u00ecm m \u0111\u1ec3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = x4<\/sup>\u00a0– 2(m + 1)x2<\/sup>\u00a0+ m\u00a02<\/sup>\u00a0c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb t\u1ea1o th\u00e0nh ba \u0111\u1ec9nh c\u1ee7a m\u1ed9t tam gi\u00e1c vu\u00f4ng<\/p>\n A. m = 0 \u00a0\u00a0\u00a0 B. m = -1<\/p>\n C. m = 0 ho\u1eb7c m = -1 \u00a0\u00a0\u00a0D. Kh\u00f4ng c\u00f3 \u0111\u00e1p \u00e1n<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 6:<\/b><\/p>\n Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n Ch\u00fa \u00fd. Ph\u00e2n bi\u1ec7t c\u00e1c kh\u00e1i ni\u1ec7m: c\u1ef1c tr\u1ecb, \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a h\u00e0m s\u1ed1, \u0111i\u1ec3m c\u1ef1c tr\u1ecb c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1.<\/p>\n C\u00e2u 7:<\/b><\/p>\n Ta c\u00f3: y’ = 2sinxcosx – \u221a3sinx, y” = 2cos2x – \u221a3cosx.<\/p>\n V\u1edbi x \u2208 (0; \u03c0) th\u00ec y’ = 2sinxcosx – \u221a3sinx = 0<\/p>\n <\/p>\n N\u00ean h\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = \u03c0\/6 Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n C\u00e2u 8:<\/b><\/p>\n <\/p>\n \u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y = |x| c\u00f3 d\u1ea1ng h\u00ecnh v\u1ebd.<\/p>\n T\u1eeb \u0111\u1ed3 th\u1ecb trong h\u00ecnh ta c\u00f3 h\u00e0m s\u1ed1 y = |x| li\u00ean t\u1ee5c t\u1ea1i x = 0 nh\u01b0ng kh\u00f4ng c\u00f3 \u0111\u1ea1o h\u00e0m t\u1ea1i \u0111i\u1ec3m \u0111\u00f3. S\u1eed d\u1ee5ng \u0111\u1ecbnh ngh\u0129a c\u1ef1c tr\u1ecb ta c\u00f3 h\u00e0m s\u1ed1 y = |x| \u0111\u1ea1t c\u1ef1c ti\u1ec3u t\u1ea1i x = 0<\/p>\n Do \u0111\u00f3 m\u1ec7nh \u0111\u1ec1 1 v\u00e0 4 \u0111\u00fang. Ch\u1ecdn \u0111\u00e1p \u00e1n C<\/p>\n C\u00e2u 9:<\/b><\/p>\n <\/p>\n Ta c\u00f3 y’ = -12x3<\/sup>\u00a0– 4x<\/p>\n X\u00e9t y’=0 => x = 0<\/p>\n H\u00e0m s\u1ed1 ch\u1ec9 c\u00f3 m\u1ed9t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 0. Ch\u1ecdn \u0111\u00e1p \u00e1n C.<\/p>\n C\u00e2u 10:<\/b><\/p>\n \u0110\u1ec3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb th\u00ec y’ = 0 c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t. Do a l\u00e0 tham s\u1ed1 nh\u1eadn gi\u00e1 tr\u1ecb \u00e2m n\u00ean ch\u1ec9 c\u00f3 y = a2<\/sup>x4<\/sup>\u00a0– 2x2<\/sup>\u00a0+ 3 th\u1ecfa m\u00e3n. Ch\u1ecdn \u0111\u00e1p \u00e1n C<\/p>\n C\u00e2u 11:<\/b><\/p>\n Ta c\u00f3: y’ = 4x3<\/sup>\u00a0– 4(m + 1)x = 4x(x2<\/sup>\u00a0– m – 1)<\/p>\n \u0110\u1ec3 h\u00e0m s\u1ed1 c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb th\u00ec y’ = 0 c\u00f3 ba nghi\u1ec7m ph\u00e2n bi\u1ec7t <=> m > – 1<\/p>\n Khi \u0111\u00f3 \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 c\u00f3 ba \u0111i\u1ec3m c\u1ef1c tr\u1ecb l\u00e0<\/p>\n <\/p>\n \u0110\u1ed1i chi\u1ebfu \u0111i\u1ec1u ki\u1ec7n t\u1ed3n t\u1ea1i c\u1ef1c tr\u1ecb ta c\u00f3 m = 0 l\u00e0 gi\u00e1 tr\u1ecb c\u1ea7n t\u00ecm. Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 6:\u00a0Cho h\u00e0m s\u1ed1 y = x3\u00a0– 2×2\u00a0+ 3. \u0110i\u1ec3m M(0; 3) l\u00e0: A. C\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 \u00a0\u00a0\u00a0 C. \u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 B. \u0110i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 \u00a0\u00a0\u00a0 D. \u0110i\u1ec3m c\u1ef1c ti\u1ec3u c\u1ee7a \u0111\u1ed3 th\u1ecb h\u00e0m s\u1ed1 C\u00e2u 7:\u00a0T\u00ecm \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i c\u1ee7a h\u00e0m s\u1ed1 y […]<\/p>\n","protected":false},"author":3,"featured_media":28065,"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-C<\/td>\n 7-C<\/td>\n 8-C<\/td>\n 9-C<\/td>\n 10-C<\/td>\n 11-A<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n