C\u00e2u 7:<\/b>\u00a0GTLN c\u1ee7a h\u00e0m s\u1ed1 y = 2sinx + cos2x tr\u00ean \u0111o\u1ea1n [0; \u03c0] l\u00e0<\/p>\n
A. 1 \u00a0\u00a0\u00a0B. 3\/2 \u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D. 7\/4<\/p>\n
C\u00e2u 8:<\/b>\u00a0Cho h\u00e0m s\u1ed1 y = f(x) x\u00e1c \u0111\u1ecbnh v\u00e0 li\u00ean t\u1ee5c tr\u00ean R v\u00e0 c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang<\/p>\n
<\/p>\n
A. H\u00e0m s\u1ed1 c\u00f3 \u0111\u00fang m\u1ed9t c\u1ef1c tr\u1ecb.<\/p>\n
B. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c ti\u1ec3u b\u1eb1ng 1.<\/p>\n
C. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t v\u00e0 gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t b\u1eb1ng -1.<\/p>\n
D. H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x = 0 v\u00e0 c\u1ef1c ti\u1ec3u t\u1ea1i x<\/p>\n
C\u00e2u 9:<\/b>\u00a0X\u00e9t h\u00e0m s\u1ed1<\/p>\n
<\/p>\n
Trong c\u00e1c kh\u1eb3ng \u0111\u1ecbnh sau, kh\u1eb3ng \u0111\u1ecbnh n\u00e0o l\u00e0 \u0111\u00fang?<\/p>\n
A. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t b\u1eb1ng 4.<\/p>\n
B. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i b\u1eb1ng 4<\/p>\n
C. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t b\u1eb1ng 0.<\/p>\n
D. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i b\u1eb1ng 0<\/p>\n
C\u00e2u 10:<\/b>\u00a0Cho t\u1ea5m nh\u00f4m h\u00ecnh vu\u00f4ng c\u1ea1nh 12cm. Ng\u01b0\u1eddi ta c\u1eaft \u1edf b\u1ed1n g\u00f3c c\u1ee7a t\u1ea5m nh\u00f4m \u0111\u00f3 b\u1ed1n h\u00ecnh vu\u00f4ng b\u1eb1ng nhau, m\u1ed7i h\u00ecnh vu\u00f4ng c\u00f3 c\u1ea1nh b\u1eb1ng x (cm), r\u1ed3i g\u1eadp t\u1ea5m nh\u00f4m l\u1ea1i nh\u01b0 h\u00ecnh v\u1ebd b\u00ean \u0111\u1ec3 \u0111\u01b0\u1ee3c m\u1ed9t c\u00e1i h\u1ed9p kh\u00f4ng n\u1eafp. V\u1edbi gi\u00e1 tr\u1ecb n\u00e0o c\u1ee7a x th\u00ec h\u1ed9p nh\u1eadn \u0111\u01b0\u1ee3c c\u00f3 th\u1ec3 t\u00edch l\u1edbn nh\u1ea5t?<\/p>\n
<\/p>\n
A. 6 \u00a0\u00a0\u00a0 B. 4 \u00a0\u00a0\u00a0 C.3 \u00a0\u00a0\u00a0 D.2<\/p>\n
C\u00e2u 11:<\/b>\u00a0Khu chung c\u01b0 Royal City c\u00f3 250 c\u0103n h\u1ed9 cho thu\u00ea. N\u1ebfu ng\u01b0\u1eddi ta cho thu\u00ea x c\u0103n h\u1ed9 th\u00ec l\u1ee3i nhu\u1eadn h\u00e0ng th\u00e1ng, t\u00ednh theo tri\u1ec7u \u0111\u1ed3ng, \u0111\u01b0\u1ee3c cho b\u1edfi:<\/p>\n
P(x) = -8x2<\/sup>\u00a0+ 3200x – 80000<\/p>\n H\u1ecfi l\u1ee3i nhu\u1eadn t\u1ed1i \u0111a h\u1ecd c\u00f3 th\u1ec3 \u0111\u1ea1t \u0111\u01b0\u1ee3c l\u00e0 bao nhi\u00eau?<\/p>\n A. 150000 \u00a0\u00a0\u00a0 B. 220000 \u00a0\u00a0\u00a0 C. 240000 \u00a0\u00a0\u00a0 D. 250000<\/p>\n C\u00e2u 12:<\/b>\u00a0M\u1ed9t nh\u00e0 m\u00e1y s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c 60000 s\u1ea3n ph\u1ea9m trong m\u1ed9t ng\u00e0y v\u00e0 t\u1ed5ng chi ph\u00ed s\u1ea3n xu\u1ea5t x s\u1ea3n ph\u1ea9m \u0111\u01b0\u1ee3c cho b\u1edfi:<\/p>\n <\/p>\n H\u1ecfi nh\u00e0 m\u00e1y n\u00ean s\u1ea3n xu\u1ea5t bao nhi\u00eau s\u1ea3n ph\u1ea9m m\u1ed7i ng\u00e0y \u0111\u1ec3 chi ph\u00ed s\u1ea3n xu\u1ea5t l\u00e0 nh\u1ecf nh\u1ea5t?<\/p>\n A. 30000 \u00a0\u00a0\u00a0 B. 40000 \u00a0\u00a0\u00a0 C. 50000 \u00a0\u00a0\u00a0D.60000<\/p>\n H\u01b0\u1edbng d\u1eabn gi\u1ea3i v\u00e0 \u0110\u00e1p \u00e1n<\/b><\/p>\n C\u00e2u 7:<\/b><\/p>\n X\u00e9t h\u00e0m s\u1ed1 y=2sin x + cos 2x tr\u00ean \u0111o\u1ea1n<\/p>\n y\u2019=2cos x- 2sin 2x = 2cos x(1- 2sin x)<\/p>\n <\/p>\n Gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a h\u00e0m s\u1ed1 n\u00e0y l\u00e0 y = 3\/2 .<\/p>\n C\u00e2u 8:<\/b><\/p>\n D\u1ef1a v\u00e0o \u0111\u1ecbnh ngh\u0129a, h\u00e0m s\u1ed1 kh\u00f4ng t\u1ed3n t\u1ea1i gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t. H\u00e0m s\u1ed1 \u0111\u1ea1t c\u1ef1c \u0111\u1ea1i t\u1ea1i x=0 v\u00e0 c\u1ef1c ti\u1ec3u t\u1ea1i x=1.<\/p>\n C\u00e2u 9:<\/b><\/p>\n H\u00e0m s\u1ed1<\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n H\u00e0m s\u1ed1 kh\u00f4ng t\u1ed3n t\u1ea1i gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t. H\u00e0m s\u1ed1 c\u00f3 gi\u00e1 tr\u1ecb c\u1ef1c \u0111\u1ea1i b\u1eb1ng 0.<\/p>\n C\u00e2u 10:<\/b><\/p>\n Th\u1ec3 t\u00edch h\u00ecnh h\u1ed9p l\u00e0 y = x(12 – 2x)2<\/sup><\/p>\n B\u00e0i to\u00e1n \u0111\u01b0a v\u1ec1 t\u00ecm x \u2208 (0; 6) \u0111\u1ec3 h\u00e0m s\u1ed1 y = x(12 – 2x)2<\/sup>\u00a0c\u00f3 gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t.<\/p>\n y’ = 12x2<\/sup>\u00a0– 96x + 144; y’ x\u00e1c \u0111\u1ecbnh \u2200x \u2208 (0; 6)<\/p>\n <\/p>\n B\u1ea3ng bi\u1ebfn thi\u00ean<\/p>\n <\/p>\n H\u00e0m s\u1ed1 \u0111\u1ea1t gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t t\u1ea1i x=2<\/p>\n C\u00e2u 11:<\/b><\/p>\n Ta c\u00f3 x \u2208 (0; 250) ,P\u2019(x) = -16x+3200.<\/p>\n Khi \u0111\u00f3 P\u2019(x)=0 <=> -8x+3200=0 <=> x = 400 (lo\u1ea1i).<\/p>\n Do \u0111\u00f3 l\u1ee3i nhu\u1eadn t\u1ed1i \u0111a h\u1ecd thu \u0111\u01b0\u1ee3c l\u00e0 P(250)=240000.<\/p>\n C\u00e2u 12:<\/b><\/p>\n Ta c\u00f3 x \u2208 (0; 250)<\/p>\n <\/p>\n Khi \u0111\u00f3<\/p>\n <\/p>\n <=> x = 50000, P”(50000) > 0<\/p>\n N\u00ean x=50000 l\u00e0 s\u1ed1 s\u1ea3n ph\u1ea9m c\u1ea7n s\u1ea3n xu\u1ea5t m\u1ed7i ng\u00e0y \u0111\u1ec3 t\u1ed1i thi\u1ec3u chi ph\u00ed.<\/p>\n","protected":false},"excerpt":{"rendered":" C\u00e2u 7:\u00a0GTLN c\u1ee7a h\u00e0m s\u1ed1 y = 2sinx + cos2x tr\u00ean \u0111o\u1ea1n [0; \u03c0] l\u00e0 A. 1 \u00a0\u00a0\u00a0B. 3\/2 \u00a0\u00a0\u00a0 C. 2 \u00a0\u00a0\u00a0 D. 7\/4 C\u00e2u 8:\u00a0Cho h\u00e0m s\u1ed1 y = f(x) x\u00e1c \u0111\u1ecbnh v\u00e0 li\u00ean t\u1ee5c tr\u00ean R v\u00e0 c\u00f3 b\u1ea3ng bi\u1ebfn thi\u00ean. Kh\u1eb3ng \u0111\u1ecbnh n\u00e0o sau \u0111\u00e2y l\u00e0 kh\u1eb3ng \u0111\u1ecbnh \u0111\u00fang A. H\u00e0m […]<\/p>\n","protected":false},"author":3,"featured_media":27968,"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 7-B<\/td>\n 8-D<\/td>\n 9-D<\/td>\n 10-D<\/td>\n 11-C<\/td>\n 12-C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n